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This problem comes from here. I am not looking for help on solving the problem, actually to understand something said in the setup:

Let $F$ be a distribution with $F(0-) =0 $ and$F(1)=1$. Let $\mu$ be the associated law. Define $$\Omega = [0,1] \times [0,1]^{\mathbb{N}},~~~ \mathcal{F} = \mathcal{B} \times \mathcal{B}^{\mathbb{N}},~~~ \mathbb{P} = \mu \times \mathrm{Leb}^{\mathbb{N}}$$ $$\Theta(\omega) = \omega_0, H_k(\omega) = I_{[0,\omega_0]}(\omega_k).$$ This models the situation in which $\Theta$ is chosen with law $\mu$, a coin with probability $\Theta$ of heads is then minted, and tossed at times $1,2,\ldots$ . The RV $H_k$ is $1$ if the $k$th toss produces heads, $0$ otherwise. Define $$ S_n := H_1 + H_2 + \cdots + H_n. $$ By the Strong Law and Fubini's Theorem, $$ S_n/n \to \Theta,~~ \mathrm{a.s.} $$ ... [Statement of the problem.]

It is this last comment that confuses me. Of course, it is intuitively obvious that $S_n/n \to \Theta$ by the Strong Law, but I don't see how to actually prove it. I assume the a.s. means $\mu \times \mathrm{Leb}^{\mathbb{N}}$ a.s. (the Leb stands for Lebesgue measure). I think I can say $E[H_i] = E[E[H_i|\Theta]] = E[\Theta]$ but this doesn't help much.

Or I could say $$ \int_\Omega H_i d\mathbb{P} = \int_{[0,1]^\mathbb{N}}\int_0^1 H_id\mu ~d\mathrm{Leb}^\mathbb{N} $$ but again I don't see how this helps.

To use the Strong Law I need IID RVs (Okay since they are uniformly bounded by $1$ independent is sufficient), but they are NOT independent. They are somehow independent once $\Theta$ is known. I don't know how to formulate this rigorously in order to obtain the result though. The author makes it sounds like it should be obvious.

How does one arrive at the conclusion $S_n/n \to \Theta$?

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Consider $\Omega'=[0,1]^{\mathbb N}$, $\mathcal F' = \mathcal B^{\mathbb N}$, $\mathbb P' = \mathrm{Leb}^{\mathbb N}$, and, for each $\theta$ in $[0,1]$, the random variables $h_n^\theta$ and $s_n^\theta$ defined on $(\Omega',\mathcal F')$ by $h_n^\theta(\omega')=H_n(\theta,\omega')$ and $s_n^\theta=h_1^\theta+\cdots+h_n^\theta$. For each fixed $\theta$, the usual strong law of large numbers shows that $\mathbb P'(A_\theta)=1$ where $A_\theta=[s_n^\theta/n\to\theta]$. Consider $A=[S_n/n\to\Theta]$, then $$ \mathbb P(A\mid\Theta=\theta)=\mathbb P'(A_\theta), $$ hence Fubini theorem applied to $\mathbb P=\mu\otimes\mathbb P'$ yields $$ \mathbb P(A)=E(G(T)),\qquad G(\theta)=\mathbb P'(A_\theta), $$ where $T$ can be any random variable with distribution $\mu$, that is, $$ \mathbb P(A)=\int\mathbb P'(A_\theta)\mathrm d\mu(\theta), $$ from which $\mathbb P(A)=1$ follows.

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