A law of large numbers for bounded processes

Question

Let $(\Omega,\mathcal F,P)$ be a probability space and let $([0,1],\mathcal B[0,1],\lambda$) be the unit interval with Lebesgue measure on the Borel subsets of $[0,1]$. Let $f:[0,1]\times \Omega \to \mathbb R$ be $\mathcal B[0,1]\otimes \mathcal F$ measurable and bounded. I want to show the existence of a subsequence $(n_k)$ such that

$$\frac{1}{n_k}\sum_{i=1}^{n_k} f(t_i,\omega) \overset{L^1(\Omega,\mathcal F,P)}\to \int f(t,\omega)d\lambda \quad \text{as } k\to \infty $$

for $\lambda^\infty$-almost every sequence $(t_i)\subset [0,1]$, where $\lambda^\infty$ denotes the product measure on the $\sigma$-algebra $\mathcal B^\infty:=\bigotimes_{n=1}^\infty \mathcal B[0,1]$.

I tried the following:

Fix $\omega\in\Omega$. For each $j\geq1$, let $((t_i),\omega)\mapsto f(t_j,\omega)$ denote the composition of the maps $((t_i),\omega)\mapsto (t_j,\omega)$ and $(t,\omega) \mapsto f(t,\omega)$. Since the $j$th projection map $(t_i)\mapsto t_j$ on $[0,1]^{\infty}:=\prod_{i=1}^{\infty} [0,1]$ is $\mathcal B^\infty$ measurable, we see that $f(t_j,\omega)$ is $\mathcal B^\infty\otimes \mathcal F$ measurable. Moreover, by construction of the product measure $\lambda^\infty$, the projection maps are i.i.d. random variables on $([0,1]^{\infty},\mathcal B^{\infty},\lambda^{\infty})$, and so by the SLLN we get

$$\frac{1}{n}\sum_{i=1}^n f(t_i,\omega) \overset{\lambda^\infty\text{-a.s}}\to \int f(t,\omega)d\lambda \quad \text{as } n\to \infty $$

for each $\omega\in \Omega$. For each $n\geq 1$, define the map $$((t_i),\omega)\mapsto S^n((t_i),\omega):=\frac{1}{n}\sum_{i=1}^n f(t_i,\omega) -\int f(t,\omega)d\lambda.$$ By Fubini's theorem $\omega\to \int f(t,\omega)d\lambda$ is $\mathcal F$-measurable, and so $S^n((t_i),\omega)$ is seen to be $\mathcal B^\infty\otimes \mathcal F$ measurable. Also $S^n((t_i),\omega)$ is bounded since $f$ is bounded. Therefore the DCT gives

$$\int |S^n((t_i),\omega)| d\lambda^{\infty} \to 0 \quad \text{as } n\to \infty$$

for each $\omega\in \Omega$. By Fubini's theorem, these integrals are $\mathcal F$-measurable functions of $\omega$, and they are bounded, and so another application of the DCT gives

$$\int \int |S^n((t_i),\omega)| d\lambda^{\infty}dP \to 0 \quad \text{as } n\to \infty$$

We can interchange the order of integration by Fubini's theorem to obtain

$$\int \int |S^n((t_i),\omega)| dPd\lambda^{\infty} \to 0 \quad \text{as } n\to \infty$$

This says that $\int |S^n((t_i),\omega)| dP \to 0 $ as $n\to \infty $ w.r.t. the $L^1([0,1]^{\infty} ,\mathcal B^{\infty},\lambda^{\infty})$ norm. We can extract a subsequence $(n_k)$ converging $\lambda^{\infty}$-almost surely, i.e. such that

$$\frac{1}{n_k}\sum_{i=1}^{n_k} f(t_i,\omega) \overset{L^1(\Omega,\mathcal F,P)}\to \int f(t,\omega)d\lambda \quad \text{as } k\to \infty $$ for $\lambda^\infty$-almost every sequence $(t_i)\subset [0,1]$.

Is this reasoning correct? Thanks a lot for your help.

Answer 1

About the almost sure convergence

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First, I restate a lemma you might have seen multiple times.
Lemma (an LLN theorem) Let $X_1,X_2,\dots$ be a sequence of centered random variables such that they are uniformly bounded from below, pairwise uncorrelated and their variances are uniformly bounded, then almost surely $$\lim_{n \rightarrow \infty} \frac{X_1+X_2+\dots+X_n}{n} =0$$

Demonstration: See here $\square$.

Back to your question, we consider the sequence of random variables $(X_n)$ defined by $$X_n(\mathbf{t},\omega)= f(\mathbf{t}_n,\omega)-\int_{[0,1]}f(s,\omega)ds $$ in the probability space $\left( [0,1]^{\mathbb{N}}\times \Omega \space\space,\space\space\mathcal{B}([0,1]^{\mathbb{N}} )\otimes \mathcal{F}\space\space,\space\space\lambda^{\infty}\otimes P\right)$ ( Note that $\mathbf{t} \in [0,1]^{\mathbb{N}}$)
So according to our lemma, there is a set $A \in \mathcal{B}([0,1]^{\mathbb{N}})\otimes \mathcal{F}$ such that $|A|=1$ and for any $(\mathbf{t},\omega) \in A$, we have: $$\lim_{n \rightarrow \infty} \frac{f(\mathbf{t_1},\omega)+\dots+f(\mathbf{t}_n,\omega)}{n}= \int_{[0,1]}f(s,\omega)ds$$

Now, for any $\mathbf{t} \in [0,1]^{\infty}$, let $$B_\mathbf{t}=\left\{ \omega \in \Omega: \frac{f(\mathbf{t_1},\omega)+\dots+f(\mathbf{t}_n,\omega)}{n} \text{ doesn't converge to } \int_{[0,1]}f(s,\omega)ds \right\}$$ So, $$\int_{[0,1]^{\mathbb{N}}} P( B_{\mathbf{t}})d\lambda^{\infty}(\mathbf{t})=\int_{[0,1]^{\mathbb{N}}}\int_{\Omega} \mathbf{1}_{\omega \in B_{\mathbf{t}}}dP(\omega)d\lambda^{\infty}(\mathbf{t}) \stackrel{(*)}{\le} \int_{[0,1]^{\mathbb{N}}}\int_{\Omega} \mathbf{1}_{ (\mathbf{t},\omega) \not \in A}dP(\omega)d\lambda^{\infty}(\mathbf{t})=|A^c|=0$$ where in (*), we use the fact that for any pair $(\mathbf{t},\omega)$ if $\omega \in B_{\mathbf{t}}$ , $(\mathbf{t},\omega)$ can not be included in $A$ (definition of $B$ and property of $A$).

So $P(B_\mathbf{t})=0$ $\lambda^{\infty}$-almost surely.

Hence for $\lambda^{\infty}$-almost surely $\mathbf{t} \in [0,1]^{\mathbb{N}}$, $$\lim_{n \rightarrow \infty} \frac{f(\mathbf{t_1},\omega)+\dots+f(\mathbf{t}_n,\omega)}{n}= \int_{[0,1]}f(s,\omega)ds$$ $P$-almost surely.

Side note If I'm not wrong, all the needed measurability is well handled by the measurability of $f$. Side note 2 By $|A|$ and $|A^c|$ , I mean $\lambda^{\infty}\otimes P( A)$ and $\lambda^{\infty}\otimes P( A^c)$, respetively.