# 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 $f$?

• I have posted a detailed answer to your question. Please, let me know if you have any question regarding my answer. – Ramiro Apr 25 '16 at 19:57

I will assume that you meant convergence globally in measure.

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 $f$.