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I am aware that this question has been asked many times, but I am still not getting it so I am asking it again.

Given a measure space $(X,M,\mu)$, let $f_n$ be a sequence of complex-valued $\mu$-measurable functions that converge pointwise to $f$ on subset $E$ of $X$ with $\mu(E^c) =0 $. Then, I know that if the measure is not complete, $f$ is not necessarily measurable. But, I read that $f$ can be redefined on $E^c$ so as to become measurable. For instance, one can define $f = 0$ on $E^c$. But, why is then this redefined function measurable?

In Folland's Real Analysis book, he refers to Proposition 2.12 in regards to this matter, but I don't understand how that proposition helps.

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  • $\begingroup$ You should make the statement more precise. Give the measure space a name, tell us where the $f_n$ are defined and where $f$ is defined. $\endgroup$ – zhw. Apr 10 '18 at 18:39
  • $\begingroup$ @zhw. I made my statement more precise $\endgroup$ – nan Apr 10 '18 at 20:25
  • $\begingroup$ I want to know precisely what's going on. Is it true that each $f_n : X\to \mathbb R$ is measurable wrt $(X\mathcal M, \mu)?$ Where is $f$ defined, and where does it map into? $\endgroup$ – zhw. Apr 10 '18 at 21:32
  • $\begingroup$ @zhw. Please see my edit. $f$ is defined on $X$ and maps into a set of complex values $\endgroup$ – nan Apr 11 '18 at 2:12
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Suppose $f_n:X\to \mathbb C$ is measurable for $n=1,2,\dots,$ $f:X\to \mathbb C,$ and $f_n \to f$ pointwise a.e. This means there exists a measurable $E,$ with $\mu(E^c)=0,$ such that $f_n \to f$ pointwise everywhere on $E.$

Define $g_n = f_n\cdot \chi_E,\, n=1,2,\dots$ Then each $g_n$ is measurable on $X,$ and $g_n$ converges pointwise everywhere on $X$ to the function $f\cdot\chi_E.$ It follows that $f\cdot\chi_E$ is measurable on $X.$

In other words, by altering the definition of $f$ on $E^c,$ if necessary, to be $0$ there, and leaving $f$ alone on $E,$ we obtain a measurable function on $X.$

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If the $\sigma$-algebra of context is $\mathcal{M}$, then we can make $\mathcal{M}^*$, the completion of $\mathcal{M}$.

Then $f$ is measurable on $\mathcal{M}^*$, not on $\mathcal{M}$.

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The function $g(x)=\lim f_n(x)$ for $x$ such that the limit exists and $0$ for all other $x$ is a measurable function for any sequence of measurable functions $\{f_n\}$. In our case $f=g$ almost everywhere and $f=g$ on $E$. This $g$ is the 'redefined' function that Folland is referring to.

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  • $\begingroup$ why is $g$ measurable? $\endgroup$ – nan Apr 11 '18 at 13:40
  • $\begingroup$ $\{x:\lim f_n(x)$ exists $\}=\cap_k \cup_m \cap_{n \geq m} \{ \{x:|f_n(x)-f_m(x)|<1/k \}$. So $\{x:\lim f_n(x)$ exists $\}$ is measurable. Measurability of $g$ now follows by a standard argument using $\limsup f_n$ and $\liminf f_n$. $\endgroup$ – Kavi Rama Murthy Apr 12 '18 at 5:11

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