Proof of Measurable Maps into a Hilbert Space with an Orthonormal Basis: Norms, Inner Products and Linear Combinations, Synthesis from Coordinates, Square-Integrability, and Almost-Everywhere Equality

Each result cited is universally quantified over the data in its own statement.

Preliminaries. Clause 2 of that setting fixes an arbitrary real Hilbert space $X$ with an arbitrary orthonormal basis $(e_{k})_{k\in\mathbb{N}}$, its norm distance $d$, and clause 4 of the same setting takes $\mathcal{B}(X)$ to be the Borel $\sigma$-algebra of $(X,d)$. Hence Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs may be applied with $E$ in place of $X$ and $(f_{k})_{k\in\mathbb{N}}$ in place of $(e_{k})_{k\in\mathbb{N}}$, the $\sigma$-algebra $\mathcal{B}(X)$ there being our $\mathcal{B}(E)$. Applying Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §measurable so instantiated, with $(\Omega,\mathcal{F})=(S,\mathcal{S})$, we obtain the coordinate test: a map $u:S\to E$ is measurable if and only if $u_{k}$ is measurable for every $k\in\mathbb{N}$. A measurable function $S\to\mathbb{R}$ with nonnegative values is also a measurable map into $[0,\infty]$, by the agreement of the two readings recorded in Measure Spaces and the Lebesgue Integral: Standing Notation §measurable; if it is moreover integrable, then by Integrable Function and the Lebesgue Integral its integral equals its integral as a nonnegative function, because its positive part is itself and its negative part is the zero function, whose integral is $0$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-integral (with the null set $\varnothing$, which is null because $\varnothing$ is a measurable set by Sigma-Algebra and Measurable Space and has measure $0$ by Measure, Measure Space, and Probability Measure).

Claim 1. Let $s\in S$ and $k\in\mathbb{N}$. By conditions (b) and (c) of the definition of an inner product, $(v+w)_{k}(s)=\langle v(s)+w(s),f_{k}\rangle_{E}=v_{k}(s)+w_{k}(s)$ and $(tv)_{k}(s)=\langle tv(s),f_{k}\rangle_{E}=t\,v_{k}(s)$. By the coordinate test the functions $v_{k}$ and $w_{k}$ are measurable for every $k$; hence $v_{k}+w_{k}$ and $tv_{k}$ are measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $v+w$ and $tv$ are measurable by the coordinate test. Now fix $s\in S$ and apply Orthonormal Expansions in a Real Hilbert Space §parseval with $H=E$, the orthonormal basis $(f_{k})_{k\in\mathbb{N}}$, $x=v(s)$ and $y=w(s)$: the series $\sum_{k=1}^{\infty}v_{k}(s)^{2}$ converges with sum $|v(s)|_{E}^{2}$, which is the last assertion of the claim, and the series $\sum_{k=1}^{\infty}v_{k}(s)w_{k}(s)$ converges with sum $\langle v(s),w(s)\rangle_{E}$. Thus the functions $\sigma_{n}=\sum_{k=1}^{n}v_{k}w_{k}$, $n\in\mathbb{N}$, which are measurable by claims 2 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, converge at every point of $S$ to $\langle v,w\rangle_{E}$, so $\langle v,w\rangle_{E}$ is measurable by claim 5 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. Taking $w=v$ and using $\langle v(s),v(s)\rangle_{E}=|v(s)|_{E}^{2}$ from Real Inner Product Space §norm, the function $|v|_{E}^{2}$ is measurable.

Claim 2. Fix $s\in S$. Since $\sum_{k=1}^{\infty}g_{k}(s)^{2}$ converges, Orthonormal Expansions in a Real Hilbert Space §riesz-fischer, applied with $H=E$, the orthonormal sequence $(f_{k})_{k\in\mathbb{N}}$ and $c_{k}=g_{k}(s)$, shows that the series $\sum_{k=1}^{\infty}g_{k}(s)f_{k}$ converges in $E$ in the sense of Series in a Real Inner Product Space §convergent, which is the sense used in that theorem, and that its sum $v(s)$ satisfies $\langle v(s),f_{j}\rangle_{E}=g_{j}(s)$ for every $j\in\mathbb{N}$. Hence $v_{j}=g_{j}$ for every $j\in\mathbb{N}$; these functions are measurable by hypothesis, so $v$ is measurable by the coordinate test.

