Almost sure convergence of posterior distribution to parameter I'm trying to solve an exercise form Rick Durrett's book on probability, following  a section on almost sure convergence of martingales:
Let $Z_1,Z_2,...$ be i.i.d standard normal random variables, let $\theta$ be an independent random variable with finite mean, and let $Y_i:=Z_i+\theta$. $\;$ Show that $\mathbb{E}[\theta \vert Y_1,...,Y_n] \overset{n\rightarrow \infty}{\rightarrow}\theta$ almost surely.
I know that $\mathbb{E}\Big[ \frac{1}{n} \sum_{i=1}^n Y_i \Big \vert Y_1,...,Y_n \Big]=  \frac{1}{n} \sum_{i=1}^nY_i$, and that by the strong law of large numbers we know that $\frac{1}{n} \sum_{i=1}^nY_i \overset{n\rightarrow \infty}{\rightarrow}\theta$ almost surely. I'm unsure as to how best to proceed from here, or in fact how to use the independence of $\theta$. $\;$ I would appreciate any help or hints.
 A: $\newcommand{\F}{\mathcal{F}}\newcommand{\P}{\mathbb P}\newcommand{\E}{\mathbb E}$Edit. First I was not very careful enough with the measurability of the almost surely limit. 
Define: 
\begin{align}
\F_n=\sigma(Y_1,...,Y_n)
\end{align}
As noted in the comments we have:
\begin{align}
\E[\theta|\F_n]\to \E[\theta|\F_\infty] \ \ \ \ \ \text{ a. s.} 
\end{align}
But unfortunately that is not enough. We don't know whether  $\E[\theta|\F_\infty]=\theta$ holds. That is what we will show. Before showing that define $$X_n:= \frac{1}{n}\sum_{k=1}^n Y_k$$
Clearly $X_n$ is $\F_\infty$-measurable. 
Conditioning on $\theta$ we know that $Y_i\sim \mathcal N(\theta,1)$ i.i.d. So:
\begin{align}
\E[X_n|\theta]=\theta \ \ \ \ \ \text{ and } \ \ \ \ \ \E[(X_n-\theta)^2|\theta]=\frac{1}{n}
\end{align} 
Taking the expectation of both sides of the latter:
\begin{align}
\E[(X_n-\theta)^2]=\frac{1}{n}
\end{align}
That implies $X_n\to \theta $ in $L^2$. One knows that there is a subsequence $X_{n_k}$ that converges to $\theta$ a. s.. There is $\hat\theta$ which is $\F_\infty $-measurable and $\theta=\hat \theta $ a. s. (why?). Hence $$\E[\theta|\F_\infty] = \E[\hat\theta |\F_\infty] =\hat\theta$$
Conclusion:
\begin{align}
\E[\theta|Y_1,...,Y_n]\to \theta \ \ \ \ \ \text{ a. s. }
\end{align} 
