Real Analysis, Folland problem 2.1.3

Question

If $\{f_n \}$ is a sequence of measurable functions on $X$, then $\{x:\lim f_n(x) \text{ exists}\}$ is a measurable set.

Corollary $2.9$: If $\{f_j\}$ is a sequence of complex-valued measurable functions and $f(x) = \lim_{j\rightarrow \infty}f_j(x)$ exists for all $x$, then $f$ is measurable.

Proof: Suppose, $\{f_n \}$ is a sequence of measurable functions on $X$. Then by Corollary $2.9$, there exists a function $$f(x) = \lim_{n\rightarrow \infty}f_n(x)$$ which is measurable thus this implies the set $\{x:\lim f_n(x) \text{ exists}\}$ is a measurable set.

Answer 1

So (checking Folland) Corollary 2.9 is the proposition:

If $f_j$ is a sequence of complex valued measurable functions, and $\lim_{j\to\infty}f_j(x)$ exists for all $x\in X$ then $f(x)=\lim_{j\to\infty} f_j(x)$ is measurable.

So unfortunately, your suggested proof doesn't quite work, since there could some places where the limit does not exist and you cannot apply the lemma. Well, you probably could, but it requires a little more care.

Here's a way to do this: taking measurable functions to be real valued measurable functions, we know that $g=\liminf_{j\to\infty}f_j$ and $h=\limsup_{j\to\infty}f_j$ are measurable since the $\liminf,\limsup$ always exist. From here, you'll need to show that if $g,h$ are measurable functions, then $$\{x:g(x)=h(x)\}$$ is measurable, which is a little difficult (it's much easier when assuming the $f_j$ are real valued; if not, you'll need to apply exercise 2.2). From here the proposition follows.