Mylee Underwood

Answered

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:

$\{x\in X\phantom{\rule{mediummathspace}{0ex}}:\phantom{\rule{mediummathspace}{0ex}}f(x)\ne g(x)\}=\bigcup _{m=1}^{\mathrm{\infty}}\{x\in X\phantom{\rule{mediummathspace}{0ex}}:\phantom{\rule{mediummathspace}{0ex}}|f(x)-g(x)|\ge \frac{1}{m}\}$

So, I don't know how to use that hint. μ convergent means (correct me if I'm wrong), that

${f}_{n}\to f\text{is}\mu \text{convergent}\phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}\mu (\{x\in X\phantom{\rule{mediummathspace}{0ex}}:\phantom{\rule{mediummathspace}{0ex}}{\mathrm{\forall}}_{\epsilon}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty}}{lim}|{f}_{n}(x)-f(x)|\epsilon \})=0$

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

Hint

Use the fact, that:

$\{x\in X\phantom{\rule{mediummathspace}{0ex}}:\phantom{\rule{mediummathspace}{0ex}}f(x)\ne g(x)\}=\bigcup _{m=1}^{\mathrm{\infty}}\{x\in X\phantom{\rule{mediummathspace}{0ex}}:\phantom{\rule{mediummathspace}{0ex}}|f(x)-g(x)|\ge \frac{1}{m}\}$

So, I don't know how to use that hint. μ convergent means (correct me if I'm wrong), that

${f}_{n}\to f\text{is}\mu \text{convergent}\phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}\mu (\{x\in X\phantom{\rule{mediummathspace}{0ex}}:\phantom{\rule{mediummathspace}{0ex}}{\mathrm{\forall}}_{\epsilon}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty}}{lim}|{f}_{n}(x)-f(x)|\epsilon \})=0$

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

Answer & Explanation

billyfcash5n

Expert

2022-07-06Added 17 answers

I would guess μ-convergent means the sequences converge in measure, which means for every $\epsilon >0$, $\mu (x\in X:|{f}_{n}(x)-f(x)|>\epsilon )\to 0$ 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 $\epsilon >0$,

$\mu (|f-g|>\epsilon )\le \mu (|f-{f}_{n}|>\frac{\epsilon}{2})+\mu (|{f}_{n}-g|>\frac{\epsilon}{2})\to 0\text{as}n\to \mathrm{\infty}.$

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 ${f}_{{{n}_{k}}_{j}}$.

$|f-g|\le |f-{f}_{n}|+|{f}_{n}-g|,$

and use this to show that for every $\epsilon >0$,

$\mu (|f-g|>\epsilon )\le \mu (|f-{f}_{n}|>\frac{\epsilon}{2})+\mu (|{f}_{n}-g|>\frac{\epsilon}{2})\to 0\text{as}n\to \mathrm{\infty}.$

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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