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.
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}