I am trying to fill in the details of Nate Eldredge's answer to "Is this set of random variables a Hilbert space?". In particular, let $(\Omega,\mathcal{F}, \mu)$ be a measure space and $H$ be a separable Hilbert space. Let $L^2(\Omega \rightarrow H) \equiv L^2(\Omega,\mathcal{F},\mu; H)$ be the space of Borel measurable functions $f : \Omega \rightarrow H$ which are square-integrable, that is,
$$
|| f ||_{L^2(\Omega \rightarrow H)} \equiv \int_\Omega | f(\omega) |_H^2 \,d\mu(\omega) < +\infty
$$
I want to prove the following:
Lemma: $L^2(\Omega \rightarrow H)$ is a Hilbert space under the inner product $$\langle f, g, \rangle_{L^2(\Omega \rightarrow H)} \equiv \int_\Omega \langle f(\omega), g(\omega) \rangle_H \,d\mu(\omega)$$
The answer to the linked question gave the following proof sketch:
Proof Sketch. The only hard part is completeness, and the proof is pretty much the same as the proof that ordinary $L^2$ spaces are complete. Take a Cauchy sequence $\{f_n\}$ and pass to a subsequence $\{f_{n_k}\}$so that $\|f_{n_k} – f_{n_{k+1}}\|_{L^2(\Omega,\mu;H)}^2 < 2^{-k}$. Use a Borel-Cantelli argument to show that $\{f_{n_k}(\omega)\}$ is a Cauchy sequence in $H$ for almost every $\omega$. By completeness of $H$, $\{f_{n_k}\}$ converges almost everywhere; call the limit $f$. Then use the triangle inequality to show that in fact $f_n \to f$ in $L^2(\Omega,\mu;H)$-norm.
I am having trouble making the completeness argument rigorous. Here is my attempt:
- Suppose $f_1,f_2,\ldots \in L^2(\Omega \rightarrow H)$ is a Cauchy sequence. Then we can choose a subsequence $(f_{n_k})_{k=1}^\infty$ such that
$$
|| f_{n_a} – f_{n_b} || < \frac{1}{2^a} \qquad \forall\, b \geq a \in \mathbb{N}
$$ - We want to show that the set $\{ \omega \in \Omega \mid (f_{n_k}(\omega))_{k=1}^\infty \text { is not Cauchy} \}$ has measure zero. Another way to write this set is
$$
\begin{align}
&\left\{ \omega \in \Omega \mid \exists\, k \in \mathbb{N}, \forall\, a \in \mathbb{N}, \exists\, b \geq a, |f_{n_a}(\omega) – f_{n_b}(\omega)| > \tfrac{1}{2^k} \right\}\\
&= \bigcup_{k = 1}^\infty \bigcap_{a = 1}^\infty \bigcup_{b \geq a} \left\{ \omega \in \Omega \mid |f_{n_a}(\omega) – f_{n_b}(\omega)| > \tfrac{1}{2^k} \right\}
\end{align}
$$
From here, I can imagine using Markov's inequality on the set $E_{abk} = \left\{ \omega \in \Omega \mid |f_{n_a}(\omega) – f_{n_b}(\omega)| > \tfrac{1}{2^k} \right\}$ to show that
$$
\mu(E_{abk}) \leq 2^k || f_{n_a} – f_{n_b} ||_{L^2(\Omega \rightarrow H)} \leq \frac{2^k}{2^a}
$$
which follows from the defining property of the subsequence. It also looks like a Borel-Cantelli argument should apply to show that the measure of the sets $\bigcap_a \bigcup_{b \geq a} E_{abk}$ is zero, but I'm not sure how to deal with the dependence of $E_{abk}$ on both $a$ and $b$. How might I be able to resovle these issues?
Best Answer
Just take $b=a+1$ in your argument. Then use the following elementary argument: if $|a_{n_{k}} -a_{n_{k+1}}| < \frac 1 {2^{k}}$ for all $k$ then $|a_{n_{k}}-a_{n_{j}}| \leq |a_{n_{k}}-a_{n_{k+1}}|+|a_{n_{k+1}}-a_{n_{k+2}}|+...+|a_{n_{j-1}}-a_{n_{j}}|<\sum_{l=k}^{j-1} \frac 1 {2^{l}} \to 0$ as $j>l \to \infty$. This shows that $\{f_{n_{k}}\}$ is almost surely Cauchy and you define $f$ as its almost sure limit. Now use Fatou's Lemma (in step 1) to to prove convergence in $L^{2}$ norm.