convergence of a subsequence of sequence converges in measure If a sequence $(f_n)$ converge in measure to a function $f$, then every subsequence of $(f_n)$ converge in measure to $f$.
Let $g_{n_k}$ a subsequence of $(f_n)$ then $|g_{n_k}(x)-f(x)|\leq |g_{n_k}(x)-f_n|+|f_n(x)-f(x)|$ so $\mu\{x\in X: |g_{n_k}(x)-f(x)|\geq \alpha\}\leq \mu\{x\in X: |g_{n_k}(x)-f_n(x)|\geq \alpha/2\}+\mu\{x\in X: |f_{n}(x)-f(x)|\geq \alpha/2\}$
taking $n\rightarrow\infty$ and  $n_k\rightarrow\infty$ we have
 $\lim \mu\{x\in X: |g_{n_k}(x)-f(x)|\geq \alpha\}\leq \lim \mu\{x\in X: |g_{n_k}(x)-f_n(x)|\geq \alpha/2\}$
How I can prove that $g_{n_k}$ converge in measure to $f$? 
 A: I will assume that you meant convergence globally in measure. 
I will answer your question, providing all the details. 
By definition, a sequence $(f_n)$ converge in measure to a function $f$ if, 
for every $ \varepsilon> 0$ , 
$$\lim_{n\to\infty} \mu(\{x \in X: |f(x)-f_n(x)|\geq \varepsilon\}) = 0$$
It means: for every $ \varepsilon> 0$ and for every $\delta > 0$ , exists $N\in\mathbb{N}$, such that for all $n>N$, 
$$\mu(\{x \in X: |f(x)-f_n(x)|\geq \varepsilon\}) < \delta$$
Let $(f_{n_k})$ be a subsequence of $(f_n)$. So, we have that, for any  $N\in\mathbb{N}$, there is $K\in \mathbb{N}$ such that, for all $k>K$, $n_k> N$.  
So, for every $ \varepsilon> 0$ and for every $\delta > 0$ , exists $N\in\mathbb{N}$ and exists $K\in \mathbb{N}$ such that for all $k>K$ we have  $n_k> N$ and 
$$\mu(\{x \in X: |f(x)-f_{n_k}(x)|\geq \varepsilon\}) < \delta$$
So, we have prove that, for every $ \varepsilon> 0$ and for every $\delta > 0$ , exists $K\in \mathbb{N}$ such that for all $k>K$, 
$$\mu(\{x \in X: |f(x)-f_{n_k}(x)|\geq \varepsilon\}) < \delta$$
which means that, for every $ \varepsilon> 0$, 
$$\lim_{k\to\infty} \mu(\{x \in X: |f(x)-f_{n_k}(x)|\geq \varepsilon\}) = 0$$
So, $(f_{n_k})$ converge in measure to $f$.
