Gregory Jones

2022-01-12

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

### Answer & Explanation

Shannon Hodgkinson

$\left\{x:f\left(x\right)<0\right\}\subseteq {\cup }_{k}\left\{x:{f}_{{n}_{k}}\left(x\right)<0\right\}\cup A$ 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

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

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