If ${f}_{n}\ge 0$ a.e. and ${f}_{n}\to f$ in measure then $f\ge 0$ a.e.

Answer & Explanation

Shannon Hodgkinson

Beginner2022-01-13Added 34 answers

$\{x:f\left(x\right)<0\}\subseteq {\cup}_{k}\{x:{f}_{{n}_{k}}\left(x\right)<0\}\cup A$ where A is the set of points where $\left({f}_{{n}_{k}}\left(x\right)\right)$ does not converge to $f\left(x\right)>$

scomparve5j

Beginner2022-01-14Added 38 answers

${f}_{n}\ge 0$ a.e., in other words, for each $n=1,2,\dots ,$ there is a set ${N}_{n}\text{}\text{with}\text{}\left|{N}_{n}\right|=0$ and that $f}_{n}\left(x\right)\ge 0\text{}\text{for all}\text{}x\in {\left({N}_{n}\right)}^{c$.
Now the subsequence $\left({f}_{{n}_{k}}\right)$ is such that ${f}_{{n}_{k}}\left(x\right)\to f\left(x\right)$ a.e., so a set N is such that ${f}_{{n}_{k}}\left(x\right)\to f\left(x\right)$ for all $x\in {N}^{c}$.
Now $|{\cup}_{n}{N}_{n}\cup N|=0$ and for all $x\in {({\cup}_{n}{N}_{n}\cup N)}^{c}$
we have ${f}_{{n}_{k}}\left(x\right)\ge 0\text{}\text{and}\text{}{f}_{{n}_{k}}\left(x\right)\to f\left(x\right)$, this implies $f\left(x\right)\ge 0$ and hence $f\ge 0$ a.e. The later almost everywhere is due to ${\cup}_{n}{N}_{n}\cup N$.