We say that $(X_n)_n$ converges in probability to $+\infty$ if $$\forall \epsilon>0,\lim_n P(X_n<\epsilon)=0.$$
Prove that there exist a subsequence $(X_{\phi(n)})_n$ which converges to $+\infty$ a.s.
From the definition, we have $$\forall \epsilon>0,\forall \eta>0,\exists n_0 \in \mathbb{N};\forall n \geq n_0,P(X_n<\epsilon)\leq\eta,$$ but I don't see how to extract a subsequence which converges a.s to $+\infty.$
Convergence in probability means $$\forall \epsilon>0,\forall \eta>0,\exists n_0 \in \mathbb{N};\forall n\geq n_0,P(X_n<\epsilon)\leq\eta.$$ For all $n \in \mathbb{N},$ define $\psi(n)=\inf\left\{ k \in \mathbb{N};\forall p \geq k,P(X_p<2^n) \leq\frac{1}{2^n}\right\},\phi(0)=\psi(0)$ and $\phi(n+1)=\max(\phi(n)+1,\psi(n+1)),$ so we will have $$\sum_nP(X_{\phi(n)}<2^n)\leq \sum_n\frac{1}{2^n}<+\infty,$$ Borel-Cantelli lemma implies $P(\liminf_n\left\{X_{\phi(n)} \geq 2^n\right\})=1.$ Let $w \in \liminf_n\left\{X_{\phi(n)} \geq 2^n\right\}.$ Let $\epsilon>0.$
$\exists n_0 \in \mathbb{N};\forall n \geq n_0,X_{\phi(n)}(w) \geq 2^n$ and $\exists n_1 \in \mathbb{N};\forall n \geq n_1;2^n \geq\epsilon.$
Let $n \geq \max(n_0,n_1).$ We have $X_{\phi(n)}(w)\geq2^n \geq \epsilon,$ which means that $(X_{\phi(n)})_n$ converges a.s to $+\infty.$