If $f_n \to f$ in measure, then there is a subsequence such that $f_{n_k}\to f$ almost everywhere

Question

Suppose we are on a finite measure space and $f_n\to f$ in measure. I want to show there is a subsequence such that $f_{n_k}\to f$ a.e.

I know that $f_n\to f$ in measure $\iff$ $\int_X\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu\to0$. So if $f_n\to f$ in measure, then by Fatou's lemma, $\int_X\text{lim inf}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu\leq\text{lim inf}\int_X\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu=0$. Since $\text{lim inf}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}\geq 0$, this means $\text{lim inf}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}=0$. Can I infer from this that there is a subsequence such that $\frac{|f_{n_k}(x)-f(x)|}{1+|f_{n_k}(x)-f(x)|}\to 0$ and so $f_{n_k}\to 0$ almost everywhere?

Answer 1

No, your argument does not give a convergent subsequence. (You can only get a subsequence depending on x). Here is a valid proof: Choose $n_k$ such that $\mu \{|f_n -f| >2^{-k}\} < 2^{-k}$ for $n \geq n_k$. Since $\sum_k \mu \{|f_{n_k} -f| >2^{-k}\} < \infty$ it follows the set of points x for which $|f_{n{_k}}(x)-f(x)| > 2^{-k}$ has measure 0 which shows that $f_{n{_k}} -f \to 0$ a.e..

Answer 2

Is your function $f$ non-negative? Without that, you can not apply Fatou's lemma. The answer to your question is yes, I think you may find this post helpful Do we need to have a subsequence such that $\lim_{k\to\infty}\left\|x_{n_k}\right\|=\liminf_{n\to\infty}\left\|x_n\right\|$?