A position wavefunction is integrated over regions of space. To give that statement a precise meaning, we need a notion of measurable region, an integral that handles limits, and a space in which functions that differ only on negligible sets represent the same vector. This chapter builds those tools from elementary real analysis.
Our target is the opening framework and observables and Born probabilities pages of Alphabeta Math, checked on October 15, 2026. We assume elementary real analysis, including sequences, limits, the completeness of the real numbers, and the Riemann integral, along with the Euclidean metric, basic topology, and complex arithmetic. We use the compactness of closed bounded Euclidean boxes proved in chapter 2. Probability comes in chapter 5; Hilbert spaces and operators come in chapter 6.
4.1. Which sets can we measure?
A set is countable if it is finite or its elements can be listed as a sequence indexed by the positive integers. A countable union permits finitely many sets too, by appending empty sets. The rationals are countable: list fractions with integer numerator and positive integer denominator in increasing order of the sum of the numerator’s absolute value and the denominator; each group is finite. Finite lists of rational endpoints are countable by the same enumeration of their integer numerators and denominators.
Definition 4.1. Measurable spaces
A σ-algebra on a set \(X\) is a collection \(\Sigma\) of subsets containing \(X\), closed under complements relative to \(X\) and under countable unions. The pair \((X,\Sigma)\) is a measurable space; members of \(\Sigma\) are measurable sets. The symbol σ indicates countable closure.
The Borel σ-algebra \(\mathcal B(X)\) on a topological space is the smallest σ-algebra containing its open sets. “Smallest” means the intersection of all such σ-algebras.
Alphabeta Math: Borel σ-algebra
We need countable unions because a sequence of approximations can describe a region through infinitely many steps. Complements let us describe an excluded region. De Morgan’s laws show closure under countable intersections too; \(\varnothing=X^c\) is measurable, as are differences \(A\setminus B=A\cap B^c\). Intersections of σ-algebras have each closure property, so the Borel definition does specify a σ-algebra.
Examples. Both \(\{\varnothing,X\}\) and the collection of all subsets of \(X\) are σ-algebras. On \(\mathbb R\), intervals and closed sets are Borel. Indeed closed sets are complements of open sets, and \([a,b]=\bigcap_{n\ge1}(a-1/n,b+1/n)\). Every open subset of \(\mathbb R^d\) is a countable union of boxes with rational endpoints lying inside it: a sufficiently small box can be placed about each point, and there are only countably many rational endpoint lists. To see the countability, list rational numbers by integer numerator and positive integer denominator, ordering the pairs by the sum of their absolute sizes; list finite tuples by the same diagonal procedure. Thus these boxes generate \(\mathcal B(\mathbb R^d)\).
Definition 4.2. Measures and null sets
A measure on \((X,\Sigma)\) is a function \(\mu:\Sigma\to[0,\infty]\) with \(\mu(\varnothing)=0\) and countable additivity:
\[\mu\!\left(\bigcup_{n=1}^{\infty}A_n\right)=\sum_{n=1}^{\infty}\mu(A_n)\quad\text{if the measurable }A_n\text{ are pairwise disjoint}.\]The sum is the limit of its nonnegative partial sums. A measurable \(N\) is null if \(\mu(N)=0\). A statement holds almost everywhere, abbreviated a.e., if it holds outside a measurable null set. The measure is complete if every subset of a measurable null set is measurable. A measure space is \((X,\Sigma,\mu)\).
Examples. On any set with all subsets measurable, counting measure assigns the number of elements to a finite set and \(\infty\) to an infinite one. Disjoint unions count each element once, proving the axiom. For a fixed \(x_0\), the point mass \(\delta_{x_0}(A)=1\) if \(x_0\in A\), and zero otherwise, is a measure: at most one member of a disjoint family contains \(x_0\). Length on the real line will be constructed below. Nullness depends on the measure: a singleton has length zero but counting measure one.
Proposition 4.3. Basic measure rules
A measure is monotone, \(A\subseteq B\Rightarrow\mu(A)\le\mu(B)\), and countably subadditive, \(\mu(\bigcup_n A_n)\le\sum_n\mu(A_n)\). If \(A_n\subseteq A_{n+1}\), then \(\mu(\bigcup_n A_n)=\lim_n\mu(A_n)\). If \(A_{n+1}\subseteq A_n\) and \(\mu(A_1)<\infty\), then \(\mu(\bigcap_n A_n)=\lim_n\mu(A_n)\).
Worked example for Proposition 4.3
Counterexample: Continuity from above needs finite initial measure
Proof.
Write \(B=A\sqcup(B\setminus A)\), where \(\sqcup\) means disjoint union, to obtain monotonicity. Replace a general family by disjoint pieces \(B_n=A_n\setminus\bigcup_{j<n}A_j\). They have the same union, and \(\mu(B_n)\le\mu(A_n)\), proving subadditivity. For an increasing family, its disjoint increments are \(A_1,A_2\setminus A_1,\ldots\). Their first \(n\) measures sum to \(\mu(A_n)\), so countable additivity proves continuity from below. For a decreasing family apply this to \(A_1\setminus A_n\); subtract their measures from the finite number \(\mu(A_1)\). This gives continuity from above without an undefined \(\infty-\infty\).
□Examples and a failure. Countable unions of null sets are null by subadditivity. For length, \((0,1-1/n)\) increases to \((0,1)\); its lengths tend to one. The finite-measure hypothesis for decreasing sets is necessary: \([n,\infty)\) has infinite length for each \(n\), because it contains intervals of arbitrarily large length, while its intersection is empty.
4.2. Constructing length and volume
It would be circular to integrate before defining the measure used by the integral. Here is a direct construction. Read it in dimension \(d=1\) first: boxes then mean bounded intervals and volumes mean lengths. The same proof works in every positive integer dimension, including the \(d=3\) used for particle position.
