$V\in L^2$, $U$ is standard normal, $U,V$ independent$\implies$ $\lim_{\sigma \to \infty} E[V| V+\sigma U]=E[V]$ a.s. How to show that 
\begin{align}
\lim_{\sigma \to \infty} E[V| V+\sigma U]=E[V],  a.s.
\end{align} 
where $E[V^2]<\infty$ and $U$ is standard normal and $U$ and $V$ are independent. 
I was thinking that one option is to use that 
\begin{align}
 E[V| V+\sigma U=t]= \frac{ E[ V \phi_\sigma(  t-V)]}{  E[  \phi_\sigma(  t-V)]}
\end{align}
where  $\phi_\sigma(  t)=\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{t^2}{2 \sigma^2}}$. 
Another option is to use some Martingale theorem. However, I am not sure how to show all the requirements. 
 A: Using the formula you provided, we calculate 
\begin{align}
\lim_{\sigma\to\infty} \sqrt{2\pi\sigma^2}\mathbb E(V\phi_\sigma(t-V)) & =\lim_{\sigma\to\infty} \int_\Omega V(\omega) \exp\left(-\frac{(t-V(\omega)) ^2} {2\sigma^2}\right)\,d\mathbb P(\omega)\\
 & =\int_\Omega\lim_{\sigma\to\infty} V(\omega) \exp\left(-\frac{(t-V(\omega)) ^2} {2\sigma^2}\right)\,d\mathbb P(\omega)\\
&=\int_\Omega V(\omega) \cdot 1 \, d\mathbb P(\omega)\\
& =\mathbb E(V) 
\end{align}
We still need to justify interchanging limit and integration. 
We have a dominating integrable random variable 
$$\left|V(\omega) \exp\left(-\frac{(t-V(\omega)) ^2} {2\sigma^2}\right)\right|\leq |V(\omega) |$$
Moreover the LHS converges to $V(\omega) $. Hence the claim follows by Dominating convergence theorem. 
Similarly we have using DCT
\begin{align}
\lim_{\sigma\to\infty} \sqrt{2\pi\sigma^2}\mathbb E(\phi_\sigma(t-V)) =\lim_{\sigma\to\infty} \int_\Omega  \exp\left(-\frac{(t-V(\omega) )^2} {2\sigma^2}\right)\,d\mathbb P(\omega) =\mathbb E(1)=1
\end{align}
In total 
\begin{align} 
\lim_{\sigma\to\infty} \mathbb E(V\mid V+\sigma U=t) =\lim_{\sigma\to\infty}\frac{\sqrt{2\pi\sigma^2}\mathbb E(V\phi_\sigma(t-V))} {\sqrt{2\pi\sigma^2}\mathbb E(\phi_\sigma(t-V))} =\frac{\mathbb E(V)} {1}=\mathbb E(V) 
\end{align} 
