Part IV · Quantum Mechanics — Chapter 4.3

Function Spaces: Measure, L², and Completeness

Three promises come due together: the integral is thrown away and rebuilt, the space of states becomes a space, and completeness stops being an assumption.

Tools you'll need  — Chapter 0.2 §1.1 above all, for the Riemann integral as a limit of tagged sums, for the function $\chi$, and for the sequence of functions its warning box constructed and then set aside. Also its §4.4 for differentiation under the integral sign, and the grind box there that named the hypothesis this chapter proves. Chapter 0.5 §1 for the three inner-product axioms and Cauchy–Schwarz, §2 for orthonormal bases, Parseval and the resolution of the identity, and §3.1 for the fact that a projection is the closest point in a subspace. Its §1.2 table has the row this chapter exists to make legitimate. Chapter 0.9 §1 for the Fourier modes and their orthogonality, §1.3 for the completeness assumption and its mark, §1.4 for the square wave and the Gibbs overshoot, §2.3 for Plancherel, and §5.2 for the argument about a narrowing bump that §8 reuses almost word for word. Chapter 0.3 §3 for convergence and the tests that decide it without naming a limit. Chapter 0.4 §2 for basis and dimension, which is the argument §7 has to replace. Chapter 0.8 §6.4 for the Lorentzian, which Worked example 3 turns into a state. Chapter 4.2 §8 for the three lines that force infinite dimension, and its closing brick for the division of labour between this chapter and the next. Chapter 4.1 §5 for the integral whose term-by-term evaluation §4.1 here makes airtight.

1 · Why the Riemann integral has to go

Here is the debt this section collects, and it is Chapter 0.2's. We look again at the function that chapter could not integrate, define the one piece of vocabulary the book has been using without ever writing down, and then exhibit the failure that actually matters. The failure is not that some function is not integrable. It is that the class of integrable functions is not closed under the limits we need to take, and by the end of the section you will have a sequence with nowhere to converge to.

1.1 · The function Chapter 0.2 left on the table

Chapter 0.2 defined the integral as a limit of tagged sums and then, in a warning box attached to that definition, told you the definition would not survive. The words were these: "In Chapter 4.3 we will throw away [the definition] and rebuild the integral from scratch (the Lebesgue integral), and the reason is visible already." We are going to do exactly that, so let's start by looking again at what was visible already.

The function is Chapter 0.2's $\chi$, and we keep its name. It takes the value $1$ at every rational number and $0$ at every irrational one. Now recall why the definition of the integral cannot cope with it. A tagged sum lets you choose the sample point in each cell, and every cell, however short, contains rationals and irrationals both. So you may tag entirely at rationals, or entirely at irrationals, and on $[0,1]$ the two choices give

and both are available at every mesh, however fine. The definition asks for a single number that every tagging approaches, and there are two numbers that every mesh can produce. So $\chi$ is not Riemann integrable, and no refinement helps, because refining does not remove the choice.

Now let's be honest about how much that proves, because on its own it proves very little. A single function that cannot be integrated is an annoyance rather than a crisis. Physics is not short of integrals it declines to attempt, and a pathology you can name and step around costs nothing. If this were the whole case against the Riemann integral, the sensible response would be to note $\chi$ as a curiosity and carry on.

The real case is about limits, and to state it we need one word the book has been leaning on without ever defining.

1.2 · Cauchy sequences, defined here

Chapter 0.3 §3 gave you tests that decide whether a series converges. The ratio test is the one you use most, and notice what it does: it settles convergence without producing the limit. That is the whole trick, and it is worth naming, because everything in this chapter turns on the difference between a sequence that is heading somewhere and a sequence that has somewhere to go.

Work in a space with a norm, which for now means Chapter 0.5's $\lVert v\rVert=\sqrt{\avg{v,v}}$. A sequence $v_1,v_2,\ldots$ converges to $v$ when $\lVert v_n-v\rVert\to0$, and that definition names the limit. The alternative names nothing outside the sequence itself. Call $(v_n)$ a Cauchy sequence when its terms eventually huddle together:

The two conditions are not symmetric, and the asymmetry is the subject of this chapter. One direction is immediate. If $v_n\to v$ then, given $\varepsilon$, take $M$ beyond which every term is within $\varepsilon/2$ of $v$, and the triangle inequality of Chapter 0.5 §1.4 closes it:

Every convergent sequence is therefore Cauchy. The converse is the interesting question, and the first thing to see about it is that it is not a question about the sequence at all. It is a question about the space the sequence lives in.

Take the rational numbers with the ordinary absolute value, and take the decimal truncations $1,\,1.4,\,1.41,\,1.414,\,\ldots$ of $\sqrt2$. Any two terms beyond the $M$-th differ by less than $10^{-M}$, so the sequence is Cauchy. It does not converge in $\mathbb{Q}$, because the number it is closing in on is not a rational number. Nothing is wrong with the sequence. Something is missing from the space, and the missing thing has a name.

A space in which every Cauchy sequence converges to a point of the space is called complete. The real numbers are what you get by insisting on exactly that: you take the Cauchy sequences of rationals and declare the ones that ought to have the same limit to be the same number. Completeness enters this book at that point, as part of what a real number is rather than as a theorem about them, and Chapter 0.3's convergence tests all quietly stood on it. A test that certifies convergence without exhibiting a limit is only worth having in a space where the limit is guaranteed to be there.

That is the property we are going to need for functions, and §6 is where it is established. Keep (4.3.2) in view, because the rest of this section is one sequence of functions that satisfies it and has nowhere to land.

1.3 · Chapter 0.2's sequence, and the two failures it already exhibits

The same warning box in Chapter 0.2 that promised the rebuild also built the sequence that forces it, and we use that sequence rather than a new one. Enumerate the rationals as $q_1,q_2,q_3,\ldots$, which can be done, and let $f_n$ be the function on $[0,1]$ equal to $1$ at $q_1,\ldots,q_n$ and $0$ everywhere else. Each $f_n$ differs from the zero function at finitely many points, so each is Riemann integrable, and the value is

Chapter 0.2 then observed that $f_n\to\chi$ pointwise and stopped there. Let's carry it further, because the sequence shows two distinct defects and they need separating.

The first defect is that the class is not closed under limits. The sequence $f_1\le f_2\le f_3\le\cdots$ is increasing, every term is Riemann integrable, and every integral is $0$, which is as tame as a sequence can be. Its pointwise limit is $\chi$, which is not Riemann integrable at all. So the natural statement that any theory of integration wants to make,

fails here for the worst possible reason. It is not that the two sides differ. It is that the right-hand side does not refer to anything. Any convergence theorem is a statement of the form (4.3.5), so within the Riemann theory there can be no convergence theorem worth the name.

The second defect is that the norm is not a norm. Chapter 0.5 §1.2's table proposed $\avg{f,g}=\int\overline{f}g$ as an inner product on functions, and said in the next paragraph that the row "is the entire reason Chapter 4.3 is possible." Test the axioms on the sequence above. For $n\gt m$ the difference $f_n-f_m$ is $1$ at finitely many points and $0$ elsewhere, so

and yet $f_n\ne f_m$. That is a direct violation of Chapter 0.5's axiom (iii), the demand that $\avg{v,v}\gt0$ for every non-zero $v$. Without axiom (iii) there is no distance, because two different objects sit at distance zero, and Chapter 0.5's whole apparatus rests on all three axioms holding. Section 5 repairs this, and the repair is not cosmetic: it changes what a vector is.

Notice also that (4.3.6) makes $(f_n)$ Cauchy in the strongest imaginable sense, every pair being at distance exactly zero. So the sequence is Cauchy and its pointwise limit is not integrable. That sounds like the third failure, and it is not, which is worth saying out loud rather than gliding over. Distance zero from $f_1$ also means distance zero from the zero function, so in this norm the sequence does have a limit sitting inside the class. Chapter 0.2's sequence is too thin to show that anything is missing. We fatten it.

1.4 · The same rationals, thickened, and the hole that is left

Keep Chapter 0.2's enumeration $q_1,q_2,\ldots$ and give each rational a little room. Around $q_k$ put the open interval $I_k$ of length $2^{-k-2}$, and let $U_n$ be the part of $I_1\cup\cdots\cup I_n$ lying in $(0,1)$. That is a finite union of intervals, so its indicator function

is a step function, perfectly Riemann integrable, whose integral is the total length $\abs{U_n}$ of the intervals making up $U_n$. Those totals increase with $n$ and never exceed the sum of all the lengths, which is $\sum_{k\ge1}2^{-k-2}=\tfrac14$. An increasing sequence of real numbers bounded above converges, so $\abs{U_n}$ tends to a limit $\lambda\le\tfrac14$.

Now measure the distance between two terms. For $n\gt m$ the difference $g_n-g_m$ is the indicator of $U_n$ with $U_m$ removed, so its square is itself, and

because $\abs{U_n}$ converges and a convergent sequence of numbers is Cauchy. So $(g_n)$ is a Cauchy sequence of Riemann-integrable functions. The question is what it converges to, and the answer is the point of the section.

Grind box — no Riemann-integrable function is the limit, in full

Suppose $h$ is Riemann integrable on $[0,1]$ and $\int_0^1\abs{g_n-h}^{2}\dd x\to0$, the integrals existing. We derive a contradiction in four steps.