Definition 4.4. Lebesgue outer measure
A bounded coordinate box \(Q\subset\mathbb R^d\) is a product of intervals with endpoints \(a_j\le b_j\), with any choice of including endpoints. Its volume is \(v(Q)=\prod_{j=1}^d(b_j-a_j)\). Empty boxes have volume zero. For any subset \(E\subseteq\mathbb R^d\), define
\[\lambda_d^*(E)=\inf\left\{\sum_{n=1}^{\infty}v(Q_n): E\subseteq\bigcup_{n=1}^{\infty}Q_n\right\}.\]This measures the least total volume of a countable box cover. A finite cover is included by appending empty boxes. Covers always exist, since \(\mathbb R^d\) is the union of countably many bounded boxes.
Proof of the outer-measure rules and the volume of a box
The empty cover gives \(\lambda_d^*(\varnothing)=0\). A cover of a larger set covers a smaller one, giving monotonicity. For subadditivity, if \(\sum_n\lambda_d^*(E_n)=\infty\) there is nothing to prove. Otherwise choose a box cover of \(E_n\) costing at most \(\lambda_d^*(E_n)+\varepsilon2^{-n}\). Their combined countable cover costs at most \(\sum_n\lambda_d^*(E_n)+\varepsilon\). Let \(\varepsilon\downarrow0\). These three rules define an outer measure; they do not yet give additivity.
For a closed box \(K\), its own cover gives \(\lambda_d^*(K)\le v(K)\). For the opposite inequality, take any countable box cover. Enlarge the \(n\)-th box to a bounded open box with additional volume at most \(\varepsilon2^{-n}\), possible by continuity of a finite product as the endpoints move. The compact-box theorem proved in chapter 2 gives a finite subcover of \(K\). For a finite box cover, partition a large surrounding box using all coordinate endpoints of the cover and of \(K\). Each grid cell inside \(K\) is inside at least one covering box, apart from its boundary. Summing the products of the cell side lengths gives \(v(K)\); summing over each covering box counts each cell at least once and gives the sum of those box volumes. The grid identities are just the distributive law for products of sums of lengths. Hence \(v(K)\) is at most the total cost of the original cover plus \(\varepsilon\). Let \(\varepsilon\downarrow0\) and take the infimum. Any other box with the same endpoints contains closed boxes with endpoints approaching its endpoints from inside, unless its volume is zero. Monotonicity therefore gives the same lower bound; in the zero-volume case nonnegativity suffices. Consequently \(\lambda_d^*(Q)=v(Q)\) for every box.
Definition 4.5. Lebesgue measurable sets
A set \(A\subseteq\mathbb R^d\) is Lebesgue measurable if, for every subset \(E\subseteq\mathbb R^d\),
\[\lambda_d^*(E)=\lambda_d^*(E\cap A)+\lambda_d^*(E\setminus A).\]Write \(\mathcal L_d\) for these sets and \(\lambda_d(A)=\lambda_d^*(A)\) for their measure. The condition says that cutting a test set along \(A\) loses no outer volume. Subadditivity always gives the ≤ direction; measurability requires the other direction too.
Theorem 4.6. The construction gives a complete Borel-containing measure
\(\mathcal L_d\) is a σ-algebra, contains all Borel sets, and \(\lambda_d\) is a complete countably additive measure. It assigns every box its volume and every countable set measure zero.
Outer measure and Carathéodory’s theorem; Euclidean Lebesgue measure
Proof of the σ-algebra, additivity, completeness, and Borel claims
The following argument works for any outer measure \(m^*\). The empty set satisfies the cutting condition, and the condition is symmetric in \(A,A^c\). If \(A,B\) satisfy it, split \(E\) first along \(A\) and then split \(E\cap A\) along \(B\). Subadditivity for the two pieces outside \(A\cap B\) gives
\[m^*(E)\ge m^*(E\cap A\cap B)+m^*(E\setminus(A\cap B)).\]The reverse inequality is automatic. Thus intersections, and then finite unions and differences, are measurable. If \(A_n\) are disjoint measurable sets, successive cutting gives, for every \(k\),
\[m^*(E)\ge\sum_{n=1}^k m^*(E\cap A_n)+m^*\!\left(E\setminus\bigcup_{n\ge1}A_n\right).\]Let \(k\to\infty\). Subadditivity bounds \(m^*(E\cap\bigcup_n A_n)\) by the infinite sum, so the union satisfies the cutting condition. Disjointifying arbitrary measurable unions proves countable closure. Taking \(E=\bigcup_n A_n\) in the same finite inequalities, followed by subadditivity, proves countable additivity. If \(m^*(N)=0\), every subset \(S\subseteq N\) has outer measure zero. For every \(E\), monotonicity gives \(m^*(E)\ge m^*(E\setminus S)\), which is the cutting condition because \(m^*(E\cap S)=0\). This proves completeness.
For our box outer measure, the half-space \(H=\{x:x_j\le c\}\) is measurable. Cut every box of a cover into its intersections with \(H\) and \(H^c\); both pieces are boxes or empty, and their volumes sum to the original volume. Therefore their covers give \(\lambda_d^*(E\cap H)+\lambda_d^*(E\setminus H)\le\sum_n v(Q_n)\). Take the infimum. Complements give \(\{x:x_j>c\}\), and \(\{x:x_j<c\}=\bigcup_{n\ge1}\{x:x_j\le c-1/n\}\). Finite intersections therefore give all open boxes, and the rational-box description of open sets gives all Borel sets. Box volumes were proved above. A singleton lies in boxes of arbitrarily small volume, hence is null; countable subadditivity proves the countable-set claim.
