If for every subsequence $\{X_{n_m}\}$ there is a further subsequence $\{X_{n_{m_k}}\}$ that converges a.e. to $X$, then $X_n \to X$ in probability?
Above statement is part of Theorem 2.3.2 of Durrett's Probability, Theory and Examples.
Assume for every subsequence $\{X_{n_m}\}$ there is a further subsequence $\{X_{n_{m_k}}\}$ that converges a.e. to $X$. Following his idea, fix $\delta >0$,let $y_n =P\{ \vert X_n - X \vert > \delta \}$. I think he thinks that $\{X_{n_{m_k}}\}$ $\to$ $X$ a.e. implies $y_{n_{m_k}}\to 0$ so that $y_n \to 0$ so that $X_n \to X$ in measure.
I wonder if this step: $\{X_{n_{m_k}}\}$ $\to$ $X$ a.e. ⇒ $y_{n_{m_k}}\to 0$ is true or not?
If $X_{n_{m_k}} \rightarrow X$ a.e., then we have $1_{\{|X_{n_{m_k}}-X| > \delta\}} \rightarrow 0$ a.e. ($1_A$ denotes the indicator function). Therefore we get
$$y_{n_{m_k}}:=P({\{|X_{n_{m_k}}-X| > \delta\}}) = \int 1_{{\{|X_{n_{m_k}}-X| > \delta\}}} dP \rightarrow 0$$
Where in the last step we can use Lebesgue's dominated convergence theorem, since $(|1_{{\{|X_{n_{m_k}}-X| > \delta\}}}|)_{k \in \mathbb{N}}$ is bounded by $1 \in L^1(P)$.