1. The integral of $h$ is at most $\tfrac14$. Apply Cauchy–Schwarz to $\abs{g_n-h}$ and the constant function $1$ on $[0,1]$. Chapter 0.5 §1.4 proved that inequality from all three axioms and §1.3 has just withdrawn the third for this pairing, so the citation needs a word. Its proof survives on this pair. The opening inequality asks only that $\avg{u,u}$ never be negative, which is still true here, and every later appeal to (iii) is a division by $\avg{v,v}$, which for $v=\mathbf 1$ is $\avg{\mathbf 1,\mathbf 1}=1\gt0$. So

$$\left|\int_0^1 (h-g_n)\right| \;\le\; \int_0^1\abs{h-g_n} \;\le\; \lVert h-g_n\rVert_2\cdot\lVert 1\rVert_2 \;=\; \lVert h-g_n\rVert_2 \;\to\; 0 .$$

So $\int_0^1 h=\lim_n\int_0^1 g_n=\lambda\le\tfrac14$ by (4.3.7).

2. A non-negative Riemann-integrable function with zero integral has infimum zero on every subinterval. If $\phi\ge0$ had $\inf_J\phi=c\gt0$ on a subinterval $J$ of length $\ell$, then any partition with the endpoints of $J$ among its points has lower sum at least $c\ell\gt0$, so $\int\phi\ge c\ell\gt0$.

3. On every subinterval, $h$ comes arbitrarily close to $1$. Let $I\subseteq(0,1)$ be any subinterval. Since $U$ contains every rational of $(0,1)$ it meets $I$, and $U$ is open, so $I\cap U$ contains a closed subinterval $J$. Now $U_1\subseteq U_2\subseteq\cdots$ are open sets whose union is $U\supseteq J$, so by compactness $J\subseteq U_n$ for some $n$, and then $g_m=1$ on $J$ for every $m\ge n$. Hence

$$\int_J (1-h)^{2}\,\dd x \;=\; \int_J\abs{g_m-h}^{2}\dd x \;\le\; \int_0^1\abs{g_m-h}^{2}\dd x \;\to\;0,$$

so $\int_J(1-h)^{2}=0$. By step 2 the infimum of $(1-h)^{2}$ over $J$ is zero, so $h$ takes values as close to $1$ as you please inside $J$, and therefore $\sup_I h\ge\sup_J h\ge1$.

4. Contradiction. Take any partition of $[0,1]$. Step 3 was proved for subintervals of the open interval $(0,1)$, and the first and last cells are not subsets of that, so the transfer needs one word. Every cell contains a subinterval of $(0,1)$, and a supremum over a cell is at least the supremum over a piece of it, so step 3 gives $\sup_{C_i}h\ge1$ on every cell. Now fix $\varepsilon\gt0$ and tag each cell at a point where $h(x_i^{*})\gt1-\varepsilon$, which the supremum makes possible. The tagged sum is then $\sum_i h(x_i^{*})\Delta x_i\gt(1-\varepsilon)\sum_i\Delta x_i=1-\varepsilon$, and such a tagging exists at every mesh. Riemann integrability is the statement that every tagging converges to $\int_0^1h$, so $\int_0^1 h\ge1-\varepsilon$ for every $\varepsilon$, hence $\int_0^1 h\ge1$. Step 1 gave $\int_0^1 h\le\tfrac14$. $\blacksquare$

Nothing there needed the fact that the integral is the infimum of the upper sums. Chapter 0.2 defined the integral by tagged sums and never proved that characterisation, and choosing the tags by hand is both shorter and closer to the definition the book actually has.

The two steps that did the work are worth naming. Density of $U$ forced the tagged sums up. Smallness of $U$ held the integral down. A theory of integration that could see the difference between "meets every interval" and "has length $\tfrac14$" would have no trouble here, and the Riemann integral cannot see it, because a cell width and a sample point are the only instruments it has.

So we have a Cauchy sequence of perfectly ordinary step functions with no limit in the class. This is the third failure and it is the fatal one. Chapter 4.2 §7 built time evolution as a limit, and Chapter 0.9 §1.3 asserted that partial Fourier sums converge to the function they came from. Both are statements that a Cauchy sequence has a limit, and in this class they are false.

What we need is now precise rather than vague, and it is a list of four items.

• A notion of the size of a set that can tell a dense set of total length $\tfrac14$ from the whole interval. Section 2.
• An integral built on that notion, defined on a class closed under limits, and agreeing with Chapter 0.2's wherever Chapter 0.2's exists. Section 3.
• Convergence theorems that let a limit pass through the integral sign, so that (4.3.5) becomes a theorem with hypotheses instead of a wish. Section 4.
• A repair of axiom (iii), so that distance zero means equality and Chapter 0.5's geometry transfers intact. Section 5.

Section 6 then shows that the four together buy completeness, and hands back the sequence $(g_n)$ with a limit attached.

2 · Measure, with one thing quoted

This section replaces one word. Chapter 0.2's integral was built on the length of an interval, and length is the thing that could not see the difference between a dense set and a fat one. We write down what a general notion of size would have to satisfy, and we find that three demands fix almost everything. Then one construction is quoted rather than built, because building it is a chapter of its own and this book is not going to pretend otherwise. What the quotation buys is stated precisely, and §8 comes back to collect part of it.

2.1 · The three demands, and why one of them is countable

We want to attach a number $\mu(A)$ to a set $A$ of real numbers, and the number should behave the way length behaves. Three demands do it.

• It extends length. For an interval, $\mu$ is the length: $\mu\big((a,b)\big)=b-a$, and the same for the closed and half-open versions.
• It is countably additive. If $A_1,A_2,\ldots$ are disjoint then the size of the union is the sum of the sizes.
• It is translation invariant. Sliding a set along the line does not change its size.

The first and third are obvious requests. The second is the one carrying the weight, and the word countably is doing all of the carrying, so let's see why finite additivity would not be enough. Every failure in §1 was a countable process. The set of rationals is a countable union of single points, the set $U$ was a countable union of intervals, and the sequences $(f_n)$ and $(g_n)$ were indexed by the integers. A notion of size that only adds finitely many pieces cannot say anything about any of them.

Countable additivity has one consequence we will use so often that it is worth extracting now. Suppose $A_1\subseteq A_2\subseteq\cdots$ is an increasing family with union $A$. Write the union as a disjoint one by peeling off shells, $A=A_1\cup(A_2\setminus A_1)\cup(A_3\setminus A_2)\cup\cdots$, and apply countable additivity to the shells. The partial sums of the resulting series are exactly the $\mu(A_n)$, so

This is called continuity from below, and it is the exact point at which the word "countably" gets spent. Section 4's monotone convergence theorem is (4.3.9) promoted from sets to functions, and it has no other engine.

2.2 · Which sets, and the shape the collection has to have

The demands above say nothing about which sets get a size, and §2.5 will show that not all of them can. So the collection of sets we can measure is part of the structure, and its shape is forced by the operations we intend to perform on it.

A collection $\mathcal M$ of subsets of $\R$ is a $\sigma$-algebra when it contains $\R$, is closed under taking complements, and is closed under countable unions. Countable intersections come free, because $\bigcap_n A_n$ is the complement of $\bigcup_n A_n^{c}$, and set differences come free for the same reason. Those are precisely the operations a countable limit process performs, which is why the definition looks the way it does rather than some other way.

The smallest $\sigma$-algebra containing every open interval is called the collection of Borel sets, and it contains everything you are likely to write down: open sets, closed sets, countable unions of closed sets, countable intersections of those, and so on upward. The set $U$ of §1.4 is a countable union of open intervals, so it is Borel, and so is the set of rationals.

2.3 · The construction, quoted as a package

Everything so far is a specification. Meeting it is a genuine construction, and here is the first of the two places where this chapter takes something on trust. The construction starts by defining an outer measure, which covers a set as economically as possible with intervals and takes the cheapest cover:

That definition applies to every set whatever, which is why it cannot be the end of the story: something so generous is not going to be countably additive. Carathéodory's criterion then selects the sets on which it behaves, by keeping $A$ exactly when $A$ splits every other set additively.

2.4 · Sets of measure zero, and the phrase "almost everywhere"

Call $N$ a null set when $\mu(N)=0$. The most useful fact about null sets is how easy they are to come by, and the proof is a construction you have already seen in this chapter.

Every countable set is null. Let the set be $\{p_1,p_2,\ldots\}$, fix $\varepsilon\gt0$, and cover $p_k$ by an interval of length $\varepsilon2^{-k}$. That is a countable cover of total length $\varepsilon$, so $\mu^{*}\le\varepsilon$ by (4.3.10), and $\varepsilon$ was arbitrary. In particular

Look at what that construction was. It is §1.4's thickening of the rationals, with the thickness sent to zero instead of held at $\tfrac14$. The same cover that produced a dense set of length $\tfrac14$ produces, in the limit, the statement that the rationals have no length at all. Chapter 0.2 could not write down (4.3.11), and it is the whole reason $\chi$ is about to become integrable.

A property holds almost everywhere, abbreviated a.e., when the set where it fails is null. So $\chi=0$ almost everywhere, and the functions $f_n$ of §1.3 are all equal to each other almost everywhere. That phrase is going to reorganise the whole subject in §5, where two functions agreeing almost everywhere stop being two functions.