Examples. \(\lambda_1([a,b])=b-a\); changing either endpoint does not affect length, because singletons are null. The rationals in \([0,1]\) have measure zero, although every subinterval contains a rational. Thus “null” does not mean “empty” or “far from every point.” A coordinate hyperplane is also null: its intersection with \([-m,m]^d\) fits in a box of thickness \(2\varepsilon\) in one direction and side lengths \(2m\) in the others, with volume tending to zero. The whole hyperplane is a countable union of these null intersections.
Lebesgue measurability includes all subsets of Lebesgue null sets, by the completeness proof. Below we work with a stated measure space, and with this complete Lebesgue measure when integrating over \(\mathbb R^d\). We do not assume every subset of Euclidean space is measurable.
4.3. Measurable functions and simple approximation
Definition 4.7. Measurable functions
A real function \(f:X\to\mathbb R\) is measurable if \(\{x:f(x)>t\}\in\Sigma\) for every real \(t\). We also allow extended real values in \([-\infty,\infty]\), meaning the real numbers together with a value below every real number and one above every real number, using the same threshold condition. A nonnegative extended function takes values in \([0,\infty]\). A complex function is measurable when its real and imaginary parts are measurable. For a measurable set \(A\), its indicator \(\mathbf1_A\) equals one on \(A\) and zero elsewhere. A simple function takes finitely many finite values and has measurable level sets; it can be written \(s=\sum_{j=1}^k a_j\mathbf1_{A_j}\) with disjoint measurable \(A_j\).
Alphabeta Math: Measurable functions and simple approximation
The inverse image \(f^{-1}(B)=\{x:f(x)\in B\}\) is measurable for every real Borel set \(B\). To prove this equivalence, the sets \(B\) whose inverse images are measurable form a σ-algebra. It contains the rays \((t,\infty)\), hence intervals and all open sets by the rational basis. The converse follows by taking a ray. This explains why the definition uses thresholds: they already test every Borel condition on a function’s value.
Examples. Continuous Euclidean functions are Borel measurable: inverse images of open sets are open. The indicator of the rationals is measurable because its two level sets are Borel. It is discontinuous everywhere, so measurable functions are a larger class than continuous functions. With \(X=\{a,b\}\) and \(\Sigma=\{\varnothing,X\}\), a function taking distinct values at \(a,b\) is not measurable: a threshold between its values has a nonmeasurable inverse image.
Proposition 4.8. Operations and limits preserve measurability
Sums, products, absolute values, maxima and minima of finite real measurable functions are measurable; the same applies to complex sums, products, conjugates and moduli. Countable suprema and infima, and pointwise limits when they exist, are measurable as extended real functions. Every nonnegative measurable \(f\) has nonnegative simple approximations \(s_n\uparrow f\).
Proof.
For real \(f,g\), the inverse image of an open rectangle under \(x\mapsto(f(x),g(x))\) is the intersection of two measurable inverse images. Every open subset of \(\mathbb R^2\) is a countable union of rational rectangles. Therefore composing this pair with any continuous real function is measurable. This proves sum, product, max and min; the single-function argument proves absolute value. Apply it to real and imaginary components to obtain the complex statements. For a supremum, \(\{\sup_n f_n>t\}=\bigcup_n\{f_n>t\}\). Infima follow from negation, or directly from countable intersections of threshold sets. Thus \(\liminf f_n=\sup_N\inf_{n\ge N}f_n\) and \(\limsup f_n=\inf_N\sup_{n\ge N}f_n\) are measurable, proving the limit statement. Finally put
\[s_n(x)=2^{-n}\left\lfloor 2^n\min(f(x),n)\right\rfloor,\]where \(\lfloor r\rfloor\) is the greatest integer at most \(r\), and \(\min(\infty,n)=n\). Each level set is a measurable inverse image of an interval. The finer dyadic grid and larger cap make \(s_n\le s_{n+1}\le f\). At a finite value, eventually the cap does not act and \(0\le f-s_n<2^{-n}\); at infinity, \(s_n=n\). Hence \(s_n\uparrow f\).
□Example. On \([0,1]\), these approximations to \(f(x)=x\) are staircases of heights in multiples of \(2^{-n}\). Unlike the Riemann method, we group points by function value; a level set can be scattered throughout the domain.
4.4. Defining the integral
Definition 4.9. Nonnegative Lebesgue integral
For nonnegative simple \(s=\sum_j a_j\mathbf1_{A_j}\) on disjoint measurable sets, define \(\int_X s\,d\mu=\sum_j a_j\mu(A_j)\), using \(0\cdot\infty=0\). For a nonnegative measurable \(f\), define
\[\int_X f\,d\mu=\sup\left\{\int_X s\,d\mu:0\le s\le f,\ s\text{ simple}\right\}\in[0,\infty].\]Write \(\int_A f\,d\mu=\int_X\mathbf1_A f\,d\mu\). For Lebesgue measure we also write \(\int f(x)\,dx\).
This definition never subtracts infinities. Its meaning is the total mass assigned by measurable pieces of the domain, with each piece weighted by the function’s value. Common refinement of the sets in two simple representations gives the same integral, by finite additivity of \(\mu\); it also proves additivity, nonnegative homogeneity, and monotonicity for simple functions. In particular the supremum definition agrees with the simple integral: every simple function below a simple \(f\) has integral at most that of \(f\), and \(f\) itself is among the candidates. The supremum immediately gives monotonicity for all nonnegative functions.
Examples. \(\int\mathbf1_A\,d\mu=\mu(A)\). Under counting measure on \(\mathbb N=\{1,2,\ldots\}\), \(\int f\,d\mu=\sum_{n\ge1}f(n)\) for \(f\ge0\): finite truncations give the lower bound, and each simple function below \(f\) has integral bounded by the sum. Under \(\delta_{x_0}\), the integral is \(f(x_0)\), by the same simple-function definition.
