Proof of Completeness of the Lebesgue Space of Square-Integrable Vector-Valued Functions

Write $\lambda=\lambda_{[0,T]}$, $\mathcal{B}=\mathcal{B}_{[0,T]}$ and $P_{T}=T^{-1}\lambda$, so that $([0,T],\mathcal{B},P_{T})$ is a probability space and, by claim 7 of the inner-product lemma, a map $w:[0,T]\to\mathbb{R}^{d}$ lies in $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ exactly when each component $w^{i}$ is a square-integrable random variable on that probability space, in which case

Claim 1. Let $(\xi_{n})_{n\in\mathbb{N}}$ be a Cauchy sequence in $\bigl(L^{2}([0,T];\mathbb{R}^{d}),d_{L^{2}}\bigr)$, and for each $n$ choose a representative $u_{n}\in\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ with $\xi_{n}=[u_{n}]$.

Fix $i\in\{1,\dots,d\}$. Since $(u_{n}-u_{m})^{i}=u_{n}^{i}-u_{m}^{i}$, identity $(\dagger)$ applied to $u_{n}-u_{m}$ gives

all terms of the sum being nonnegative. Given a real $\varepsilon>0$, choosing $N$ with $d_{L^{2}}(\xi_{n},\xi_{m})<\varepsilon\sqrt{T}$ for $n,m\ge N$ therefore yields $\lVert u_{n}^{i}-u_{m}^{i}\rVert_{2}<\varepsilon$ for $n,m\ge N$; that is, $(u_{n}^{i})_{n}$ is Cauchy in mean square. By mean-square completeness, applied on $([0,T],\mathcal{B},P_{T})$ with the sub-$\sigma$-algebra taken to be $\mathcal{B}$ itself, there is a square-integrable random variable $X^{i}$ on that space with $\lVert u_{n}^{i}-X^{i}\rVert_{2}\to0$.

Define $u:[0,T]\to\mathbb{R}^{d}$ by $u(t)=\bigl(X^{1}(t),\dots,X^{d}(t)\bigr)$. Its components are the square-integrable random variables $X^{i}$, so $u\in\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ by $(\dagger)$. Applying $(\dagger)$ to $u_{n}-u$,

a finite sum of $d$ real sequences each with limit $0$, hence with limit $0$. So $\xi_{n}\to[u]$ in $\bigl(L^{2}([0,T];\mathbb{R}^{d}),d_{L^{2}}\bigr)$. Every Cauchy sequence converges, so the space is complete.

Claim 2. Put $w_{n}=u_{n}-u$, which lies in $\mathcal{L}^{2}([0,T];\mathbb{R}^{d})$ by claim 2 of the inner-product lemma, and $\varepsilon_{n}=\lVert[w_{n}]\rVert_{L^{2}}$, so $\varepsilon_{n}\to0$ by hypothesis.

Choice of subsequence. Construct $n_{1}<n_{2}<\dots$ recursively: having chosen $n_{1}<\dots<n_{j-1}$ (with no constraint when $j=1$), use $\varepsilon_{n}\to0$ to pick $n_{j}>n_{j-1}$ with $\varepsilon_{n_{j}}\le4^{-j}$.

A Chebyshev bound. For each $j$ the function $|w_{n_{j}}|^{2}$ is $\mathcal{B}$-measurable by claim 1 of the inner-product lemma, so

belongs to $\mathcal{B}$. Pointwise $4^{-j}\mathbf{1}_{A_{j}}\le|w_{n_{j}}|^{2}$, where $\mathbf{1}_{A_{j}}$ is the indicator of $A_{j}$; so by the integral of a simple function and monotonicity,

whence $\lambda(A_{j})\le4^{-j}$.

Countable subadditivity. If $E_{1},E_{2},\dots\in\mathcal{B}$, put $F_{j}=E_{j}\setminus\bigcup_{i<j}E_{i}\in\mathcal{B}$; these are pairwise disjoint with $\bigcup_{j}F_{j}=\bigcup_{j}E_{j}$ and $F_{j}\subseteq E_{j}$. Since $\lambda$ is a measure, additivity applied to $E_{j}=F_{j}\cup(E_{j}\setminus F_{j})$ gives $\lambda(F_{j})\le\lambda(E_{j})$, and countable additivity gives $\lambda\bigl(\bigcup_{j}E_{j}\bigr)=\sum_{j}\lambda(F_{j})$.

The exceptional set. Let $N=\bigcap_{J\ge1}\bigcup_{j\ge J}A_{j}$, a member of $\mathcal{B}$. Fix $J\ge1$. By the previous paragraph, $\lambda\bigl(\bigcup_{j\ge J}A_{j}\bigr)$ is the sum of the nonnegative series $\sum_{j\ge J}\lambda(F_{j})$ whose partial sums satisfy, for every $K\ge J$,

hence the sum itself is at most $\tfrac{4}{3}4^{-J}$. Since $N\subseteq\bigcup_{j\ge J}A_{j}$, monotonicity of $\lambda$ gives $\lambda(N)\le\tfrac{4}{3}4^{-J}$ for every $J\ge1$. As $4^{-J}$ has limit $0$, $\lambda(N)=0$.

Convergence off $N$. Let $t\in[0,T]\setminus N$. By the definition of $N$ there is $J$ with $t\notin A_{j}$ for every $j\ge J$, that is $|w_{n_{j}}(t)|^{2}\le4^{-j}$, so $|u_{n_{j}}(t)-u(t)|=|w_{n_{j}}(t)|\le2^{-j}$ for all $j\ge J$. Given a real $\eta>0$, choose $j_{0}\ge J$ with $2^{-j_{0}}<\eta$; then $|u_{n_{j}}(t)-u(t)|\le2^{-j}\le2^{-j_{0}}<\eta$ for every $j\ge j_{0}$. Hence $\bigl(u_{n_{j}}(t)\bigr)_{j}$ converges to $u(t)$ in $\mathbb{R}^{d}$.