2.5 · Not every set is measurable

The flag in §2.3 restricted $\mu$ to a collection $\mathcal M$ rather than to all subsets of $\R$, and a reader is entitled to ask whether that restriction is a real one or a technician's caution. It is real, and the demonstration is short enough to run here. It is due to Vitali.

The point of running it is to make the flag honest. If $\mathcal M$ could be all subsets, the quoted package would be quoting a formality. It cannot, and the three demands of §2.1 are exactly what makes it impossible.

Grind box — the Vitali set, built

Work on $[0,1)$ and declare $x\sim y$ when $x-y$ is rational. This is an equivalence relation, so it cuts $[0,1)$ into disjoint classes, each class being one number plus every rational, cut down to $[0,1)$. One class is the rationals of $[0,1)$ on their own. Every other class is an irrational offset plus all the rationals. Choose one representative from each class and collect the choices into a set $V$. That step is the axiom of choice, and it is where the whole construction lives.

Now translate $V$ around the circle. For $q\in\mathbb{Q}\cap[0,1)$ let

$$V\oplus q \;=\; \{v+q : v\in V,\ v+q\lt1\} \;\cup\; \{v+q-1 : v\in V,\ v+q\ge1\},$$

which is $V$ shifted by $q$ with the overhang wrapped back to the start.

They are disjoint. If $v_1+q_1$ and $v_2+q_2$ agree modulo $1$ then $v_1-v_2$ is rational, so $v_1$ and $v_2$ lie in the same class, so $v_1=v_2$ by the choice, so $q_1=q_2$.

They cover. Given $x\in[0,1)$, let $v$ be the representative of $x$'s class. Then $x-v$ is rational and lies in $(-1,1)$, so it is either $q$ or $q-1$ for some $q\in\mathbb{Q}\cap[0,1)$, and $x\in V\oplus q$.

The contradiction. Suppose $V\in\mathcal M$ with $\mu(V)=m$. Each $V\oplus q$ is two translated pieces of $V$, so by (c) and (b) of §2.3 it is measurable with $\mu(V\oplus q)=m$. The rationals in $[0,1)$ are countable, so countable additivity applies to the whole family:

$$1 \;=\; \mu\big([0,1)\big) \;=\; \sum_{q\in\mathbb{Q}\cap[0,1)}\mu(V\oplus q) \;=\; \sum_{q} m .$$

If $m=0$ the sum is $0$. If $m\gt0$ the sum is $+\infty$. Neither is $1$, so $V$ is not in $\mathcal M$. $\blacksquare$

Every ingredient was one of the three demands, plus the axiom of choice. Drop countable additivity for finite additivity and the argument collapses, which is one reason the countable version is not a stylistic preference.

So $\mathcal M$ is a genuine restriction, and the sets it leaves out are the ones you cannot write down. Every set arising from a formula, a limit, or a physical construction is Borel, hence in $\mathcal M$, and §3 never has to think about it again.

a natural place to stop  ·  the problem is set, and so is what "size" must mean

3 · The Lebesgue integral, built

With size in hand, the integral takes one paragraph to define and one section to check. The whole idea is a single change of direction, which §3.1 states in a sentence. Section 3.2 says which functions are eligible and proves the property that the Riemann class lacked. Section 3.3 does the construction. Section 3.5 shows the new integral agrees with the old one wherever the old one works, so that nothing in Parts 0 to III has to be relearned, and §3.6 is honest about the one thing that is lost.

3.1 · One change of direction

Here is the whole idea in a sentence. Riemann slices the domain. Lebesgue slices the range.

Chapter 0.2 chopped the $x$-axis into cells, took the value of $f$ somewhere in each cell, and added up value times width. The new definition chops the $y$-axis instead. For each level $y$ it asks how big the set of points where $f$ is near $y$ is, and adds up level times size.

The picture that makes it stick is counting money. Given a pile of coins, Riemann's method is to go along the pile in order and add each coin's value as you reach it. Lebesgue's method is to sort the pile into denominations first, count how many coins are in each pile, and add up denomination times count. Three terms map exactly: the value of a coin is the value of $f$, the position of a coin in the pile is the point $x$, and the number of coins of a given denomination is the size of the set where $f$ takes that value.

One term has no counterpart, and it is the one §1 turned on. A coin has a single definite value, so there is no cell, no width, and no freedom about where to sample. Both methods therefore return the same total on every pile, which is §3.5 in miniature, and neither of them can fail. So the picture cannot show you why one method survives where the other does not, and that claim is left where it is proved, in §4.

3.2 · Which functions, and the property the Riemann class lacked

Slicing the range means asking for the size of the set where $f$ exceeds a level, so that set had better be measurable. Call $f$ measurable when

That is the entire condition, and it is weak. Every continuous function passes, since $\{f\gt a\}$ is then open. Every step function passes. So does $\chi$, for which $\{\chi\gt a\}$ is $\R$ when $a\lt0$, the rationals when $0\le a\lt1$, and empty when $a\ge1$. Three sets, all of them in $\mathcal M$, and nothing you can write down fails.

Now the property that Riemann's class did not have, and the reason the $\sigma$-algebra was defined with countable operations in §2.2. Let $f_1,f_2,\ldots$ be measurable and suppose $f_n(x)\to f(x)$ at every $x$. Then $f(x)\gt a$ exactly when the values $f_n(x)$ eventually stay above some level strictly greater than $a$, and "some level", "eventually" and "stay above" are three countable operations, written below in that order:

Every set on the right is in $\mathcal M$ by (4.3.12), and $\mathcal M$ is closed under countable unions and intersections, so the set on the left is in $\mathcal M$ too. A pointwise limit of measurable functions is measurable.

Compare that with §1.3, where a monotone limit of Riemann-integrable functions was not Riemann integrable. The defect is gone, and it is gone by construction rather than by luck: the class was defined by a condition phrased in countable operations precisely so that countable operations could not escape it.

3.3 · Simple functions, then the supremum

The construction goes in two stages, and the first stage is the one where the definition actually happens. A simple function is a finite sum of indicators of measurable sets,

and its integral is defined to be the obvious thing, level times size, added up. Since the $E_k$ are disjoint there is no ambiguity about what to write:

That is the range-slicing of §3.1 with only finitely many levels. Two things follow from finite additivity of $\mu$ and nothing else, and both are used constantly below: the integral of a sum of simple functions is the sum of the integrals, and $\varphi\le\psi$ implies $\int\varphi\le\int\psi$.

The second stage extends this to any non-negative measurable $f$ by approaching it from below with simple functions and taking the best available answer:

a supremum that may be $+\infty$. Monotonicity is immediate from (4.3.16), since $f\le g$ means every $\varphi$ competing for $f$ also competes for $g$, so the supremum can only grow. That one-line fact is used in almost every proof that follows.

It is worth seeing that the supremum is always attained in the limit by an explicit staircase, because §4 will apply its theorems to exactly this sequence. Cut the range into steps of height $2^{-n}$ up to a ceiling of $n$ and read off which step $f$ is on:

Each $\varphi_n$ is simple by (4.3.12), the sequence increases because halving the step can only refine the estimate, and $\varphi_n(x)\to f(x)$ at every $x$. So every non-negative measurable function is an increasing limit of simple ones, which is the range being sliced ever more finely.

Finally, drop the requirement that $f$ be non-negative. Write $f=f^{+}-f^{-}$ with $f^{+}=\max(f,0)$ and $f^{-}=\max(-f,0)$, both non-negative and measurable, and define $\int f=\int f^{+}-\int f^{-}$. Call $f$ integrable when both pieces have finite integral, which is the same as saying $\int\abs f\lt\infty$. For a complex-valued $f$, integrate the real and imaginary parts separately. That is the whole definition.

3.4 · The first two answers, one of them a debt

Chapter 0.2's warning box promised that the new integral "gives $\int_0^1\chi=0$ into the bargain", and the promise is now one line. The function $\chi$ restricted to $[0,1]$ is the indicator of $\mathbb{Q}\cap[0,1]$, which is simple, so (4.3.15) applies directly and (4.3.11) supplies the size:

The function that no partition could pin down has an integral, and getting it took one multiplication. The same applies to §1.4's thickened set: $\mathbf 1_U$ is a limit of the simple functions $g_n$, and its integral will turn out to be $\mu(U)$ once §4 licenses passing to the limit. That is the loose end §6 ties.

3.5 · It agrees with Riemann wherever Riemann works

If the new integral disagreed with the old one anywhere the old one was defined, every calculation in Parts 0 to III would need checking, so the following matters more than its proof suggests.

If $f$ is Riemann integrable on $[a,b]$, then $f$ is measurable, it is Lebesgue integrable, and the two integrals are equal.

The idea is that upper and lower sums are integrals of simple functions, so the Riemann machinery is already inside the Lebesgue machinery, viewed correctly. Take partitions $P_n$ with mesh going to zero, each refining the last, and build the two staircases

the sums running over the cells $C_i$ of $P_n$. These are simple functions whose Lebesgue integrals, by (4.3.15), are exactly the lower and upper Riemann sums of $f$ over $P_n$. Riemann integrability says those two sequences of numbers close on each other, and that squeeze is what the grind box turns into the theorem.

Grind box — Riemann implies Lebesgue, with the same value

Shift $f$ by a constant so that $f\ge0$, which changes both integrals by the same amount. Write $L_n$ and $U_n$ for the lower and upper Riemann sums over $P_n$, and $R$ for the Riemann integral.