Theorem 4.10. Monotone convergence
If \(0\le f_n\uparrow f\) pointwise, with all functions measurable, then
\[\lim_n\int f_n\,d\mu=\int f\,d\mu.\]Proof.
Monotonicity gives a limit \(L\le\int f\). Fix a nonnegative simple \(s\le f\) and \(0<c<1\). The sets \(E_n=\{f_n\ge cs\}\) increase, and every point where \(s>0\) belongs to their union, since \(f_n\uparrow f\ge s>cs\). On \(E_n\), \(f_n\ge cs\), so \(\int f_n\ge c\int\mathbf1_{E_n}s\). On each positive level set of \(s\), continuity from below of measure proves \(\int\mathbf1_{E_n}s\to\int s\). Thus \(L\ge c\int s\). Let \(c\uparrow1\) and take the supremum over \(s\). The argument also covers infinite integrals.
□Example. \(\mathbf1_{[-n,n]}\uparrow1\) on \(\mathbb R\), and its integral \(2n\) tends to \(\infty=\int1\). Monotone convergence does not require a finite total measure or finite integrals.
Proposition 4.11. Integral algebra and null sets
For nonnegative measurable \(f,g\), \(\int(f+g)=\int f+\int g\), and \(\int af=a\int f\) for \(a\ge0\). For a sequence \(f_n\ge0\), \(\int\sum_n f_n=\sum_n\int f_n\). A nonnegative function has integral zero exactly when it is zero a.e. Changing values on a measurable null set, while preserving measurability, does not change its integral.
Proof.
Choose the simple approximations from Proposition 4.8 for \(f,g\); their sums increase to \(f+g\). Simple-function additivity and monotone convergence give additivity. Scaling simple approximations gives homogeneity, with the zero case handled separately. Apply finite additivity to partial sums and monotone convergence to obtain the series formula. If \(\int f=0\), then \(\int f\ge n^{-1}\mu(\{f>1/n\})\), so each threshold set is null; their union is \(\{f>0\}\). Conversely, if \(f\) vanishes off a null set, every simple function below it has integral zero. For functions agreeing off a null set \(N\), split each integral into the parts on \(N\) and \(N^c\); the first part is zero and the second agrees.
□Example. On \([0,1]\), \(\int\mathbf1_{\mathbb Q}=0\) and \(\int\mathbf1_{\mathbb R\setminus\mathbb Q}=1\). The first function has no Riemann integral: every upper sum is one and every lower sum is zero. The Lebesgue integral admits functions that the Riemann integral cannot handle.
Definition 4.12. Real and complex integrability
For a finite real measurable function, put \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\), so \(f=f^+-f^-\) and \(|f|=f^++f^-\). It is integrable if \(\int|f|<\infty\); then \(\int f=\int f^+-\int f^-\). A measurable complex \(f=u+iv\) is integrable if \(\int|f|<\infty\); define \(\int f=\int u+i\int v\). Indeed \(|u|,|v|\le|f|\), so both component integrals exist.
Some texts also define a signed extended integral when only one of the two parts is infinite. We use “integrable” only for finite absolute integral, which is the class needed below. The expression \(\infty-\infty\) has no assigned value.
Proposition 4.13. Linearity and the integral triangle inequality
Integrable complex functions form a complex vector space, and integration is linear. For every integrable \(f\),
\[\left|\int f\,d\mu\right|\le\int|f|\,d\mu.\]Proof.
The pointwise bound \(|af+bg|\le|a||f|+|b||g|\) gives integrability. For real functions, \((f+g)^++f^-+g^-=(f+g)^-+f^++g^+\); integrating this equality of nonnegative functions and subtracting finite terms proves additivity. Positive scaling follows from nonnegative homogeneity; negative scaling exchanges positive and negative parts. Componentwise addition and multiplication now give complex linearity. If \(z=\int f\ne0\), choose \(\alpha=\overline z/|z|\), so \(\alpha z=|z|\). Then \(|z|=\int\operatorname{Re}(\alpha f)\le\int|f|\), since \(\operatorname{Re}(\alpha f)\le|f|\). For \(z=0\) the inequality is immediate.
□Example. \(f(x)=(1+i)\mathbf1_{[0,1]}(x)\) has integral \(1+i\) and \(\int|f|=\sqrt2\), giving equality. For \(f=\mathbf1_{[0,1]}-\mathbf1_{[1,2]}\), the integral is zero but the absolute integral is two. Cancellation and small absolute size are different statements.
4.5. Passing an integral through a limit
Theorem 4.14. Fatou’s lemma
For measurable \(f_n\ge0\),
\[\int\liminf_n f_n\,d\mu\le\liminf_n\int f_n\,d\mu.\]Here \(\liminf_n a_n=\sup_N\inf_{n\ge N}a_n\), including extended values.
Proof.
Let \(g_N=\inf_{n\ge N}f_n\). These measurable nonnegative functions increase to \(\liminf f_n\). Since \(g_N\le f_n\) for \(n\ge N\), \(\int g_N\le\inf_{n\ge N}\int f_n\). Let \(N\to\infty\) and use monotone convergence on the left.
□Strict inequality. On \([0,1]\), let \(f_n=n\mathbf1_{(0,1/n)}\). At each point \(f_n\to0\), but \(\int f_n=1\). Thus Fatou gives \(0\le1\), and pointwise convergence alone cannot justify exchanging integral and limit. On \(\mathbb R\), \(\mathbf1_{[n,n+1]}\to0\) gives the same failure with uniformly bounded functions: the domain has infinite measure.