Claim 3. Put $a=|v|_{E}^{2}$ and $b=|w|_{E}^{2}$, nonnegative measurable functions by claim 1 with $\int_{S}a\,d\mu<\infty$ and $\int_{S}b\,d\mu<\infty$. Fix $s\in S$ and write $\alpha=|v(s)|_{E}$, $\beta=|w(s)|_{E}$. By The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle, $|v(s)+w(s)|_{E}\le\alpha+\beta$; both sides being nonnegative, $|v(s)+w(s)|_{E}^{2}\le(\alpha+\beta)^{2}=2\alpha^{2}+2\beta^{2}-(\alpha-\beta)^{2}\le2a(s)+2b(s)$. By Elementary Identities in a Real Inner Product Space §homogeneity, $|tv(s)|_{E}^{2}=t^{2}a(s)$. Since $|f_{k}|_{E}=1$ by Orthogonality, Orthogonal Complement and Orthonormal Families in a Real Inner Product Space §orthonormal, The Cauchy-Schwarz Inequality in a Real Inner Product Space gives $|v_{k}(s)|\le\alpha$, hence $v_{k}(s)^{2}\le a(s)$, and it also gives $|\langle v(s),w(s)\rangle_{E}|\le\alpha\beta\le\frac{1}{2}(\alpha^{2}+\beta^{2})=\frac{1}{2}(a(s)+b(s))$, using $(\alpha-\beta)^{2}\ge0$. The maps $v+w$ and $tv$ are measurable by claim 1, so $|v+w|_{E}^{2}$ and $|tv|_{E}^{2}$ are nonnegative measurable functions by claim 1 again, and by the monotonicity, additivity and homogeneity in Linearity and Monotonicity of the Lebesgue Integral §nonnegative,

$$\int_{S}|v+w|_{E}^{2}\,d\mu\le2\int_{S}a\,d\mu+2\int_{S}b\,d\mu<\infty,\qquad\int_{S}|tv|_{E}^{2}\,d\mu=t^{2}\int_{S}a\,d\mu<\infty .$$

The function $v_{k}$ is measurable by claim 1, and the map $|v_{k}|^{2}$ of Power-Integrable Functions and the p-Seminorm §measurable-power sends $s$ to $(|v_{k}(s)|)^{2}$, which equals the natural square $|v_{k}(s)|^{2}=v_{k}(s)^{2}$ by Properties of Real Powers of Nonnegative Real Numbers §agreement; hence $\int_{S}|v_{k}|^{2}\,d\mu\le\int_{S}a\,d\mu<\infty$ by Linearity and Monotonicity of the Lebesgue Integral §nonnegative, and $v_{k}$ is $2$-integrable in the sense of Power-Integrable Functions and the p-Seminorm §space. The function $\langle v,w\rangle_{E}$ is measurable by claim 1 and $|\langle v,w\rangle_{E}|$ is measurable by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, with $\int_{S}|\langle v,w\rangle_{E}|\,d\mu\le\frac{1}{2}\bigl(\int_{S}a\,d\mu+\int_{S}b\,d\mu\bigr)<\infty$ by Linearity and Monotonicity of the Lebesgue Integral §nonnegative; so $\langle v,w\rangle_{E}$ is integrable by the criterion recorded in Integrable Function and the Lebesgue Integral. For the inequality, let $\varphi,\psi:S\to\mathbb{R}$ be $\varphi(s)=|v(s)|_{E}$ and $\psi(s)=|w(s)|_{E}$. The norm map $E\to\mathbb{R}$ is Lipschitz from $(E,d_{E})$ to $(\mathbb{R},d_{\mathbb{R}})$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §lipschitz, hence continuous by A Lipschitz Map is Uniformly Continuous, hence measurable with respect to $\mathcal{B}(E)$ and the Borel $\sigma$-algebra of $(\mathbb{R},d_{\mathbb{R}})$, which is that of the real line, by claims 3 and 2 of Borel Measurability and Bounded Integration on a Metric Space; so $\varphi$ and $\psi$, its compositions with $v$ and $w$, are measurable by claim 4 of Borel Measurability and Bounded Integration on a Metric Space. As above, $(|\varphi(s)|)^{2}=\varphi(s)^{2}=a(s)$ and $(|\psi(s)|)^{2}=b(s)$, so $\varphi,\psi$ are $2$-integrable with $\lVert\varphi\rVert_{2}=(\int_{S}a\,d\mu)^{1/2}$ and $\lVert\psi\rVert_{2}=(\int_{S}b\,d\mu)^{1/2}$ by Power-Integrable Functions and the p-Seminorm §seminorm. The exponent $2$ is conjugate to itself by Conjugate Exponents and Young's Inequality §conjugate, since $\frac{1}{2}+\frac{1}{2}=1$, so Hoelder's Inequality, for Two and for Finitely Many Factors §holder gives $\int_{S}|\varphi\psi|\,d\mu\le\lVert\varphi\rVert_{2}\lVert\psi\rVert_{2}$. Since $|\langle v(s),w(s)\rangle_{E}|\le\varphi(s)\psi(s)=|\varphi(s)\psi(s)|$ for every $s$, the bound $|\int f\,d\mu|\le\int|f|\,d\mu$ of Linearity and Monotonicity of the Lebesgue Integral §integrable for $f=\langle v,w\rangle_{E}$ (there $\int_{S}|f|\,d\mu$ is the integral of $|f|$ as an integrable function, $|f|$ being integrable by the finiteness shown above and the criterion of Integrable Function and the Lebesgue Integral; by the preliminaries it equals the integral of $|f|$ as a nonnegative function, which is the reading used below) and the monotonicity in Linearity and Monotonicity of the Lebesgue Integral §nonnegative give

$$\Bigl|\int_{S}\langle v,w\rangle_{E}\,d\mu\Bigr|\le\int_{S}|\langle v,w\rangle_{E}|\,d\mu\le\int_{S}|\varphi\psi|\,d\mu\le\Bigl(\int_{S}|v|_{E}^{2}\,d\mu\Bigr)^{1/2}\Bigl(\int_{S}|w|_{E}^{2}\,d\mu\Bigr)^{1/2}.$$