One line first, because Chapter 0.2 defined that integral by tagged sums and not by upper and lower ones. On a fixed partition the tagged sums come as close as you like to $L_n$ and to $U_n$, by tagging near the infimum or near the supremum in each cell, and Chapter 0.2 §1.1 observed that every tagged sum is trapped between the two. Integrability sends every tagged sum to $R$ as the mesh shrinks, so $L_n$ and $U_n$ are squeezed onto $R$ with them. Refinement makes the first increase and the second decrease, so $L_n\uparrow R$ and $U_n\downarrow R$.

The same refinement condition makes $\ell_n$ increasing and $u_n$ decreasing pointwise, so both have pointwise limits

$$\ell=\lim_n \ell_n, \qquad u=\lim_n u_n, \qquad \ell\;\le\; f\;\le\; u,$$

and both are measurable by §3.2. Now three steps.

1. $\int\ell=R$. Monotonicity of (4.3.16) against $\ell_n\le\ell\le u_m$ for every $n$ and $m$ gives $L_n\le\int\ell\le U_m$. Let $n$ and $m$ run and the two ends squeeze onto $R$.

2. $u=\ell$ almost everywhere. The function $u-\ell$ is non-negative and satisfies $u-\ell\le u_n-\ell_n$, so monotonicity and additivity on simple functions give $\int(u-\ell)\le U_n-L_n\to0$, hence $\int(u-\ell)=0$. For any non-negative measurable $\phi$ the inequality $\tfrac1k\mathbf 1_{\{\phi\gt1/k\}}\le\phi$ and monotonicity give Chebyshev's bound $\mu\{\phi\gt1/k\}\le k\int\phi$, so here every $\mu\{u-\ell\gt1/k\}$ vanishes, and the union over $k$ is the set where $u\gt\ell$. It is null.

3. Conclusion. Since $\ell\le f\le u$ and $\ell=u$ off a null set, $f=\ell$ almost everywhere, so $f$ is measurable by property (e) of §2.3, and its integral equals $\int\ell=R$, two functions agreeing off a null set having the same integral by (4.3.15) applied to the competing simple functions. $\blacksquare$

So no arithmetic in this book changes. Every Gaussian in Chapter 0.2, every Fourier coefficient in Chapter 0.9, and every action integral in Part I means the same number it always did. What has been added is a larger class of functions and, in §4, the theorems that make the enlargement worth having.

3.6 · The one thing that is lost, said plainly

The new integral is not merely more powerful than the old one, and pretending otherwise would store up a surprise. There is a class of integrals the Riemann theory handles and the Lebesgue theory declines.

4 · Monotone and dominated convergence

This is the section the rebuild was for. Section 1 showed that the Riemann integral admits no theorem of the form "the integral of the limit is the limit of the integrals". Here are the two that the Lebesgue integral does admit. The first is proved from continuity from below and nothing else. The second follows from the first through one intermediate step, and it is the one Chapter 0.2 borrowed against. Both are stated with their hypotheses in front, because in both cases the hypothesis is what the theorem is really about.

4.1 · Monotone convergence, derived

Let $0\le f_1\le f_2\le\cdots$ be measurable and let $f_n(x)\to f(x)$ at every $x$. Then $\int f_n\to\int f$. The limit function is measurable by §3.2, so both sides refer to something, and that is exactly what failed in §1.3.

One direction is free. Monotonicity of (4.3.16) gives $\int f_n\le\int f$ for every $n$, and the left side increases, so its limit $L$ satisfies $L\le\int f$. The work is the other direction, and it consists of showing that $L$ beats every simple function competing in the supremum that defines $\int f$.

So fix a simple $\varphi=\sum_k c_k\mathbf 1_{E_k}$ with $0\le\varphi\le f$, and fix a slack factor $\theta$ slightly below $1$, kept a different letter from the levels $c_k$ because the two are about to be multiplied together. Since $f_n$ climbs to $f\ge\varphi$, every point is eventually caught above $\theta\varphi$, so the sets

Write $\int_A g$ for $\int g\,\mathbf 1_A$, the integral restricted to a set. Discarding everything outside $A_n$ can only lower an integral of a non-negative function, so $\int f_n\ge\int_{A_n}f_n\ge\theta\int_{A_n}\varphi$, and the last of those is a simple function's integral, which (4.3.15) writes out explicitly:

The limit in the middle is (4.3.9), continuity from below, applied to the increasing sets $E_k\cap A_n$ inside each $E_k$. That is where the countable in "countably additive" is spent, and it is the only place the proof spends anything. So $L\ge\theta\int\varphi$ for every $\theta\lt1$, hence $L\ge\int\varphi$, and taking the supremum over $\varphi$ gives $L\ge\int f$. With the free direction, $L=\int f$. $\blacksquare$

Two corollaries follow at once, and the first of them repairs an omission. Nothing so far has established that $\int(f+g)=\int f+\int g$ for general non-negative measurable $f$ and $g$, since (4.3.15) only gave it for simple functions. Take the staircases $\varphi_n\uparrow f$ and $\psi_n\uparrow g$ of (4.3.17), note that $\varphi_n+\psi_n\uparrow f+g$, and apply monotone convergence to all three sequences. Additivity for simple functions passes to the limit, so the Lebesgue integral is linear.

The second corollary is the one that gets used in anger. Let $h_1,h_2,\ldots$ be non-negative and measurable and apply monotone convergence to the partial sums $\sum_{k\le n}h_k$, which increase because the terms are non-negative:

No convergence hypothesis is needed anywhere in (4.3.22), since both sides are allowed to be $+\infty$ and the theorem says they are infinite together. Sign is the only thing being asked for.

One interchange that looks like a neighbour of this one is not settled here, and it is worth saying so rather than leaving a reader to wonder. Exchanging a sum with an integral is (4.3.22). Exchanging two integrals is Fubini's theorem, which Chapter 0.2 quoted and marked when it squared the Gaussian, and which needs a measure on a product space rather than on the line. That is a second construction of the kind §2.3 already quoted once, this chapter does not perform it, and nothing below uses it. Chapter 0.2's mark stands where it is.

A debt, collected. Chapter 4.1 §5 evaluated the integral in the Planck spectrum by expanding the denominator as a geometric series and integrating term by term, and it wrote in parentheses that "the theorem that makes it airtight is monotone convergence, proved in Chapter 4.3." Here it is. For $y\gt0$ the expansion $1/(\ee^{y}-1)=\sum_{n\ge1}\ee^{-ny}$ has every term positive, so (4.3.22) applies with no further hypothesis, and

The Stefan–Boltzmann constant of Chapter 4.1 rests on that interchange, and the interchange now rests on a theorem rather than on an expectation. Worked example 2 derives the $\sum n^{-4}=\pi^{4}/90$ that turns the middle expression into the last.

4.2 · Fatou's lemma, and what it says when it is strict

Monotone convergence needs the sequence to climb, and most sequences do not. The next result asks for nothing but non-negativity, and pays for that generosity with an inequality instead of an equality.

For non-negative measurable $f_n$, $\ \int\liminf_n f_n\le\liminf_n\int f_n$. The proof is monotone convergence applied to the wrong sequence on purpose. Set $h_n=\inf_{k\ge n}f_k$. That sequence has to be measurable before monotone convergence can touch it, and §3.2 closed the class under pointwise limits rather than under infima, so here is the missing line. Writing $\{f\ge a\}=\bigcap_{j\ge1}\{f\gt a-1/j\}$ puts each $\{f_k\ge a\}$ in $\mathcal M$, then $\{h_n\ge a\}=\bigcap_{k\ge n}\{f_k\ge a\}$ is in $\mathcal M$ as a countable intersection, and $\{h_n\gt a\}=\bigcup_{j\ge1}\{h_n\ge a+1/j\}$ is the condition (4.3.12) asks for. The $h_n$ increase with $n$ and climb to $\liminf f_n$ by definition, so monotone convergence gives $\int h_n\to\int\liminf f_n$. Also $h_n\le f_k$ for every $k\ge n$, so monotonicity gives $\int h_n\le\inf_{k\ge n}\int f_k$. Let $n$ run and the right side becomes $\liminf\int f_n$. $\blacksquare$

An inequality is only interesting when you know how it fails to be an equality, and Chapter 0.2 already built the example. Its grind box on differentiating under the integral sign used

while $h(x,a)\to0$ as $a\to0$ for every fixed $x\gt0$. Take $a=1/n$ and you have a sequence with $\int\liminf=0$ and $\liminf\int=1$. Chapter 0.2 said of it that "the mass has not vanished. It has escaped into an ever narrower, ever taller spike near the origin", and Chapter 0.9 §5.2 identified the same object as the Dirac delta being born. Fatou's lemma is the exact statement that mass can escape but cannot appear from nowhere, and the direction of the inequality is the direction mass can go.

4.3 · Dominated convergence, and Chapter 0.2's outstanding hypothesis

Now the theorem the book has been waiting on. It buys back the equality, and the price is a single function that holds the whole sequence down.

Let $f_n$ be measurable with $f_n\to f$ pointwise almost everywhere, and suppose there is one integrable $g$ with $\abs{f_n}\le g$ for every $n$. Then $f$ is integrable and $\int\abs{f_n-f}\to0$, so in particular $\int f_n\to\int f$.