Theorem 4.15. Dominated convergence
Suppose measurable real or complex \(f_n\to f\) a.e., and \(|f_n|\le g\) a.e. for one nonnegative integrable \(g\). Then \(f\) is integrable,
\[\int|f_n-f|\,d\mu\longrightarrow0,\qquad \int f_n\,d\mu\longrightarrow\int f\,d\mu.\]Here \(f\) is taken to be measurable. On a complete space it can be assigned arbitrary finite values on the exceptional null set and remains measurable.
Alphabeta Math: dominated convergence
Worked example for Theorem 4.15
Counterexample: A finite measure space does not by itself justify exchanging limit and integral
Proof.
Remove the union of the countably many exceptional null sets by assigning all functions zero there; integral values are unchanged. Taking limits gives \(|f|\le g\), so \(f\) is integrable. Set \(h_n=|f_n-f|\le2g\); then \(h_n\to0\). Fatou applied to \(2g-h_n\ge0\) gives \(2\int g\le\liminf_n(2\int g-\int h_n)=2\int g-\limsup_n\int h_n\). Hence \(\int h_n\to0\). Linearity and the integral triangle inequality give convergence of the integrals. On a complete space, changing a measurable function on a null set preserves measurability: each inverse image changes by subsets of that set, and those subsets are measurable.
□Example. On \([0,1]\), \(x^n\to0\) except at the endpoint one; the bound \(g=1\) is integrable, and \(\int_0^1x^n\,dx=1/(n+1)\to0\). In the shrinking-spike example, no integrable function can dominate all \(f_n\): if one did, dominated convergence would force their integrals to tend to zero, contradicting the computed value one.
The monotone convergence and Fatou statements also hold with a.e. hypotheses: set the functions to zero on the union of their exceptional null sets. This uses only measurable null sets and does not require completeness if the replacement is zero on the whole exceptional set.
Why the Lebesgue integral agrees with the Riemann integral
Let \(f\) be bounded and Riemann integrable on \([a,b]\). Choose successively refined finite partitions \(P_n\) whose upper and lower Riemann sums differ by less than \(2^{-n}\), using the Riemann integrability criterion and common refinements. Off the countable set of partition endpoints, define lower and upper step functions \(\ell_n,u_n\) by the infimum and supremum of \(f\) on its partition interval. Assign the values \(-M,M\) at endpoints, where \(|f|\le M\). These are measurable and satisfy \(\ell_n\uparrow\ell\le f\) and \(u_n\downarrow u\ge f\). Their Lebesgue integrals are the lower and upper Riemann sums, since interval endpoints are null. Dominated convergence gives \(\int(u-\ell)=\lim_n\int(u_n-\ell_n)=0\). Thus \(u=\ell\) outside a measurable null set. There \(f\) equals the measurable function \(\ell\); completeness makes its remaining bounded values measurable too. It is integrable since \(|f|\le M\) on a finite-measure interval. The inequalities \(\int\ell_n\le\int f\le\int u_n\) squeeze its integral to the Riemann integral. Complex Riemann integrals agree by applying this argument to real and imaginary parts.
This proves agreement for every bounded Riemann-integrable function, without assuming a separate characterization of its discontinuities. It does not say that every Lebesgue-integrable function is Riemann integrable; the rational indicator above disproves that.
4.6. Functions as vectors: L¹ and L²
The integral cannot detect a change on a null set. Therefore an integral norm cannot distinguish every pointwise function: \(\mathbf1_{\{0\}}\) is a nonzero function with zero integral norm under length. We must identify such functions before calling the expression a norm.
Definition 4.16. L¹ and L² equivalence classes
Two measurable complex functions are equivalent if they agree a.e.; write \([f]\) for the equivalence class of \(f\). Define
\[L^1(X,\mu)=\{[f]:\int|f|\,d\mu<\infty\},\qquad L^2(X,\mu)=\{[f]:\int|f|^2\,d\mu<\infty\}.\] \[\|[f]\|_1=\int|f|\,d\mu,\qquad\|[f]\|_2=\left(\int|f|^2\,d\mu\right)^{1/2}.\]Add classes and multiply by complex scalars using representatives: \([f]+[g]=[f+g]\), \(a[f]=[af]\). We usually suppress the brackets. Whenever a pointwise expression matters, choose a measurable representative explicitly.
Worked example for Definition 4.16
Counterexample: Norm convergence need not be pointwise convergence
Agreement a.e. is an equivalence relation: reflexivity and symmetry are immediate, and transitivity follows because a union of two null exceptional sets is null. Operations are independent of representatives by the same argument. Integral invariance under null changes proves that the displayed quantities are well defined. The zero-integral criterion proves \(\|[f]\|_p=0\) precisely for the zero class.
Examples and a warning. On \(\mathbb R\), zero and \(\mathbf1_{\{0\}}\) define the same \(L^2\) vector. Evaluating an \(L^2\) vector at zero is consequently undefined without extra structure. On \((0,1)\), \(x^{-1/2}\) belongs to \(L^1\) but not to \(L^2\). On \((1,\infty)\), \(x^{-1}\) belongs to \(L^2\) but not to \(L^1\). The verification below uses only interval lengths. On a finite-measure space every \(L^2\) function is in \(L^1\), as the next theorem shows with \(g=1\).
