I have the following problem to solve:
Let $X_1$, $X_2$, ... be a sequence of random variables such that $$\lim_{k\to\infty} \sup_{n,m\geq k} \mathbb{E}[(X_n - X_m)^2] = 0$$ holds.
Proof that there exists a random variable $X$ with $\mathbb{E}[X^2]<\infty$ such that $$\lim_{n\to\infty} \mathbb{E}[(X_n - X)^2] = 0$$
I don't know where to start here. Any hint will be appreciated.
The condition $$ \lim_{k\to\infty}\sup_{n,m\geqslant k}\mathbb E[(X_n-X_m)^2] = 0 $$ implies that the sequence $(X_n)$ is Cauchy in $L^2$. So we may choose a subsequence $(X_{n_k})$ such that $\mathbb E\left[\left(X_{n_{k+1}}-X_{n_k}\right)^2\right]\leqslant 2^{-k} $ for $k=1,2,\ldots$. Set $$Y=|X_{n_1}| + \sum_{k=1}^\infty |X_{n_{k+1}}-X_{n_k}| $$ and $$Y_K = |X_{n_1}| + \sum_{k=1}^K |X_{n_{k+1}}-X_{n_k}|,$$ then Minkowsi's inequality yields \begin{align} \mathbb E\left[Y_K^2 \right] &\leqslant \mathbb E[X_{n_1}^2] + \sum_{k=1}^K\mathbb E\left[\left(X_{n_{k+1}}-X_{n_k} \right)^2\right]\\ &\leqslant E[X_{n_1}^2] + \sum_{k=1}^{K}2^{-k}\\ &=E[X_{n_1}^2] + 1 - 2^{-K}. \end{align} By monotone convergence, it follows that $Y_K\stackrel{K\to\infty}\longrightarrow$ in $L^2$. Set $$X=X_{n_1} + \sum_{k=1}^\infty (X_{n_{k+1}}-X_{n_k})$$ and $$X_K = X_{n_1} + \sum_{k=1}^K (X_{n_{k+1}}-X_{n_k}) = X_{n_{k+1}},$$ then it follows that $X_K\stackrel\longrightarrow X$ a.s. and $X\in L^2$. Now, observe that \begin{align} (X-X_K)^2 &\leqslant (2\max\{|X|, |X_K|\})^2\\ &\leqslant 4|X|^2 + 4|X_K|^2\\ &8|Y| \end{align} for all $K$. Dominated convergence yields $\mathbb E[(X_{n_K}-X)^2]\stackrel{K\to\infty}\longrightarrow 0$. Finally, given $\varepsilon>0$, we may choose $N$ so that $\mathbb E[(X_n-X_M)^2]<\frac\varepsilon2$ for $n,m>N$. Choose $n_K>N$ with $\mathbb E[(X_{n_K}-X)^2]<\frac\varepsilon2$, then $$ \mathbb E[(X_n-X)^2] \leqslant \mathbb E[(X_n-X_{n_K})^2] + \mathbb E[(X_{n_K}-X)^2<\frac\varepsilon2+\frac\varepsilon2<\varepsilon $$ for $n>N$ and hence $X_n\stackrel{n\to\infty}\longrightarrow X$ in $L^2$.