The proof is Fatou applied to a cleverly chosen non-negative sequence. Since $\abs f\le g$ as well, we have $\abs{f_n-f}\le2g$, so $2g-\abs{f_n-f}$ is non-negative and Fatou applies to it. Its pointwise lower limit is $2g$, because $\abs{f_n-f}\to0$, so

using the linearity established in §4.1. Now $\int 2g$ is finite, which is the whole content of the hypothesis, so it may be cancelled from both sides. That leaves $\limsup\int\abs{f_n-f}\le0$, and the quantity is non-negative, so it is zero. $\blacksquare$

The promise, collected verbatim. Chapter 0.2 §4.4's grind box justified swapping a derivative with an integral and wrote: "The sufficient hypothesis (dominated convergence, proved properly in Chapter 4.3) is that the $a$-derivative of the integrand be bounded, uniformly in $a$ near the point of interest, by a single fixed integrable function." That is the theorem above, and the swap it licenses is worth stating as its own result rather than left implied.

Suppose $F(x,a)$ is integrable in $x$ for each $a$ near $a_0$, that $\partial_aF$ exists, and that $\abs{\partial_aF(x,a)}\le G(x)$ for all such $a$, with $G$ integrable. Then

To see it, take any sequence $a_n\to a_0$ and form the difference quotients $Q_n(x)=[F(x,a_n)-F(x,a_0)]/(a_n-a_0)$. The mean value theorem writes each $Q_n(x)$ as $\partial_aF(x,\xi)$ for some $\xi$ between $a_n$ and $a_0$, so $\abs{Q_n}\le G$ for every $n$, which is the domination. And $Q_n\to\partial_aF(x,a_0)$ pointwise, which is the convergence. Dominated convergence therefore gives $\int Q_n\to\int\partial_aF$, and the left side is by construction the difference quotient of $a\mapsto\int F$. Every sequence gives the same limit, so the derivative exists and (4.3.26) holds.

Chapter 0.2's own case now closes. There $F(x,a)=\ee^{-ax^{2}}$ and you were asked to accept the bound

with the right-hand side a fixed function of finite integral. That is exactly the hypothesis of (4.3.26), so every Gaussian moment Chapter 0.2 extracted by differentiating under the integral sign is now derived rather than provisional. The generating function of Chapter 0.2 §4.5, and with it the whole differentiation-with-respect-to-a-source apparatus that Part V runs on, inherits the same licence.

a natural place to stop  ·  the integral is built; what follows is the space

5 · $L^{2}$, and the functions that are not functions

Now the space itself. We collect the square-integrable functions, put Chapter 0.5's inner product on them, and check its three axioms one at a time. Two hold with no work at all. The third fails, and it fails on exactly the sequence Chapter 0.2 built, so the repair is forced rather than chosen. Making the repair costs something real, and §5.3 says what. Once it is made, every theorem of Chapter 0.5 §1 arrives with no proof needed, and §5.5 records the one way this space is not like $\R^{n}$.

5.1 · The set, and that it is a vector space

Fix a measurable set $X$, which for our purposes is $\R$ or an interval, and collect the measurable complex-valued functions on it whose squared modulus has finite integral:

Two things need checking before that definition means anything, and both are one line. First, the collection is closed under addition, since $\abs{f+g}^{2}\le2\abs f^{2}+2\abs g^{2}$ pointwise and the right-hand side is integrable. Second, the inner product converges, since $\abs{\overline fg}\le\tfrac12(\abs f^{2}+\abs g^{2})$ by the arithmetic-geometric mean inequality, so the integral defining $\avg{f,g}$ is finite for every pair in the set.

That second check is the reason the exponent is $2$ rather than anything else. The square is the power for which the inner product of two members always exists, which is why quantum mechanics lives in this space and not in its neighbours.

5.2 · The three axioms, and the one that fails

Chapter 0.5 §1.1 laid out exactly three axioms, and the whole of that chapter follows from them. Take them in order against (4.3.28).

• (i) Conjugate symmetry, $\avg{f,g}=\overline{\avg{g,f}}$. Holds, because $\overline{\int\bar gf}=\int\bar fg$ and conjugation commutes with the integral.
• (ii) Linearity in the second slot. Holds, because the integral is linear, which §4.1 established for the Lebesgue integral.
• (iii) Positive definiteness, $\avg{f,f}\gt0$ for $f\ne0$. Fails.

It fails, and the counterexample is the one this chapter opened with. Chapter 0.2's $\chi$ is non-zero at every rational number, so it is not the zero function, and yet (4.3.18) gives $\avg{\chi,\chi}=\int\abs\chi^{2}=0$. The same holds for every $f_n$ of §1.3.

The general statement is worth having, because it says precisely how bad the failure is. For non-negative measurable $\phi$, Chebyshev's bound from the grind box of §3.5 gives $\mu\{\phi\gt1/k\}\le k\int\phi$, so a vanishing integral forces every one of those sets to be null, and their union is $\{\phi\gt0\}$:

So the axiom does not fail badly. It fails by exactly the width of a null set, and (4.3.29) tells you what to do about it.

5.3 · The repair, and what it costs

Since the norm cannot tell two functions apart when they agree almost everywhere, stop asking it to. Declare $f$ and $g$ to be the same vector when $f=g$ almost everywhere. That is an equivalence relation, and the space of states is the set of equivalence classes:

Axiom (iii) now holds by construction, since $\avg{f,f}=0$ says $f$ is in the class of the zero vector, which is what "$f=0$ in $L^{2}$" now means. All three axioms hold, so $L^{2}(X)$ is an inner-product space in exactly Chapter 0.5's sense, and this is the moment that chapter was pointing at when it said of its table's function row that "that row is the entire reason Chapter 4.3 is possible."

The quotient is not free, and pretending it is would set up a misunderstanding that costs a whole chapter later.

5.4 · Chapter 0.5's theorems, transferred without a word changed

Here is the return on Chapter 0.5 having been written abstractly. That chapter proved Cauchy–Schwarz from the three axioms alone, and its proof mentions no basis, no dimension and no component. Every step of it is now available here with nothing to check:

The first is Chapter 0.5's (0.5.6) with the vectors renamed, and the second is the triangle inequality that follows from it in three lines there. Both are theorems about inner-product spaces, and $L^{2}$ is one. Chapter 0.5 said this would happen, in the solution to Problem 1 of its §10, where Gram–Schmidt run on $1,x,x^{2},\ldots$ produced the Legendre polynomials: "the machinery does not care that the vectors are functions. That is the whole content of the abstraction, and it is what Chapter 4.3 will exploit." This is the exploitation, and it consists of writing nothing down.

What does not transfer is anything Chapter 0.5 proved using finite dimension, and that chapter listed the four places it used it: "in the induction, in rank–nullity, in the interchange of sums, in the claim that an injective map is surjective." Sections 6 and 7 supply replacements for as much of that as this chapter needs, and Chapters 4.4 and 4.5 handle the rest.

5.5 · Square-integrable does not mean integrable

One structural fact separates $L^{2}(\R)$ from every finite-dimensional space and from $L^{2}$ of a bounded interval, and it has physical consequences that Chapter 4.9 will need.

On a bounded interval the two conditions are nested. Cauchy–Schwarz against the constant function $1$ gives, for $f\in L^{2}[a,b]$,

so square-integrable implies integrable there. On the whole line the constant function $1$ is not in $L^{2}$, the factor $\sqrt{b-a}$ has nothing to converge to, and the implication fails. Both inclusions fail, in fact, and one example each settles it:

• $f(x)=(1+\abs x)^{-3/4}$ has $\int\abs f^{2}=\int(1+\abs x)^{-3/2}\lt\infty$ and $\int\abs f=\int(1+\abs x)^{-3/4}=\infty$. So $f\in L^{2}(\R)$ and $f\notin L^{1}(\R)$.
• $f(x)=\abs x^{-1/2}$ on $(0,1)$ and $0$ elsewhere has $\int\abs f=2$ and $\int\abs f^{2}=\int_0^1 x^{-1}\dd x=\infty$. So $f\in L^{1}(\R)$ and $f\notin L^{2}(\R)$.

The physical reading is worth having now rather than in Chapter 4.9. A wavefunction belongs to $L^{2}$ because $\int\abs\psi^{2}=1$ is what normalisation means, and that guarantees nothing about $\int\psi$, or about $\int x\abs\psi^{2}$. Worked example 3 exhibits a perfectly normalised state whose mean position does not exist, and the state is one Chapter 0.9 already introduced under another name.

6 · Completeness: the theorem, and the failure it repairs

Everything built so far exists for this section. Section 1 handed you a Cauchy sequence with nowhere to go. Sections 2 to 5 enlarged the class of functions, built an integral that survives limits, and fixed the norm so that distance zero means equality. The claim now is that the enlargement was exactly the right size: not merely bigger, but big enough that no Cauchy sequence can escape it. Section 6.2 states the theorem and marks it as quoted. Section 6.3 then takes the two sequences of §1 and shows you where they land.

6.1 · What is being claimed, and what is not

Convergence in $L^{2}$ means $\lVert f_n-f\rVert_2\to0$, which written out is

and is often called convergence in mean square. Read what it says and, more importantly, what it does not. It says the total squared discrepancy shrinks to nothing. It says nothing whatever about the value of $f_n$ at any particular point, and §5.3 already explained why it cannot: the vectors have no pointwise values to make a statement about. Section 8.4 exhibits a sequence converging in the sense of (4.3.33) whose graphs never stop overshooting.