Verification of the L¹ and L² examples
Partition \((0,1)\) into \(I_k=(2^{-(k+1)},2^{-k}]\), \(k\ge0\), ignoring the endpoint one. On \(I_k\), \(x^{-1/2}\le2^{(k+1)/2}\), so its integral is at most \(\sum_{k\ge0}2^{-(k+1)/2}<\infty\). Its square satisfies \(x^{-1}\ge2^k\) there, giving integral at least \(\sum_{k\ge0}2^k2^{-(k+1)}=\sum_{k\ge0}1/2=\infty\). On \((1,\infty)\), use \(J_k=[2^k,2^{k+1})\), \(k\ge0\). The square of \(x^{-1}\) is at most \(2^{-2k}\) on \(J_k\), giving integral at most \(\sum_{k\ge0}2^{-k}<\infty\). The function itself is at least \(2^{-(k+1)}\), giving integral at least \(\sum_{k\ge0}1/2=\infty\). Countable additivity of the nonnegative integral justifies these sums. The geometric sums are finite because \(\sum_{k=0}^N r^k=(1-r^{N+1})/(1-r)\) for \(0<r<1\).
Theorem 4.17. Integral Cauchy–Schwarz and norm inequalities
For \(f,g\in L^2\), the product \(\overline f g\) is integrable and
\[\int|fg|\,d\mu\le\|f\|_2\|g\|_2,\qquad\left|\int\overline f g\,d\mu\right|\le\|f\|_2\|g\|_2.\]The \(L^1\) and \(L^2\) norms satisfy \(\|f+g\|_p\le\|f\|_p+\|g\|_p\), for \(p=1,2\), and \(\|af\|_p=|a|\|f\|_p\). In particular both are vector spaces with norms. The expression \(\langle f,g\rangle=\int\overline f g\,d\mu\) is an inner product, conjugate-linear in the first argument and linear in the second.
Proof.
The scalar inequality \(2|fg|\le t|f|^2+t^{-1}|g|^2\), for \(t>0\), follows by expanding \((\sqrt t|f|-|g|/\sqrt t)^2\ge0\). Its right side is integrable, so the product is integrable. If both norms are positive choose \(t=\|g\|_2/\|f\|_2\); integration gives the first inequality. If either norm is zero, that function is zero a.e. and the product integral is zero. The integral triangle inequality gives the second inequality. Pointwise, \(|f+g|^2\le2|f|^2+2|g|^2\), proving closure in \(L^2\). Expand its integral and use Cauchy–Schwarz:
\[\|f+g\|_2^2=\|f\|_2^2+2\operatorname{Re}\langle f,g\rangle+\|g\|_2^2\le(\|f\|_2+\|g\|_2)^2.\]For \(L^1\), integrate the pointwise triangle inequality. Homogeneity follows from \(|af|=|a||f|\). Integral linearity and conjugation prove the inner-product algebra rules; \(\langle f,f\rangle=\|f\|_2^2\) is positive and zero only for the zero class. All formulas respect a.e. equality.
□Our convention makes the second inner-product argument linear. Alphabeta Math’s library uses the first argument linear; translate by \((f,g)_{\mathrm{AB}}=\langle g,f\rangle\). For example, \(\langle f,ig\rangle=i\langle f,g\rangle\), whereas \((if,g)_{\mathrm{AB}}=i(f,g)_{\mathrm{AB}}\). Conjugate symmetry makes the two conventions equivalent, but factors of \(i\) must be placed consistently.
Examples. On \([0,1]\), \(f=1\) and \(g=x\) give \(\int fg=1/2\le1/\sqrt3\). A multiple \(g=cf\) gives equality in the absolute-value Cauchy–Schwarz bound. Indicators of disjoint finite-measure sets are orthogonal, because their pointwise product vanishes.
4.7. Why limits stay in L²
A norm defines distance by \(d(f,g)=\|f-g\|\). A sequence is Cauchy if for each \(\varepsilon>0\), all sufficiently late pairs satisfy \(\|f_n-f_m\|<\varepsilon\). Completeness means that every Cauchy sequence has a limit in the same space. Completeness of the scalar numbers does not itself prove completeness of a function space; we need the following construction.
Theorem 4.18. L² is complete
Every Cauchy sequence in \(L^2(X,\mu)\) converges in \(L^2\). Together with the preceding inner product this makes \(L^2\) a Hilbert space, meaning a complete inner-product space.
Alphabeta Math: Riesz–Fischer completeness
Worked example for Theorem 4.18
Counterexample: Norm convergence need not be pointwise convergence
Proof: construct an absolutely convergent subsequence
Proof.
From a Cauchy sequence choose a subsequence \(f_{n_k}\) with \(\|f_{n_{k+1}}-f_{n_k}\|_2\le2^{-k}\). Choose measurable representatives and put \(h_k=f_{n_{k+1}}-f_{n_k}\). Finite triangle inequalities give \(\|\sum_{k=1}^m|h_k|\|_2\le\sum_{k=1}^m2^{-k}\le1\). Monotone convergence applied to the squares shows that \(S=\sum_{k\ge1}|h_k|\) has \(\int S^2\le1\). Consequently \(S<\infty\) a.e.: the bound \(\int S^2\ge M^2\mu(\{S=\infty\})\) for every \(M>0\) forces that set to be null.
On its complement the complex series \(\sum_k h_k\) converges absolutely, hence converges by completeness of \(\mathbb C\). Define \(f=f_{n_1}+\sum_k h_k\) there and zero on the exceptional set. It is measurable as a limit with a measurable exceptional set. The tail \(T_K=\sum_{k\ge K}|h_k|\) obeys \(\|T_K\|_2\le\sum_{k\ge K}2^{-k}\), by the same finite triangle inequalities and monotone convergence. Since \(|f-f_{n_K}|\le T_K\) a.e., the difference is square-integrable and
\[\|f-f_{n_K}\|_2\le\sum_{k\ge K}2^{-k}\longrightarrow0.\]The bound \(|f|\le|f_{n_1}|+S\) a.e. also proves \(f\in L^2\). Finally, for any sufficiently late \(n\), choose \(n_K\) late enough that both \(\|f_n-f_{n_K}\|_2\) and \(\|f_{n_K}-f\|_2\) are small. The triangle inequality proves \(f_n\to f\) for the whole original sequence.
