Mylee Underwood

2022-07-05

Show, that if ${f}_{n}\to f$ and ${f}_{n}\to g$ is $\mu$-convergent, then $f=g$ almost everywhere on $X$
Hint
Use the fact, that:
$\left\{x\in X\phantom{\rule{mediummathspace}{0ex}}:\phantom{\rule{mediummathspace}{0ex}}f\left(x\right)\ne g\left(x\right)\right\}=\bigcup _{m=1}^{\mathrm{\infty }}\left\{x\in X\phantom{\rule{mediummathspace}{0ex}}:\phantom{\rule{mediummathspace}{0ex}}|f\left(x\right)-g\left(x\right)|\ge \frac{1}{m}\right\}$
So, I don't know how to use that hint. μ convergent means (correct me if I'm wrong), that

So I don't see it how the hint should be used.
It is not written what kind of measure our $\mu$ is though, usually when there's nothing written we assume it's a Lebesgue measure, but I don't know if that has to be the case here

billyfcash5n

Expert

I would guess μ-convergent means the sequences converge in measure, which means for every $\epsilon >0$, $\mu \left(x\in X:|{f}_{n}\left(x\right)-f\left(x\right)|>\epsilon \right)\to 0$ as $n\to \mathrm{\infty }$. One way to proceed is note that
$|f-g|\le |f-{f}_{n}|+|{f}_{n}-g|,$
and use this to show that for every $\epsilon >0$,

Another approach is to use the fact that convergence in measure implies a subsequence converges almost everywhere. You can use this to get a subsequence ${f}_{{n}_{k}}\to f$ a.e., and then since ${f}_{{n}_{k}}\to g$ in measure, there is a subsequence ${f}_{{{n}_{k}}_{j}}\to g$ a.e. Therefore $f=g$ a.e, both being a.e. limits of ${f}_{{{n}_{k}}_{j}}$.

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