With that understood, here is the claim. Every sequence in $L^{2}(X)$ that is Cauchy in the sense of (4.3.2) converges, in the sense of (4.3.33), to an element of $L^{2}(X)$. An inner-product space with that property is called a Hilbert space, and that is the proper name Chapter 0.9 §1.3 promised the space of square-integrable functions would eventually be given.

6.2 · Riesz–Fischer

Two remarks before the theorem is spent. First, the sources quoted above prove the same statement for every $L^{p}$ at once, so nothing about the exponent $2$ is doing the work here. What the exponent $2$ gives is the inner product of §5.1, and it is the combination of the two that makes $L^{2}$ the space quantum mechanics needs. Second, the proof produced convergence almost everywhere only along a subsequence, and it is worth being exact about what that does and does not mean. Section 8.4's example is not evidence that the thinning is necessary. Its partial sums converge in norm and they also converge at every point, as §8.4's own theorem and Problem 5(c) both say, so what they fail is uniform convergence, which is a different statement. Whether norm convergence drags a whole sequence to a limit almost everywhere is a question this chapter does not take up, and the subsequence is all the sketch above delivers.

6.3 · The two sequences of §1, and where they land

Now collect. Both sequences from §1 are sequences in $L^{2}[0,1]$, both are Cauchy, and both now have limits. They are different kinds of repair and it is worth taking them in order.

Chapter 0.2's sequence. The functions $f_n$ of §1.3 were $1$ at $q_1,\ldots,q_n$ and $0$ elsewhere. In $L^{2}$ each of them is the zero vector, by (4.3.29) and the vanishing of $\lVert f_n\rVert_2$. Their pointwise limit $\chi$ is also the zero vector, for the same reason. So the sequence converges, its limit is in the space, the limit's integral exists, and the integral is

Compare (4.3.34) with (4.3.5), the statement that failed in §1.3. It failed there because its right-hand side referred to nothing. Both sides now refer to something and the two agree. Chapter 0.2's warning box has been answered in full: the definition was thrown away, the integral was rebuilt, and $\int_0^1\chi=0$ came with it.

The thickened sequence. This is the one that showed a genuine hole, so it is the one worth watching close. The step functions $g_n=\mathbf 1_{U_n}$ of §1.4 increase to $\mathbf 1_U$ at every point, and $\mathbf 1_U$ is measurable because $U$ is open. Its square integral is finite, so it is a vector of $L^{2}[0,1]$, and the distance from the sequence to it is computed by (4.3.9):

So the sequence converges in $L^{2}$, its limit is a genuine element of the space, and the limit has an integral, namely $\int_0^1\mathbf 1_U=\mu(U)$, a number at most $\tfrac14$.

Now read that against the grind box of §1.4, which is where the payoff is. That box proved that no Riemann-integrable function is an $L^{2}$ limit of the $g_n$. Since $\mathbf 1_U$ is that limit, and limits in a metric space are unique up to distance zero, the box says exactly this:

The vector the sequence was reaching for is not a relabelling of anything the old theory had. It sits at positive distance from every last one of them. The hole exhibited in §1 was real, the enlargement of §2 to §5 filled it, and (4.3.36) is the receipt.

6.4 · What completeness gets spent on

Chapter 4.2 wrote its whole apparatus on the assumption that limits of states are states, and listed the assumption as owed. Here is what the theorem now pays for, named by chapter so the credit can be tracked.

• Section 7.2, immediately. A series $\sum_n c_ne_n$ with $\sum_n\abs{c_n}^{2}\lt\infty$ converges to a vector. Without completeness the coefficients would be square-summable and the sum would be nothing, which would leave "expand in a basis" meaningless.
• Chapter 4.5's spectral theorem. Every construction there produces its operator as a limit of simpler ones, and the limit has to exist in the space it started in.
• Chapter 4.6's time evolution. The operator $\ee^{-\ii\hat Ht/\hbar}$ is defined by a series, and applying it to a state produces a state only because the partial sums are Cauchy and converge.
• Chapter 4.15's perturbation series. The first correction to a state is an infinite sum over the unperturbed basis, and it names a state only because its coefficients are square-summable and §7.2 then supplies a limit.

a natural place to stop  ·  the space is complete; what follows is finding a basis in it

7 · Orthonormal bases in infinite dimensions

Completeness in hand, we can finally ask what a basis is here. In finite dimensions Chapter 0.4 settled it by counting, and the count is not available. Two tools come first. Section 7.1 shows that the coefficients of any vector against any orthonormal set are square-summable, and §7.2 shows that any square-summable list of coefficients assembles into a vector, which is where §6 gets spent. What then replaces the count is four statements which look different, are all called completeness of a basis, and turn out to be the same statement. Section 7.3 proves that. Section 7.4 then shows that a basis of this space can always be listed, which is what every physics calculation has been assuming.

One older promise is being kept here too. Chapter 1.1 §4.3 observed that the configuration of a field is a function rather than a list of numbers, that the space of such functions is therefore infinite-dimensional, and that "Chapter 4.3 will take its infinite-dimensionality seriously." Taking it seriously is exactly this section, and the cost is now visible. The dimension count that settled every question in Chapter 0.4 is gone, and four separate statements have to be proved equivalent to replace it.

7.1 · Bessel's inequality

Let $e_1,e_2,\ldots$ be an orthonormal set in $L^{2}$, meaning $\avg{e_m,e_n}=\delta_{mn}$, with no assumption yet that there are enough of them. For any $f$ write $c_n=\avg{e_n,f}$ and look at how far $f$ is from $\sum_{n\le N}c_ne_n$. That combination is the projection of $f$ onto the span of the first $N$, which by Chapter 0.5 §3.1 is the closest point there, and that is why it is the combination worth measuring against. The inequality itself uses none of it. All it uses is that a squared norm can never be negative, so expand the squared distance and let the cross terms collapse:

That is Bessel's inequality, and the derivation is Chapter 0.5's Parseval computation with the equality weakened to an inequality because the set may not be big enough. Its content is small and its consequence is not: whatever else is true, the coefficients of any $f$ against any orthonormal set are square-summable.

7.2 · Square-summable coefficients build a vector

Bessel says the coefficients are well behaved. Completeness turns that into an actual vector, and this is the first place §6 gets spent. Suppose $\sum_n\abs{c_n}^{2}\lt\infty$ and look at the partial sums $S_N=\sum_{n\le N}c_ne_n$. Orthonormality makes the distance between two of them a tail of a convergent series:

So $(S_N)$ is Cauchy, and by Riesz–Fischer it converges to some vector of $L^{2}$. The series $\sum_n c_ne_n$ therefore means something, for any square-summable coefficients whatever. Without §6 the partial sums would be closing on a gap and the notation would name nothing.

One small tool is needed repeatedly below and is worth extracting. Cauchy–Schwarz gives $\abs{\avg{u,v}-\avg{u,v'}}=\abs{\avg{u,v-v'}}\le\lVert u\rVert\lVert v-v'\rVert$, so the inner product is continuous in each slot. Norm convergence may therefore be taken inside any inner product, which is used at every step of §7.3.

7.3 · Four statements, and the proof that they are one

The word "complete" is used of a basis as well as of a space, and it means something different there. Worse, it means four different-looking things, and physics texts move between them without comment. Here they are, for an orthonormal set $\{e_n\}$ in a Hilbert space $H$.

• (a) Spanning. The finite linear combinations of the $e_n$ come arbitrarily close to every vector of $H$.
• (b) Expansion. Every $f$ equals $\sum_n\avg{e_n,f}e_n$, the series converging in norm.
• (c) Parseval. Every $f$ satisfies $\lVert f\rVert^{2}=\sum_n\abs{\avg{e_n,f}}^{2}$.
• (d) No blind spot. The only vector orthogonal to every $e_n$ is the zero vector.

An orthonormal set satisfying these is an orthonormal basis. The equivalence is what makes the definition checkable, because (d) is the one you can usually verify and (b) is the one you want to use.

(b) implies (a), since the partial sums of the expansion are finite combinations and they converge to $f$.

(a) implies (d). Let $f$ be orthogonal to every $e_n$, hence to every finite combination of them. By (a) there are finite combinations $g_k\to f$ in norm. Continuity of the inner product from §7.2 then gives

because every term in the sequence is zero. So $f$ is the zero vector.

(d) implies (b), and this is the step that uses everything. Given $f$, its coefficients $c_n=\avg{e_n,f}$ are square-summable by Bessel, so §7.2 says $g=\sum_nc_ne_n$ exists in $H$. Pair any $e_m$ against the difference $f-g$, moving the limit inside by continuity, and only one term survives orthonormality:

So $f-g$ is orthogonal to every $e_m$, and (d) forces $f=g$, which is (b).

(b) implies (c), by taking the norm of the expansion. Pythagoras from Chapter 0.5 §1.3 gives $\lVert S_N\rVert^{2}=\sum_{n\le N}\abs{c_n}^{2}$, and the norm is continuous because $\big|\lVert u\rVert-\lVert v\rVert\big|\le\lVert u-v\rVert$, so letting $N$ run turns that into (c). And (c) implies (d) immediately: a vector orthogonal to every $e_n$ has all coefficients zero, so (c) makes its norm zero, so it is the zero vector.

