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if $f_n(x)$ is sequence of measurable functions,all defined on some measurable set $D$. and we have$$ f(x)=\lim_{n\to \infty} f_n(x)$$ almost every where. then show $f$ is measurable function.

now this mean there is set say $E$ such that $f(x)=\lim_{n\to \infty} f_n(x)$ for all $x\in D/E$ with $m(E)=0$

now what is domain of $f$ ??and how do we show $f$ is measurable ? please help

paarth
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3 Answers3

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First show that $g_1(x)=\sup_j f_j(x)$, $g_2(x)=\inf_j f_j(x)$, $g_3(x)=\limsup_{j\rightarrow \infty} f_j(x)$, $g_4(x)=\liminf_{j\rightarrow \infty} f_j(x)$ are all measurable. Then $f=g_3=g_4$.

Dunham
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assuming domain of $f$ to be $D$.

then let $$A=\{x\in D:f(x)> \alpha\}$$ then$$A=\{x\in D/E:f(x)> \alpha\}\ \cup \{x\in E:f(x)> \alpha\}$$. now $$m(\{x\in E:f(x)> \alpha\})=0$$ as being a subset of set of measure zero. now $$\{x\in D/E:f(x)> \alpha\}=\cup_{n=1}^\infty \{x\in D/E:f_n(x)> \alpha\}$$

now since $f_n$ are measurable function on $D$ hence they are measurable on $D/E$ and hence right hand side is measurable set. which prove the result.

is it correct ?? or any better method? what about the domain of $f$??

paarth
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  • I think your argument is false. Let $y \notin {D/E : f(x) > \alpha}$, then let $f_0(x) =2|\alpha|$ for all $x \in D/E$. It is measurable almost everywhere in $D$. Hence $y \in \cup_{n = 0}^\infty {x \in D/E : f_n(x) > \alpha}$ but not in ${f(x) > \alpha}$ when restrict to $D/E$. Your argument is true for $\sup_n f_n(x)$ – yuanming luo Apr 15 '25 at 01:37
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Assuming that $(D,\mathcal D,\mu)$ is a complete measure space, that each $f_n:D\to\Bbb R$ is $\mathcal D$-measurable, and that a.e. refers to $\mu$: there exists $E\in\mathcal D$ such that $f(x):=\lim_nf_n(x)$ exists for each $x\in E$ and $\mu(D\setminus E)=0$.

We can now define $f$ arbitrarily on $D\setminus E$, and the resulting function will be $\mathcal D$-measurable, because of the assumed completeness. To see this define the function $g$ by $g(x):=\liminf_n f_n(x)$; $g$ is $\mathcal D$-measurable. Given a Borel subset $B$ of $\Bbb R$, we have the disjoint union $$ \{x\in D: f(x)\in B\}=\{x\in E: g(x)\in B\}\cup\{x\in D\setminus E: f(x)\in B\}. $$ The first set on the right is an element of $\mathcal D$ because $g$ is $\mathcal D$-measurable. The second set on the right is a subset of $D\setminus E$, which is $\mu$-null. By completeness, any subset of $D\setminus E$ is $\mathcal D$-measurable. It follows that $\{x\in D: f(x)\in B\}\in\mathcal D$, so $f$ is $\mathcal D$-measurable.

John Dawkins
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