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Let $\mathrm{X}$ and $\mathrm{(X_n)_{n\in\mathbb{N}}}$ be a sequence of random variables such that $\mathrm{X_n}$ converges to $\mathrm{X}$ in probability. Prove that there is a subsequence $\mathcal({n_k})_{k\in\mathbb{N}}$ such that $\mathrm{X_{n_{k}}}$ converges almost surely to $\mathrm{X}$.

My thoughts so far: If I let $\mathcal{n_k}$ be such that $\forall$ $\epsilon$ $\gt$ 0: P($\vert$$\mathrm{X_{n_{k}}}$ - $\mathrm{X}$$\vert$$\ge$$\epsilon$)$\le$$\frac{1}{k^2}$, then the assumption would hold with the following lemma: If $\sum_{n=1}^\infty$P($\vert$$\mathrm{X_n}$-$\mathrm{X}$$\vert$$\gt$$\epsilon$)$\lt$$\infty$ for all $\epsilon$$\gt$0, then $\mathrm{X_n}$ converges to $\mathrm{X}$ almost surely.

For some reason I am not entirely convinced by this. Is there another way to prove the assumption.

Did
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fidel castro
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    Your idea is good, see this answer https://math.stackexchange.com/a/1006122/36150 – saz Jun 17 '17 at 17:35
  • If $\sum P(\vert X_n-X\vert\gt\epsilon)$ is finite then $\limsup\vert X_n-X\vert\leqslant\epsilon$ almost surely. Thus, $\limsup\vert X_n-X\vert\leqslant\frac1k$ almost surely, for every $k$, hence $\limsup\vert X_n-X\vert\leqslant0$ almost surely, that is, $X_k\to X$ almost surely. – Did Jun 17 '17 at 17:36

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