□Examples of different limits. On \([0,1]\), \(\mathbf1_{[0,1/n]}\to0\) in \(L^2\), because its squared norm is \(1/n\), but its pointwise value at zero remains one. Conversely, the shrinking spikes \(n\mathbf1_{(0,1/n)}\) converge pointwise to zero but their squared norms are \(n\), so they do not converge to zero in \(L^2\).
L² convergence need not give pointwise convergence of the whole sequence
List in order the functions \(\mathbf1_{[j2^{-k},(j+1)2^{-k})}\), first for \(k=1\), \(j=0,1\), then for \(k=2\), \(j=0,1,2,3\), and so on. Work on \([0,1)\). Each squared norm is \(2^{-k}\), so the list converges to zero in \(L^2\). At each point, exactly one function in every block equals one and the other functions in that block equal zero. Thus the numerical values fail to converge at every point. The completeness proof constructs an a.e. convergent subsequence; it does not assert pointwise convergence of the entire sequence.
For one spatial coordinate, a normalized wavefunction is a vector \(\psi\in L^2(\mathbb R,\lambda_1)\) with \(\int|\psi|^2\,dx=1\). For three coordinates replace the line by \(\mathbb R^3\) and the length measure by \(\lambda_3\). The measure of a region weighted by \(|\psi|^2\) will become its Born probability in the next chapter. Different representatives of \(\psi\) give the same integrals over every measurable region. Chapter 6 uses the inner product and completeness just proved to explain state geometry and observables.
We have not needed product integration or an interchange of integration order. The box construction supplies Euclidean volume directly. For calculations that do interchange orders, the additional hypotheses and proofs belong to Fubini–Tonelli theory.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. A finite measure space and continuity of measure
Applies to: Definition 4.1, Definition 4.2, Proposition 4.3.
On \(X=\{a,b,c\}\) take all subsets as \(\Sigma\) and weights \(\mu(\{a\})=1,\mu(\{b\})=2,\mu(\{c\})=0\). All complements and countable unions remain subsets, so this is a sigma-algebra. Define \(\mu(A)\) by summing its weights; in a disjoint family each point occurs at most once, proving countable additivity. The null sets are \(\varnothing,\{c\}\); all their subsets are measurable, so the measure is complete, and changing a function at \(c\) changes it only a.e. The increasing sets \(\{a\},\{a,b\},X,X,\ldots\) have measures \(1,3,3,\ldots\) converging to the measure of their union. The decreasing sets \(X,\{b,c\},\{c\},\{c\},\ldots\) have measures \(3,2,0,\ldots\), tending to the measure of their intersection. Monotonicity and subadditivity follow here by retaining or counting each nonnegative point weight, respectively.
Worked example 2. A null set from explicit box covers
Applies to: Definition 4.4, Definition 4.5, Theorem 4.6.
Enumerate \(N=\mathbb Q\cap[0,1]=\{q_1,q_2,\ldots\}\). For any \(\varepsilon>0\), cover \(q_j\) by an interval of length \(\varepsilon2^{-j}\). The total length is \(\varepsilon\), so \(\lambda^*(N)=0\). For every \(A\subset N\) and any test set \(E\), monotonicity gives \(\lambda^*(E\cap A)=0\), while subadditivity and monotonicity give \(\lambda^*(E)\le\lambda^*(E\setminus A)\le\lambda^*(E)\). This proves Carathéodory’s splitting equation for \(A\), even if \(E\) has infinite outer measure. Hence every subset of this null set is measurable and null, illustrating completeness. The interval \([0,1]\) is Borel and has measure 1 by the box-volume assertion; removing \(N\) leaves measure 1. Thus dense countably many points have zero measure while their containing interval does not.
Further proof details
The interval-volume lower bound used here can be checked from the outer-measure definition. Given a countable interval cover of \([0,1]\) with total length \(L<\infty\), enlarge interval \(j\) to an open interval by less than \(\varepsilon2^{-j}\) in total length. Compactness supplies a finite subcover. Ordering its endpoints and counting each covered subinterval once shows that its total length is at least the length 1 of \([0,1]\): at every gap between consecutive endpoints inside \([0,1]\), at least one covering interval contributes that gap length. Thus \(L+\varepsilon\ge1\) for every \(\varepsilon>0\), so \(L\ge1\). The one-interval cover gives the opposite bound, hence outer measure 1. Borel measurability is established by the chapter’s construction; the explicit null-cover argument then proves the stated removal and completion properties.
Worked example 3. Simple approximations with calculated integrals
Applies to: Definition 4.7, Proposition 4.8, Definition 4.9, Theorem 4.10.
On \([0,1)\) with Lebesgue measure, let \(f(x)=x\) and \(s_n(x)=2^{-n}\lfloor2^nx\rfloor\). Threshold sets of \(f\) are intervals and hence measurable. Each \(s_n\) is simple on dyadic intervals; \(s_n\le s_{n+1}\le f\) and \(0\le f-s_n<2^{-n}\), proving pointwise monotone convergence. Its integral is
\[\int s_n=2^{-2n}\sum_{j=0}^{2^n-1}j=\frac{1-2^{-n}}2\longrightarrow\frac12=\int f.\]This evaluates the simple-integral definition and its supremum, as well as monotone convergence. Sums, squares, maxima, and minima of these functions are piecewise polynomial on measurable intervals; their level sets are Borel. Their pointwise supremum is \(f\), illustrating the algebra and limit assertions for measurable functions.
Worked example 4. Signed and complex integrals
Applies to: Proposition 4.11, Definition 4.12, Proposition 4.13.
