$\liminf_n f_n=f $ $a.e.$

Question

Let $(f_n)_n$ be a sequence of measurable functions, such that $(f_n)_n$ converges in measure to $f$. I want to prove that $\liminf_n f_n=f$ $a.e.$

I tried to prove it, but I get nothing, I am thankful if someone can give me a hint.

Answer 1

No, this is not correct. For all $n\in\mathbb N$ set

$$f_n\colon[0,1]\to\mathbb R, \quad x\mapsto(-1)^k1_{[l\cdot 2^{-k},(l+1)\cdot 2^{-k}]}(x),$$

where $k= \lfloor\log_2(n)\rfloor$ and $l=n-2^k$. Then $$f_n \xrightarrow{n\to\infty} 0 \quad \text{in measure (with respect to Lebesgue's measure),}$$ but $$\liminf_{n\to\infty} f_n(x) = -1 \quad\text{and}\quad \limsup_{n\to\infty} f_n(x) = 1 \quad \text{for all $x\in[0,1]$.}$$

However, if each $f_n$ is non-negative and the limit in measure is 0 almost everywhere, the statement is true. In this case, $\liminf_{n\to\infty}f_n\ge 0$ a.e. is trivial and it remains to prove the converse, which follows if we find a subsequence that converges to 0 almost everywhere. But such a subsequence has to exist thanks to the subsequence characterisation of convergence in measure.