Rick Durrett Probability Theory 5th edition exercise 4.6.1
Sophia Terry 
Ex 4.6.1: Let $Z_1,Z_2, ...,$ be independent and identically distributed with $E|Z_i|<\infty$, let $\theta$ be an independent random variables with finite mean, and let $Y_i=Z_i+\theta$. If $Z_i$ is normal$(0,1)$ then in statistical terms we have a sample from a normal population with variance 1 and unknown mean. The distribution of $\theta$ is called the prior distribution, and $P(\theta \in \dot\mid Y_1,\ldots,Y_n)$ is called the posterior distribution after n observations. Show $E(\theta \mid Y_1,\ldots,Y_n) \rightarrow \theta$ almost surely.
The problem is from Rick Durrett's Probability Theory 5th edition.
Theorem 4.6.8: Suppose $\mathcal{F}_n \uparrow \mathcal{F}_{\infty}$, i.e., $\mathcal{F}_n$ is an increasing sequence of $\sigma$-fields and $\mathcal{F}_{\infty}=\sigma(\cup_n \mathcal{F}_n)$. As $n \rightarrow \infty,$ $E(X\mid \mathcal{F}_n) \rightarrow E(X\mid \mathcal{F}_{\infty})$ almost surely and in $L^1$.
Attempt: Let $\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)$, then by theorem 4.6.8, we have $E(\theta \mid \mathcal{F}_n) \rightarrow E(\theta\mid \mathcal{F}_{\infty})$. Then I am told to show $\theta \in \mathcal{F}_{\infty}$ using strong law of large number.
My question lies in using the strong law of large number. Since $Z_i \sim \text{normal}(0,1)$, $E Z_i=0$, then $EY_i=E \theta$. By strong law of large number, $\frac{Y_1+\cdots+Y_n}{n} \rightarrow E\theta $ almost surely, not $\rightarrow \theta$. That's what got me here. Thanks in advance!
Update: https-//
Based on the link above and Eric (Thanks!)'s comment, I think I got it.
Let $EZ_i=0$, by strong law of large number,
$\begin{align} \lim_{n \rightarrow \infty} \frac{Z_1+...+Z_n}{n}=EZ_i=0 \\ \Rightarrow \theta+\lim_{n \rightarrow \infty} \frac{Z_1+...+Z_n}{n}=\theta \\ \Rightarrow \frac{n\theta}{n}+\lim_{n \rightarrow \infty} \frac{Z_1+...+Z_n}{n}=\theta \\ \Rightarrow \lim_{n \rightarrow \infty} \frac{n\theta+Z_1+...+Z_n}{n}=\theta \\ \Rightarrow \lim_{n \rightarrow \infty} \frac{(Z_1+\theta)+...+(Z_n+\theta)}{n}=\theta \\ \Rightarrow \lim_{n \rightarrow \infty} \frac{Y_1+...+Y_n}{n}=\theta \end{align} $.
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