Proof of The Lebesgue Space of Square-Integrable Vector-Valued Functions is a Real Inner Product Space

Throughout write $\lambda=\lambda_{[0,T]}$ and $\mathcal{B}=\mathcal{B}_{[0,T]}$, and let every integral be over $[0,T]$ with respect to $\lambda$ unless another measure is displayed. All appeals to linearity and monotonicity of the integral are to linearity and monotonicity of the Lebesgue integral. We use throughout that $\mathbb{R}^{d}$ is a real vector space, that the dot product is symmetric and linear in each argument, and that $|x|^{2}=x\cdot x=\sum_{i=1}^{d}(x^{i})^{2}$, by claim 1 of the elementary properties of the Euclidean norm.

Claim 1. Let $u,v\in\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$, so all components $u^{i},v^{i}$ are $\mathcal{B}$-measurable. The maps $x\mapsto|x|^{2}$ from $\mathbb{R}^{d}$ to $\mathbb{R}$ and $(x,y)\mapsto x\cdot y$ from $\mathbb{R}^{2d}$ to $\mathbb{R}$ are polynomial in the coordinates, hence continuous and in particular sequentially continuous, so $t\mapsto|u(t)|^{2}$ and $t\mapsto u(t)\cdot v(t)$ are $\mathcal{B}$-measurable by measurability of sequentially continuous functions of measurable Euclidean maps.

For every $t\in[0,T]$, expanding the nonnegative quantities $|u(t)\pm v(t)|^{2}$ by bilinearity of the dot product gives

and taking whichever sign makes the middle term equal to $-2|u(t)\cdot v(t)|$ yields the pointwise bound $2|u\cdot v|\le|u|^{2}+|v|^{2}$. The right-hand side is measurable with finite integral, so by monotonicity and linearity

so $u\cdot v$ is integrable.

Claim 2. The components of $u+v$ and of $cu$ are $u^{i}+v^{i}$ and $cu^{i}$, which are $\mathcal{B}$-measurable by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable applied to the continuous maps $(s,r)\mapsto s+r$ and $s\mapsto cs$. Pointwise $|u+v|^{2}=|u|^{2}+2\,u\cdot v+|v|^{2}\le2|u|^{2}+2|v|^{2}$ by the bound of claim 1, so $\int|u+v|^{2}\le2\int|u|^{2}+2\int|v|^{2}<\infty$. Also $|cu|^{2}=c^{2}|u|^{2}$ by claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, whose integral is $c^{2}\int|u|^{2}<\infty$. Hence $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ is closed under the pointwise operations. Because those operations are performed in $\mathbb{R}^{d}$ separately at each $t$, every vector-space axiom holds in $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ because it holds in $\mathbb{R}^{d}$ at each $t$; the constant map $0$ is square-integrable and is the zero element.

Claim 3. Finite unions of null sets are null. Let $N_{1},N_{2}\in\mathcal{B}$ with $\lambda(N_{1})=\lambda(N_{2})=0$. Since $\lambda$ is a measure, applying countable additivity to the disjoint decomposition $N_{2}=(N_{2}\setminus N_{1})\cup(N_{2}\cap N_{1})$ gives $\lambda(N_{2}\setminus N_{1})\le\lambda(N_{2})=0$, and applying it to $N_{1}\cup N_{2}=N_{1}\cup(N_{2}\setminus N_{1})$ gives $\lambda(N_{1}\cup N_{2})=\lambda(N_{1})+\lambda(N_{2}\setminus N_{1})=0$. Induction extends this to any finite union.

$\sim$ is an equivalence relation. Reflexivity holds with $N=\emptyset$, symmetry with the same $N$, and transitivity by the preceding paragraph applied to the union of the two exceptional sets.

