Let $(X_n)_{n \geq 1}$ be a sequence of random variables. We know that $\lim_{n \to \infty} X_n = X$ (a.s.), where $X$ is a deterministic value.
I want to show that $\lim_{n \to \infty} \mathbb{E}[X_n] = X$.
For this I am trying to show that $X_n$'s are uniformly integrable (u.i.), so that the limit also holds for the expectation.
To show that $X_n$ is u.i., I'm using the fact that for each $n$ $X_n \leq Y_n$, where $Y_n$'s are random variables, and $\sqrt{n} Y_n$'s converge in distribution to a standard Gaussian distribution , so $Y_n$'s are u.i. in the limit $n \to \infty$.
Is this argument sound?
You argument cannot rest on the uniform integrability of $(Y_n)$, as it may not hold under the current assumptions. Indeed, let $(\xi_i)_{i\geqslant 1}$ be an i.i.d. sequence where $\xi_i$ is centered and has variance one and let $(A_n)$ be a sequence of events such that $\mathbb P(A_n)=1/n$. Let $$ Y_n=\frac 1n\sum_{i=1}^n\xi_i +2^{2^n}\mathbf{1}_{A_n}. $$ By the central limit theorem and the convergence in probability of $\sqrt n2^{2^n}\mathbf{1}_{A_n}$ to $0$, we get that $(\sqrt nY_n)$ converges in distribution to a standard normal random variable.
But the sequence $(Y_n)$ is not uniformly integrable, otherwise, uniform integrability of $\left(\frac 1n\sum_{i=1}^n\xi_i\right)$ would imply that $(2^{2^n}\mathbf{1}_{A_n})$ is uniformly integrable. This is not the case, since this sequence is not bounded in $\mathbb L^1$.