On \([0,1]\), let \(f=2\mathbf1_{[0,1/2]}-\mathbf1_{(1/2,1]}\). Its positive and negative parts have integrals 1 and \(1/2\), so \(\int f=1/2\) and \(\int|f|=3/2<\infty\). For \(g=f+i\), linearity gives \(\int g=1/2+i\). Its absolute integral is \((\sqrt5+\sqrt2)/2\), which is at least \(|\int g|=\sqrt5/2\). Changing either function at zero changes no integral. For a countable nonnegative sum, the disjoint intervals \(A_j=(2^{-j},2^{1-j}]\) cover \((0,1]\); therefore \(\int\sum_j\mathbf1_{A_j}=1=\sum_j2^{-j}\). A nonnegative function supported just at zero has integral zero while being zero a.e., checking the null-set assertion explicitly.
Worked example 5. Escape of mass and a valid domination
Applies to: Theorem 4.14, Theorem 4.15.
On \(\mathbb R\), \(f_n=\mathbf1_{[n,n+1]}\) are nonnegative measurable, tend pointwise to zero, and have integral 1. Fatou therefore reads \(0\le1\), strictly. There can be no integrable dominating function: domination a.e. for countably many \(n\) would force it to be at least 1 a.e. on \([1,\infty)\). On \([0,1]\), however, \(g_n(x)=x^n\to0\) except at the null singleton 1 and \(|g_n|\le1\), with integrable dominating function 1. Direct integration gives \(\int|g_n|=1/(n+1)\to0\). This verifies the full dominated-convergence conclusion and illustrates exactly the hypothesis absent from the escaping intervals.
Worked example 6. A Cauchy sequence of square-integrable classes
Applies to: Definition 4.16, Theorem 4.17, Theorem 4.18.
On \([0,1]\), the functions 1 and \(1+\mathbf1_{\{0\}}\) represent the same \(L^1\) and \(L^2\) classes, each of norm 1. For \(f=1,g=x\), \(\|g\|_2=1/\sqrt3\), \(\int\bar f g=1/2\le1/\sqrt3\), and \(\|f+g\|_2=\sqrt{7/3}\le1+1/\sqrt3\). This checks the integral inner product, Cauchy–Schwarz, and triangle inequality. Their \(L^1\) norms give \(\|1+x\|_1=3/2=1+1/2\).
For \(h_n(x)=\sum_{j=0}^n(x/2)^j\), the limit \(h(x)=1/(1-x/2)\) is measurable and at most 2, hence in \(L^2\). The geometric tail is at most \(2^{-n}\), so \(\|h-h_n\|_2\le2^{-n}\) and \(\|h_m-h_n\|_2\le2^{-m}+2^{-n}\). Thus these classes are Cauchy and their limit belongs to the same space, an explicit instance of Riesz–Fischer rather than merely pointwise convergence.
Counterexamples and limits of the hypotheses
Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.
Counterexample 1. Continuity from above needs finite initial measure
Related statements: Proposition 4.3.
Claim refuted. For every decreasing sequence of measurable sets, the measure of the intersection equals the limit of their measures.
Proof.
On the real line with Lebesgue measure, let \(A_n=[n,\infty)\). They are Borel and decreasing. Their intersection is empty: a real \(x\) is less than some positive integer \(n\). Yet every \(A_n\) has infinite measure, since it contains \([n,n+L]\) of measure \(L\) for every \(L>0\). Thus \(\lambda(\bigcap_nA_n)=0\), while \(\lim_n\lambda(A_n)=\infty\). The omitted condition \(\mu(A_1)<\infty\) is precisely what excludes this example.
Counterexample 2. A finite measure space does not by itself justify exchanging limit and integral
Related statements: Theorem 4.15.
Claim refuted. Pointwise convergence to zero on a probability space forces integrals to converge to zero.
Proof.
On \((0,1)\) with Lebesgue measure of total mass 1, let \(f_n=n\mathbf1_{(0,1/n)}\). These are measurable, nonnegative, and eventually zero at each fixed \(x>0\), so \(f_n\to0\) pointwise. However \(\int f_n=n(1/n)=1\) for every \(n\). There is no integrable common dominator: if \(g\ge f_n\) a.e. for every \(n\), outside a single null set these inequalities hold simultaneously. For \(0<x<1/4\), choose \(n=\lfloor1/(2x)\rfloor\). Then \(x<1/n\) and \(n\ge1/(4x)\), so \(g(x)\ge1/(4x)\) a.e. near zero. The integral of this lower bound diverges, since every dyadic interval \((2^{-j-1},2^{-j})\) contributes at least \(1/8\). Finite total measure does not replace the dominated-convergence hypothesis.
Counterexample 3. Norm convergence need not be pointwise convergence
Related statements: Definition 4.16, Theorem 4.18.
Claim refuted. A sequence converging in L² must converge pointwise almost everywhere as the whole sequence.
Proof.
For \(k\ge0\) and \(0\le j<2^k\), define \(f_{2^k+j}=\mathbf1_{[j2^{-k},(j+1)2^{-k})}\) on \([0,1)\). Its squared \(L^2\) norm is \(2^{-k}\), tending to zero as its index tends to infinity. Thus \(f_n\to0\) in \(L^2\). But every \(x\in[0,1)\) belongs to exactly one interval in each dyadic block, so \(f_n(x)=1\) infinitely often. From every block with \(k\ge1\) at least one other interval misses \(x\), so \(f_n(x)=0\) infinitely often too. The sequence fails to converge pointwise at every point, not just on a null exceptional set. Completeness is a statement about norm convergence of equivalence classes, not a claim about the pointwise behavior of every sequence of representatives.
Comments