Vanishing criterion. Let $w=u-v$, which lies in $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ by claim 2, and put $g=|w|^{2}$, a nonnegative real-valued $\mathcal{B}$-measurable function by claim 1. Suppose first that $u\sim v$, so that $g$ vanishes off some $N\in\mathcal{B}$ with $\lambda(N)=0$. For each natural number $n$ the function $g_{n}=\min(g,n)$ is $\mathcal{B}$-measurable and satisfies $g_{n}\le n\mathbf{1}_{N}$ pointwise, since both sides vanish off $N$ while $g_{n}\le n$ everywhere. Now $n\mathbf{1}_{N}$ is a nonnegative simple function of integral $n\lambda(N)=0$, so $\int g_{n}=0$ by monotonicity. The sequence $(g_{n})_{n}$ is nondecreasing with pointwise limit $g$, so $\int g=\lim_{n}\int g_{n}=0$ by monotone convergence. Conversely, suppose $\int g=0$. By claim 5 of the restricted Lebesgue toolkit, the set $N=\{t\in[0,T]:g(t)>0\}$ satisfies $\lambda(N)=0$; it belongs to $\mathcal{B}$ because $g$ is measurable. For $t\notin N$ we have $|w(t)|^{2}=0$, hence $w(t)=0$ by claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, that is $u(t)=v(t)$. So $u\sim v$.

Compatibility and the quotient. If $u=u'$ off a null $N_{1}$ and $v=v'$ off a null $N_{2}$, then $u+v=u'+v'$ off the null set $N_{1}\cup N_{2}$ and $cu=cu'$ off $N_{1}$; hence $[u]+[v]=[u+v]$ and $c[u]=[cu]$ are independent of the representatives chosen. The vector-space axioms transfer from $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ to the quotient by applying the class map to each axiom, and $[0]$ is the zero element.

Claim 4. Let $u\sim u'$ and $v\sim v'$, with exceptional null sets $N_{1}$ and $N_{2}$. The function $h=u\cdot v-u'\cdot v'$ is integrable by claim 1 and linearity, and vanishes off the null set $N_{1}\cup N_{2}$. The first half of the vanishing criterion of claim 3 used only that the function in question is nonnegative, measurable, and vanishes off a null set; the function $|h|$ has all three properties, being nonnegative, $\mathcal{B}$-measurable by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable applied to the continuous map $s\mapsto|s|$ on $\mathbb{R}$ and to the measurable function $h$, and zero off $N_{1}\cup N_{2}$; applying that argument to $|h|$ gives $\int|h|=0$. Since $h\le|h|$ and $-h\le|h|$ pointwise, monotonicity and linearity give $\bigl|\int h\bigr|\le\int|h|=0$, so $\int u\cdot v=\int u'\cdot v'$. Thus $\langle\cdot,\cdot\rangle_{L^{2}}$ is well defined on $L^{2}([0,T];\mathbb{R}^{d})$, and taking $v=u$, $v'=u'$ shows the same for $\lVert\cdot\rVert_{L^{2}}$.

Symmetry of the pairing is pointwise symmetry of the dot product. For linearity in the first argument, let $u,u''\in\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ and $c$ real; then $(cu+u'')\cdot v=c(u\cdot v)+u''\cdot v$ pointwise, all three products are integrable by claim 1, and linearity of the integral gives $\langle c[u]+[u''],[v]\rangle_{L^{2}}=c\langle[u],[v]\rangle_{L^{2}}+\langle[u''],[v]\rangle_{L^{2}}$. Linearity in the second argument follows by symmetry. Finally $\langle[u],[u]\rangle_{L^{2}}=\int|u|^{2}=\lVert[u]\rVert_{L^{2}}^{2}\ge0$ because the integrand is nonnegative, and by the vanishing criterion of claim 3 applied with $v=0$, $\lVert[u]\rVert_{L^{2}}=0$ holds exactly when $\int|u|^{2}=0$, exactly when $u\sim0$, exactly when $[u]=[0]$.

Claim 5. Absolute homogeneity: $\lVert c[u]\rVert_{L^{2}}^{2}=\int|cu|^{2}=c^{2}\int|u|^{2}$, so $\lVert c[u]\rVert_{L^{2}}=|c|\,\lVert[u]\rVert_{L^{2}}$ by uniqueness of the nonnegative square root.

