Let $\{f_n\}$ be a sequence of Lebesgue measurable functions defined on a Lebesgue measurable set $E\subset \mathbb{R}$. If we define $E_0:=\{x\in E: \{f_n(x)\} \ \rm converges\}$, then the set $E_0$ is Lebesgue measurable. There are a couple of ways to show this fact. One of them would be the following:
Since the functions $\liminf f_n$ and $\limsup f_n$ are Lebesgue measurable, and since for any Lebesgue measurable function $h:E\longrightarrow \mathbb{R}\cup\{\pm\infty\}$ the set $\{x\in E: h(x)=c\}$ is Lebesgue measurable, if we define $g(x):=\liminf f_n(x)-\limsup f_n(x)$ then one would conclude that the set $E_0=\{x\in E: g(x)=0\}$ is Lebesgue measurable.
However I think that there is a gap in this (possible) solution. Because we cannot guarantee that the function $g$ is well-defined, as for some $x\in E$ we may have $\lim f_n(x)=\infty$ and the Lebesgue measure of the set of such points may be positive, that is, $m(\{x\in E : \lim f_n(x)=\infty\})>0$ (similarly, one would have $m(\{x\in E : \lim f_n(x)=-\infty\})>0$), where $m$ is the Lebesgue measure.
My question is: Is it possible to fix the gap in this (possible) solution? (Or, Is there a gap really?) Thanks a lot in advance for any answer/comment/hint.
Best Answer
You may have a problem because both $u=\liminf f_n$ and $v=\limsup f_n$ are measurable, but their values may be $\pm\infty$ and $\infty-\infty$ is not well defined. However, we do not really care about those points because if either $u$ or $v$ are infinite at $x$, then clearly $x\notin E_0$, so filter them out by letting $A=u^{-1}(-\infty,\infty)\cap v^{-1}(-\infty,\infty)$. Then $A$ is measurable and contains $E_0$. Furthermore, because $u,v$ are finite on $A$, $g:A\to\Bbb R:x\mapsto v(x)-u(x)$ is well defined and measurable in $A$. It follows that $E_0=g^{-1}(0)$ is measurable.