Claim 4. For measurable $u,u':S\to E$ put $D(u,u')=\{s\in S:u(s)\ne u'(s)\}$. For $s\in S$ and $k\in\mathbb{N}$, Elementary Identities in a Real Inner Product Space §bilinear gives $\langle u(s)-u'(s),f_{k}\rangle_{E}=u_{k}(s)-u'_{k}(s)$. If $u(s)\ne u'(s)$ then $u(s)-u'(s)\ne0_{E}$ (indeed, if $x-y=0_{E}$ for $x,y\in E$, then by condition 2 of Vector Space over a Field both $w=x$ and $w=y$ satisfy $(-y)+w=0_{E}$, so $x=y$ by the uniqueness in claim 2 of Elementary Identities in a Vector Space), so by Orthonormal Basis of a Real Hilbert Space §basis there is $k$ with $u_{k}(s)\ne u'_{k}(s)$; if $u(s)=u'(s)$ then $u_{k}(s)=u'_{k}(s)$ for every $k$. Hence $D(u,u')=\bigcup_{k\in\mathbb{N}}\{s\in S:0<|u_{k}(s)-u'_{k}(s)|\}$. Each function $|u_{k}-u'_{k}|$ is measurable by claim 1 and claims 2 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, so each set in the union belongs to $\mathcal{S}$ by the criterion of Measure Spaces and the Lebesgue Integral: Standing Notation §measurable with $c=0$, and so does the countable union, $\mathcal{S}$ being a $\sigma$-algebra. We use the following fact (U): if $A,B,C\in\mathcal{S}$, $C\subseteq A\cup B$ and $\mu(A)=\mu(B)=0$, then $\mu(C)=0$; indeed, by Basic Properties of a Measure §monotone and Basic Properties of a Measure §subadditivity applied to the sequence $A,B,\varnothing,\varnothing,\dots$, and $\mu(\varnothing)=0$ from Measure, Measure Space, and Probability Measure, $\mu(C)\le\mu(A\cup B)\le0$. Now $D(v,v)=\varnothing$ has measure $0$, $D(v,v')=D(v',v)$, and $D(v,v'')\subseteq D(v,v')\cup D(v',v'')$ for measurable $v,v',v''$, so $\sim_{\mu}$ is reflexive, symmetric and, by (U), transitive. If $v\sim_{\mu}v'$ and $w\sim_{\mu}w'$, then $v+w,v'+w',tv,tv'$ are measurable by claim 1, $D(v+w,v'+w')\subseteq D(v,v')\cup D(w,w')$ and $D(tv,tv')\subseteq D(v,v')$, so $v+w\sim_{\mu}v'+w'$ by (U) and $tv\sim_{\mu}tv'$ by Basic Properties of a Measure §monotone. Assume now in addition $\int_{S}|v|_{E}^{2}\,d\mu<\infty$ and $\int_{S}|w|_{E}^{2}\,d\mu<\infty$, and let $N=D(v,v')\cup D(w,w')$; then $N\in\mathcal{S}$ and $\mu(N)=0$ by (U), so every subset of $N$ is null. For $s\in S\setminus N$ we have $v'(s)=v(s)$ and $w'(s)=w(s)$, hence $|v'(s)|_{E}^{2}=|v(s)|_{E}^{2}$, $|w'(s)|_{E}^{2}=|w(s)|_{E}^{2}$, $\langle v'(s),w'(s)\rangle_{E}=\langle v(s),w(s)\rangle_{E}$ and $v'_{k}(s)=v_{k}(s)$ for every $k$; each of these equalities therefore holds almost everywhere. The functions involved are measurable by claim 1, so the nonnegative case of The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison gives $\int_{S}|v'|_{E}^{2}\,d\mu=\int_{S}|v|_{E}^{2}\,d\mu<\infty$ and $\int_{S}|w'|_{E}^{2}\,d\mu=\int_{S}|w|_{E}^{2}\,d\mu<\infty$; and since $\langle v,w\rangle_{E}$ is integrable by claim 3, the real case of The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison shows that $\langle v',w'\rangle_{E}$ is integrable with $\int_{S}\langle v',w'\rangle_{E}\,d\mu=\int_{S}\langle v,w\rangle_{E}\,d\mu$. Finally, by claim 3 applied to $v$ and to $v'$, both $v_{k}$ and $v'_{k}$ are $2$-integrable; they agree almost everywhere, so they are related by the equivalence relation of The Lebesgue Space of Power-Integrable Functions §equivalence, and their equivalence classes $[v_{k}]$ and $[v'_{k}]$ coincide.