Write $A=\lVert[u]\rVert_{L^{2}}^{2}$, $B=\langle[u],[v]\rangle_{L^{2}}$ and $C=\lVert[v]\rVert_{L^{2}}^{2}$. For every real $s$ the element $u+sv$ lies in $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ by claim 2, and expanding by claim 4,

If $C=0$ then $[v]=[0]$ by claim 4, whence $B=0$ and $B^{2}\le AC$ holds trivially. If $C>0$, the choice $s=-B/C$ gives $0\le A-B^{2}/C$, that is $B^{2}\le AC$. Taking nonnegative square roots yields $|B|\le\lVert[u]\rVert_{L^{2}}\lVert[v]\rVert_{L^{2}}$, the Cauchy-Schwarz inequality. Consequently

and taking nonnegative square roots gives the triangle inequality.

Claim 6. The quantity $d_{L^{2}}([u],[v])=\lVert[u]-[v]\rVert_{L^{2}}$ is a nonnegative real number, and it vanishes exactly when $[u]-[v]=[0]$, that is when $[u]=[v]$, by claim 4. Symmetry follows from absolute homogeneity with $c=-1$, and the triangle inequality for $d_{L^{2}}$ from that of claim 5 applied to $[u]-[w]=([u]-[v])+([v]-[w])$. These are precisely the axioms of a metric space.

Claim 7. Scaling of the integral. Let $f:[0,T]\to[0,\infty]$ be $\mathcal{B}$-measurable. If $s=\sum_{r=1}^{p}c_{r}\mathbf{1}_{E_{r}}$ is a nonnegative simple function with $E_{r}\in\mathcal{B}$, then by the integral of a simple function,

Whether a nonnegative simple function $s$ satisfies $s\le f$ pointwise does not depend on the measure, so by the definition of the integral of a nonnegative measurable function the set of numbers $\{\int s\,dP_{T}\}$, taken over nonnegative simple $s\le f$, is exactly the set $\{T^{-1}\int s\,d\lambda\}$ taken over the same family. Since $T^{-1}>0$, multiplication by $T^{-1}$ carries upper bounds to upper bounds in both directions, so the least upper bound of the first set is $T^{-1}$ times that of the second, with the convention that both are $\infty$ together. Hence $\int f\,dP_{T}=T^{-1}\int f\,d\lambda$.

Characterisation of $\mathcal{L}^{2}$. By claim 1 of the restricted Lebesgue toolkit, $([0,T],\mathcal{B},P_{T})$ is a probability space, and a random variable on it is exactly a $\mathcal{B}$-measurable real-valued function, so the measurability requirement is the same in both descriptions. Each $(u^{i})^{2}$ is nonnegative and $\mathcal{B}$-measurable, and $|u|^{2}=\sum_{i=1}^{d}(u^{i})^{2}$ pointwise, so linearity gives $\int|u|^{2}\,d\lambda=\sum_{i=1}^{d}\int(u^{i})^{2}\,d\lambda$. By the scaling identity, the expectation on $([0,T],\mathcal{B},P_{T})$ satisfies $\mathbb{E}[(u^{i})^{2}]=T^{-1}\int(u^{i})^{2}\,d\lambda$. As all terms are nonnegative, $\int|u|^{2}\,d\lambda$ is finite if and only if $\mathbb{E}[(u^{i})^{2}]$ is finite for every $i$, that is if and only if each $u^{i}$ is a square-integrable random variable. In that case

Almost-sure equality. If $u\sim v$ with exceptional set $N$, then $P_{T}(N)=T^{-1}\lambda(N)=0$ and $u^{i}=v^{i}$ off $N$ for each $i$, so each $u^{i}$ equals $v^{i}$ almost surely. Conversely, suppose for each $i$ that $u^{i}=v^{i}$ almost surely. The set $N_{i}=\{t\in[0,T]:u^{i}(t)\ne v^{i}(t)\}$ lies in $\mathcal{B}$, being the set where the measurable function $u^{i}-v^{i}$ is nonzero, and $P_{T}(N_{i})=0$, hence $\lambda(N_{i})=T\,P_{T}(N_{i})=0$. By the first paragraph of claim 3 the finite union $N=\bigcup_{i=1}^{d}N_{i}$ is null, and $u=v$ off $N$, so $u\sim v$.