If $\theta$ ~ $U(0,1)$ (Uniform random variable taking values between $0$ and $1$), and conditioned on $\theta$, $X_1,X_2,..$ are identically distributed independent random variables each having distribution Ber($\theta$). Then I need to show
$\lim_{n \to \infty} \mathbf{E}(\theta|X_1,X_2,...X_n) = \theta$ almost surely
I tried using levy's upward convergence theorem, which would result in $\lim_{n \to \infty} \mathbf{E}(\theta|X_1,X_2,...X_n) = \mathbf{E}(\theta|\mathcal{F})$, where $\mathcal{F} = \sigma(\cup \mathcal{F_i})$ and $\mathcal{F_i} =\sigma(X_1,...X_i)$.
So I need to show $\mathbf{E}(\theta|\mathcal{F}) = \theta$, which is where I am having trouble.
Any hints on how to proceed would be helpful.
It suffices to show that $\theta$ is $\mathcal F$ measurable. But this is the case since almost surely $$\theta = \lim_{n\to\infty} \frac 1 n \sum_{i=1}^n X_i$$ and the right hand side is $\mathcal F $ measurable since $\frac 1 n \sum_{i=1}^n X_i$ is $\mathcal F$ measurable for every $n\in\Bbb N$.