The implications proved are (b) to (a) to (d) to (b), which is a closed cycle containing three of the four, and (b) to (c) to (d), which brings in the fourth. All four are therefore equivalent. $\blacksquare$

Two things are worth noticing about which chapter supplied which step. The finite-dimensional proof in Chapter 0.5 §2.3 needed none of this, because there the four statements are all immediate consequences of counting dimensions. Here the count is unavailable and completeness of the space is what replaces it, entering at §7.2 and nowhere else. That is the precise sense in which Chapter 4.2's basis expansions were on credit.

7.4 · A basis can always be listed

Everything in §7.3 assumed the orthonormal set was indexed by the integers, and that assumption needs justifying, because a Hilbert space could in principle need an uncountable basis. It cannot, for this space, and the reason is that $L^{2}$ is not too large to be approximated from a list.

A space is separable when it contains a countable dense subset. Building one for $L^{2}(\R)$ takes three steps, and the second is where §2.3's regularity clause is spent. It is spent once, here, and §8.1 re-uses the same step rather than paying for it again.

• Simple functions are dense. For $f\ge0$ in $L^{2}$ take the staircase (4.3.17). Then $\abs{f-\varphi_n}^{2}\le\abs f^{2}$, which is integrable, and $\abs{f-\varphi_n}^{2}\to0$ pointwise, so dominated convergence gives $\lVert f-\varphi_n\rVert_2\to0$. A general complex $f$ splits into four non-negative pieces.
• Indicators are nearly intervals. For measurable $E$ of finite measure, regularity supplies an open $O\supseteq E$ with $\mu(O\setminus E)$ as small as desired. Any open subset of $\R$ is a countable disjoint union of open intervals, and the reason is worth a line: take for each point the largest open interval inside $O$ containing it, two such intervals are equal or disjoint, and each one holds a rational, so there are countably many. Countable additivity then lets finitely many of them carry all but a little of the measure, and keeping those leaves a set $A$ with $\mu(E\triangle A)$ small, so $\lVert\mathbf 1_E-\mathbf 1_A\rVert_2^{2}=\mu(E\triangle A)$ is small.
• Count them. Perturbing the endpoints of those finitely many intervals to rationals, and the coefficients of the simple function to rationals with rational imaginary parts, changes the norm by as little as you like and leaves a countable collection.

So $L^{2}(\R)$ is separable. Now the consequence, which is three lines. Let $\{e_\alpha\}$ be any orthonormal set, and note that distinct members are far apart, since orthonormality and Pythagoras give

Put an open ball of radius $\tfrac12\sqrt2$ around each $e_\alpha$, and by (4.3.41) the balls are disjoint. A countable dense set must place at least one of its points in every one of them, so the balls can be labelled injectively by a countable set, and there are at most countably many of them. Every orthonormal set in a separable Hilbert space is countable, so every orthonormal basis of $L^{2}$ can be written as a sequence $e_1,e_2,e_3,\ldots$ and expansions can be written as ordinary series.

Existence is settled the same way. Run Chapter 0.5 §2.1's Gram–Schmidt on a countable dense set, discarding any vector that is already in the span of its predecessors. The output is a countable orthonormal set whose finite combinations reach everything the dense set reached, which is statement (a), so it is a basis. The recipe is the one from Chapter 0.5 with no modification, which is a third instance of that chapter's machinery not caring what the vectors are.

One consequence deserves stating, with a caution attached to it. Given two separable infinite-dimensional Hilbert spaces with bases $\{e_n\}$ and $\{e_n'\}$, the map sending $\sum c_ne_n$ to $\sum c_ne_n'$ is defined on everything by (b), lands everywhere by (b), and preserves norms by (c). So it is unitary in Chapter 0.5 §4.3's sense. Up to a unitary map there is only one separable infinite-dimensional Hilbert space, and every question about which one to use is a question about which basis is convenient.

Here is the caution, because that result is easy to over-spend. It depends on a choice of basis at each end and singles out no particular map, so it can never tell you that some given map is the right one. Section 8.5 needs exactly that, and it gets it from Plancherel instead. Nothing later in this chapter uses the statement just proved.

8 · The Fourier basis is a basis

Chapter 0.9 proved that the Fourier modes are orthonormal, said in a marked box that it had not proved there were enough of them, and named this chapter. Section 8.1 shows that continuous functions are dense, using nothing new. Section 8.2 proves, constructively, that a continuous periodic function is a uniform limit of trigonometric polynomials. Section 8.3 puts the two together and the mark comes off. Section 8.4 then measures the two kinds of convergence against each other on Chapter 0.9's own square wave, and §8.5 collects the last debt of the chapter.

8.1 · Continuous functions are dense, on the regularity step §7.4 already paid for

This subsection supplies the first of the two links §8.3 needs. Every vector of $L^{2}$ sits as close as you like to a continuous function, and on $[-\pi,\pi]$ to a continuous periodic one, which is what will let §8.2 work only with continuous functions and still reach the whole space.

Nothing here is quoted, and it is worth being explicit about why, because the result looks like the kind of thing that ought to need a flag. Section 2.3's package included regularity, clause (d), and §7.4 spent it. This section re-uses that step and adds one link of its own. Watch the clause do the second of the two jobs it was bought for.

The chain has three links and the first two are already built. Simple functions are dense in $L^{2}$, by the staircase argument of §7.4. An indicator $\mathbf 1_E$ is close in $L^{2}$ to the indicator of a finite union of intervals, which is §7.4's regularity step verbatim. So it is enough to approximate the indicator of a single interval by a continuous function, and that is done by hand.

Replace the vertical sides of $\mathbf 1_{[a,b]}$ by straight ramps of horizontal width $\delta$, giving a continuous trapezoid $T_\delta$. The two functions agree except on the two ramps, where they differ by at most $1$, so

which goes to zero with $\delta$. Composing the three links, the continuous functions are dense in $L^{2}$. On $[-\pi,\pi]$ one extra turn of the same screw is needed for §8.2: place the ramps so that the trapezoid vanishes near both endpoints, and it extends to a continuous $2\pi$-periodic function on the line with no jump at the seam. So the continuous $2\pi$-periodic functions are dense in $L^{2}[-\pi,\pi]$ as well.

8.2 · Fejér's theorem, built from a kernel you have already met

Work on $[-\pi,\pi]$, which is Chapter 0.9's box with $L=2\pi$ and therefore $k_n=n$, and write Chapter 0.9's coefficients and partial sums:

Direct attack on $S_Nf$ is the route that fails, and the reason is worth naming in advance. The argument below works by writing the approximation as an integral of $f$ against a weight, and it needs that weight to be non-negative. The weight belonging to $S_Nf$ is the Dirichlet kernel, written out below when the averaging is performed, and it is a ratio of two sines, which changes sign. The route that works replaces the partial sums by their running average, which turns the weight into a square. Define the Cesàro means

To see what averaging has done, substitute (4.3.43) into (4.3.44) and pull the sum inside the integral, which is legal because it is finite. What comes out is a convolution against a single function of $t=x-y$:

The inner sum is a geometric series, and summing it gives the Dirichlet kernel $D_M(t)=\sin\big((M+\tfrac12)t\big)/\sin(t/2)$. Averaging those over $M$ is a telescoping exercise. Multiply by $2\sin^{2}(t/2)$, which turns each $D_M$ into $2\sin(t/2)\sin\big((M+\tfrac12)t\big)=\cos(Mt)-\cos\big((M+1)t\big)$, and everything cancels except the ends, leaving $1-\cos\big((N+1)t\big)=2\sin^{2}\big((N+1)t/2\big)$. So

Now read off three properties of (4.3.46), because they are the whole theorem. It is non-negative, being a square, and that is exactly what the averaging bought. Its integral is one, since $\frac{1}{2\pi}\int_{-\pi}^{\pi}D_M=1$ for every $M$ as only the $n=0$ term survives, and the average of ones is one. And it concentrates at the origin, since for $\delta\le\abs t\le\pi$ the denominator is bounded below and

A non-negative bump, of unit integral, narrowing at the origin. That is Chapter 0.9 §5.2's delta-sequence, and the argument that follows is that section's proof with the Gaussian replaced by (4.3.46).

Fejér's theorem. If $f$ is continuous and $2\pi$-periodic then $\sigma_Nf\to f$ uniformly. Since $\int K_N/2\pi=1$, the error is

and it is split at $\abs t=\delta$. On the inside, $f$ is uniformly continuous, being continuous on a closed bounded interval, so $\delta$ can be chosen once and for all to make $\abs{f(x-t)-f(x)}\lt\varepsilon$ for every $x$ at once. That part of the integral is then below $\varepsilon$, because $K_N$ is non-negative with unit integral. On the outside, the bracket is at most $2\max\abs f$ and (4.3.47) drives $K_N$ to zero uniformly, so that part is below $\varepsilon$ for $N$ large. The bound is uniform in $x$ throughout. $\blacksquare$

Two notes on what that proof used. The uniform continuity is Heine–Cantor, which Chapter 0.2 §1.1 quoted and marked when it proved that continuous functions are Riemann integrable; this chapter leans on that standing mark and does not raise a new one. And the theorem is constructive: it produces the approximating trigonometric polynomial explicitly, as an average of partial sums, rather than asserting one exists. Since $\sigma_Nf$ is a finite combination of the $\ee^{\ii nx}$, what the theorem says is that the trigonometric polynomials are dense, in the supremum norm, in the continuous $2\pi$-periodic functions, and that a member of the dense set can be written down from $f$ by averaging.

