Let $\{X_n\}$ be a sequence of real random variables, such that $$\displaystyle\liminf_{n\to \infty}E({X_n}^2)<\infty$$
Then show there exists an integrable random variable $X$ and a subsequence $\{n_k\}$ such that $$X_{n_k}\implies X\quad \text{as }k\to \infty$$
and $$\lim_{k\to\infty}E(X_{n_k})=E(X)$$
where $\implies$ denotes weak convergence, or convergence in distribution.
I'm completely stuck in this problem. How does one begin with looking for such a subsequence. I tried to consider the subsequence $\{n_k\}$ such that $$\lim_{k\to \infty}E({X_{n_k}}^2)=\ell<\infty$$ but I couldn't proceed anywhere from here.
Best Answer
The subsequence you have chosen is bounded in $\mathbb L^2$. This sequence is tight, because $$\mathbb P\left(\left\lvert X_{n_k}\right\rvert \gt R\right)\leqslant R^{-2}\mathbb E\left[X_{n_k}^2\right]\leqslant R^{-2}\sup_{i\geqslant 1}\mathbb E\left[X_{n_i}^2\right].$$ This means that (from Prokhorov's theorem) we can extract from $\left(X_{n_k}\right)_{k\geqslant 1}$ a subsequence which converges in distribution to some random variable $X$. This subsequence is also uniformly integrable, which guarantees the converges of the expectations.