Let $(\Omega,\mathcal{F},P)$ be a probability space, with $u \in L^2( \Omega \times\mathbb{R}^d)$. Let $\rho$ be the standard mollifier approximation to Dirac's delta dunction defined on $\mathbb{R}^d$ supported in $\{ y \,|\, |y|\le 1\} \subset \mathbb{R}^d$. I would like to show $$\lim_{\delta \to 0} E\int_{|y| \le 1}\int_{\mathbb{R}^d}| u(x)-u(x+\delta y)|^2\rho(y) \,dx\,dy = 0$$
I know $\int_{\mathbb{R}^d}| u(x)-u(x+\delta y)|^2 \to 0$ as $\delta \to 0$ uniformly using transalation continuity in $L^p$ , and thus we have $\lim_{\delta \to 0} \int_{|y| \le 1}\int_{\mathbb{R}^d}| u(x)-u(x+\delta y)|^2\rho(y) \,dx\,dy = 0$. After this how can I interchange the limit with expection? Maybe somehow Dominated convergence theorem can be used.
Using
$$|u(x)-u(x+\delta y)|^2 \leq 2|u(x)|^2 + 2|u(x+\delta y)|^2$$
we find
$$\int_{\mathbb{R}^d} |u(x)-u(x+\delta y)| \, dx \leq 2 \int_{\mathbb{R}^d} |u(x)|^2 \, dx + 2 \int_{\mathbb{R}^d} |u(x+\delta y)|^2 \, dx = 4 \int_{\mathbb{R}^d} |u(x)|^2 \, dx.$$
Thus,
$$\begin{align*} \int_{|y| \leq 1} \int_{\mathbb{R}^d} |u(x)-u(x+\delta y)|^2 \varrho(y) \, dx \, dy \leq 4 \left( \int_{\mathbb{R}^d} |u(x)|^2 \, dx\right) \int_{|y| \leq 1} \varrho(y) \,dy. \end{align*}$$
As $u \in L^2(\Omega \times \mathbb{R}^d)$ the right-hand side is an integrable dominating function (which does not depend on $\delta$). Since you have already shown that the left-hand side converges to $0$ as $\delta \to 0$ we can apply the dominated convergence theorem to conclude that
$$\mathbb{E} \int_{|y| \leq 1} \int_{\mathbb{R}^d} |u(x)-u(x+\delta y)|^2 \varrho(y) \, dx \, dy \xrightarrow[]{\delta \to 0} 0.$$