8.3 · The mark comes off

Now assemble, and note that each link was proved above and none is quoted.

• Uniform convergence implies $L^{2}$ convergence on a bounded interval, since $\lVert g\rVert_2\le\sqrt{2\pi}\max\abs g$. So by Fejér's theorem in §8.2 the trigonometric polynomials come arbitrarily close in $L^{2}$ to every continuous periodic function.
• Those are dense in $L^{2}[-\pi,\pi]$, by §8.1.
• Therefore the trigonometric polynomials are dense in $L^{2}[-\pi,\pi]$, which is statement (a) of §7.3 for the set $\{e_n\}$ with $e_n(x)=\ee^{\ii nx}/\sqrt{2\pi}$.
• Chapter 0.9 §1.1 proved by direct integration that this set is orthonormal.

By the equivalence of §7.3, all four statements hold. So the Fourier modes are an orthonormal basis of $L^{2}[-\pi,\pi]$, and in particular, for every $f$ with $\int\abs f^{2}\lt\infty$,

8.4 · Two kinds of convergence, measured against each other

Equation (4.3.49) is a statement about a norm, and Chapter 0.9 §1.4 exhibited a function for which it is emphatically not a statement about points. This is the last piece of that chapter's mark, and it is also the numerical confirmation this chapter owes.

First the pointwise theorem, so that the two claims can be set beside each other. It rests on one lemma about oscillating integrals, and the lemma is worth stating because it recurs throughout Part V.

Riemann–Lebesgue lemma. If $\phi$ is integrable then $\int\phi(t)\sin(\lambda t)\,\dd t\to0$ as $\lambda\to\infty$. The reason is that rapid oscillation averages a slowly varying weight against itself, and the proof turns that into three lines in the grind box.

With it, the classical statement about jumps follows. If $f$ is $2\pi$-periodic and integrable, and at the point $x$ both one-sided limits exist with bounded one-sided difference quotients, then $S_Nf(x)\to\tfrac12\big[f(x^{+})+f(x^{-})\big]$. At a point of continuity that is $f(x)$, and at a jump it is the midpoint, exactly as Chapter 0.9 stated without proof.

One word about the hypothesis, because Chapter 0.9 gave it a name and this one does not match it. That chapter asked for the Dirichlet conditions, piecewise smooth with finitely many extrema and jumps, imposed on $f$ across the whole interval. The statement above asks less. It is a condition at the single point $x$, and away from $x$ it wants nothing beyond integrability. The same conclusion from a weaker hypothesis is the stronger theorem, so the Dirichlet conditions are superseded here rather than merely met.

Grind box — Riemann–Lebesgue, and convergence to the midpoint of a jump

Riemann–Lebesgue. For $\phi=\mathbf 1_{[a,b]}$ the integral is computed outright,

$$\left|\int_a^b\sin(\lambda t)\,\dd t\right| \;=\; \left|\frac{\cos\lambda a-\cos\lambda b}{\lambda}\right| \;\le\; \frac{2}{\lambda} \;\to\;0,$$

so it holds for step functions by linearity. The regularity argument of §7.4, run with the $L^{1}$ norm in place of the $L^{2}$ norm, makes step functions dense in $L^{1}$. So given $\varepsilon$ pick a step function $s$ with $\lVert\phi-s\rVert_1\lt\varepsilon$, and

$$\left|\int\phi\sin\lambda t\right| \;\le\; \int\abs{\phi-s} \;+\; \left|\int s\sin\lambda t\right| \;\lt\; \varepsilon + \frac{C}{\lambda},$$

which is below $2\varepsilon$ for large $\lambda$. $\blacksquare$

The jump. Summing the geometric series in (4.3.43) gives $S_Nf(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-t)D_N(t)\dd t$ with $D_N(t)=\sin\big((N+\frac12)t\big)/\sin(t/2)$. The kernel is even and $\frac{1}{2\pi}\int_{-\pi}^{\pi}D_N=1$, so $\frac{1}{2\pi}\int_0^{\pi}D_N=\frac12$ and therefore

$$\begin{aligned} S_Nf(x)-\frac{f(x^{+})+f(x^{-})}{2} \;=\;& \frac{1}{2\pi}\int_0^{\pi}\!\big[f(x-t)-f(x^{-})\big]D_N(t)\dd t\\[4pt] +\;& \frac{1}{2\pi}\int_0^{\pi}\!\big[f(x+t)-f(x^{+})\big]D_N(t)\dd t. \end{aligned}$$

Take the first. Writing $\phi(t)=\big[f(x-t)-f(x^{-})\big]/\sin(t/2)$ turns it into $\frac{1}{2\pi}\int_0^{\pi}\phi(t)\sin\big((N+\frac12)t\big)\dd t$. Near $t=0$ the numerator is at most a constant times $t$ by the bounded difference quotient, and $\sin(t/2)$ behaves like $t/2$, so $\phi$ is bounded near the origin and integrable on $[0,\pi]$. Riemann–Lebesgue kills it, and the second term goes the same way. $\blacksquare$

Now put the two convergences on the same function and watch them separate. Chapter 0.9 §1.4's square wave on $[-\pi,\pi]$ is $f=+1$ on $(0,\pi)$ and $f=-1$ on $(-\pi,0)$, with the series it derived there,

Write $S_N$ for the partial sum keeping every odd harmonic up to $N$. Parseval, which is now a theorem rather than an assumption, turns the norm of the error into a tail of a series. Using $\int_{-\pi}^{\pi}\sin^{2}nx\,\dd x=\pi$ and squaring the coefficients of (4.3.50),

the tail being estimated by an integral because the odd integers past $N$ are spaced by two. Taking the square root gives the rate this chapter promised to confirm:

Against that, set the pointwise behaviour. Chapter 0.9's grind box computed the peak of the partial sum near the jump and found it converging to $\tfrac2\pi\mathrm{Si}(\pi)=1.178980$ rather than to $1$. The jump has size $2$, so the overshoot, as a fraction of the jump, is

Two numbers, both exact, both about the same sequence of functions, and they do opposite things. Equation (4.3.52) falls to zero. Equation (4.3.53) settles on a constant and stays there forever.

They are consistent, and the reconciliation is one sentence. The overshoot keeps its height and loses its width. The first peak sits at $x=\pi/(N+1)$, so the whole ear is squeezed into a window shrinking like $1/N$, and a fixed height on a vanishing width contributes a vanishing amount to $\int\abs{f-S_N}^{2}$. A norm integrates and cannot see it. A maximum does not integrate and sees nothing else. The figure below is that sentence made visible, and it is worth using the zoom control, because the ear is invisible at any fixed magnification once $N$ is large and is exactly the same height at every magnification that tracks it.

highest harmonic N

N = 17

whole period track the jump

L2: || f - S_N ||_2 / || f ||_2 = 0.149976 times sqrt(N) = 0.618366 against 2/pi = 0.636620 (exact, from the coefficient tail)

pointwise: peak of S_N = 1.180010 overshoot = 0.090005 of the jump against (1/pi)Si(pi) - 1/2 = 0.089490 (measured on the plotted curve)

slopes over the last decade of N: L2 error -0.4981 overshoot -0.0001 -- one converges like N^(-1/2), the other does not converge at all

The physical reading matters more than the arithmetic, and it is the reason §5.3 warned about pointwise values in advance. Every statement quantum mechanics makes about a continuous coordinate is an integral over a region, so the convergence that has physical content is the one in (4.3.52). A claim about the value of a wavefunction at a point is not a claim $L^{2}$ can support, and the Gibbs ear is what that looks like when you insist on making one.

8.5 · Position and momentum are one space, joined by a unitary map

The last debt of the chapter is Chapter 0.9's closing brick, which listed where its results would be spent and sent the first entry here: "The Fourier basis and Plancherel → Chapter 4.3 (where completeness is finally proved and the position and momentum representations become two bases for one Hilbert space)."

Assemble what is now available, and it is one thing rather than two. Chapter 0.9 §2.3 proved Plancherel, that the transform $\tilde f(k)=\frac{1}{\sqrt{2\pi}}\int f\,\ee^{-\ii kx}\dd x$ preserves the norm, and read it back as the statement that the transform is unitary. That is the whole of what is needed here, and it is a statement about this one map. Section 7.4's result, that any two separable infinite-dimensional Hilbert spaces are unitarily equivalent, singles out no map at all and does no work in this section.

So what Plancherel says is that $\psi$ and $\tilde\psi$ are not two objects but two descriptions of one. Square-integrable functions of $x$ and square-integrable functions of $k$ are the same Hilbert space, and the transform is a unitary map between the two descriptions. Chapter 0.9's phrase two bases for one Hilbert space is the loose reading of that, and the box below says exactly how loose.

Writing it with $\hbar$ in place, as Part IV does everywhere, the momentum-space wavefunction is $\tilde\psi$ rescaled so that its own integral is one, with the $\sqrt\hbar$ visible:

The normalisations agree because the change of variable $p=\hbar k$ contributes exactly the $\hbar$ that the prefactor removes, which is the reason the symmetric convention of Chapter 0.9 §2.2 was chosen there. So a state has one norm, and the Born rule computes the same total probability in either description.

9 · Worked examples

10 · Your turn