Convergence almost surely for a subsequence

Question

I am trying to prove that for a sequence ${X_m^n}$

if $X_m^n \overset{a.s}{\to} X_m$ as $n \to \infty$

and $X_m \overset{a.s}{\to} X$

Then there are sub-sequences $\{m_k\}, \{n_k\}$ such that $X_{m_k}^{n_k} \overset{a.s}{\to} X$

Here is how I am going about the proof:

$A_m = \{\omega: X_m^n[\omega] {\to} X_m[\omega]$ as $n \to \infty\}$

$B = \{\omega: X_m[\omega] {\to} X[\omega]$ as $n \to \infty\}$

Then if I can prove there are sub-sequences $\{m_k\}, \{n_k\}$ such that $X_{m_k}^{n_k}[\omega] {\to} X[\omega]$ if $\omega \in A_m \cap B$Then the result follows.

Answer 1

Since a.s.-convergence implies convergence in probability, we can choose $m_k$ such that

$$\mathbb{P}(\left|X_{m_k} - X\right|>2^{-k-1})<2^{-k-1}.$$

Similarly, we can choose $n_k$'s such that

$$\mathbb{P}(\left|X_{m_k}^{n_k} - X_{m_k}\right|>2^{-k-1})<2^{-k-1}.$$

Then by the triangle inequality,

\begin{align*} \mathbb{P}(\left|X_{m_k}^{n_k} - X\right|>2^{-k}) &\leq \mathbb{P}(\left|X_{m_k} - X\right|>2^{-k-1}) + \mathbb{P}(\left|X_{m_k}^{n_k} - X_{m_k}\right|>2^{-k-1}) \\ &< 2^{-k-1} + 2^{-k-1} = 2^{-k}. \end{align*}

Therefore $\sum_{k=1}^{\infty} \mathbb{P}(\left|X_{m_k}^{n_k} - X\right|>2^{-k}) < \infty$, and so,

$$ \mathbb{P}(\left|X_{m_k}^{n_k} - X\right|>2^{-k}\text{ i.o.}) = 0 $$

by the Borel-Cantelli's lemma. This implies that $X_{m_k}^{n_k} \to X$ a.s.