Let $(X, \mathcal{M})$ be a measurable space and let $f_1, f_2, … : X \longrightarrow [-\infty, \infty]$ be measurable functions. Then $\{x \in X: \lim\limits_{n \to \infty} f_n(x) \text{ exists in } [-\infty, \infty]\} \in \mathcal{M}$.
My professor gave this as an extra problem during the spring term, and in reviewing my notes I realized I never finished this one.
I know that if for all $x \in X$, $f_n(x) \to f(x)$, then $f(x)$ is measurable, but I'm having difficulty dealing with this situation, where the $f_n(x)$ might not converge for all $x$.
My professor suggested that we start by showing that $\{x \in X: f_n(x) \to 12\} \in \mathcal{M}$, which I've done:
We know that $\limsup\limits_{n \to \infty} f_n(x)$ and $\liminf\limits_{n \to \infty} f_n(x)$ are measurable functions, and that for all $x$, $f_n(x) \to 12$ if and only if $\limsup\limits_{n \to \infty} f_n(x) = \liminf\limits_{n \to \infty} f_n(x)=12$. Thus
$$\{x \in X: f_n(x) \to 12\} = \{x \in X: \limsup\limits_{n \to \infty} f_n(x)= 12\} \cap \{x \in X: \liminf\limits_{n \to \infty} f_n(x)= 12\}$$
is in $\mathcal{M}$ since it is the intersection of two sets in $\mathcal{M}$.
I'm not sure how to adapt this to the situation at hand, though.
Best Answer
Here is an approach essentially from first principles: we can write $$ \{x\in X:\lim_{n\to\infty}f_n(x)\;\mathrm{exists}\}=A\cup A_{\infty}\cup A_{-\infty} $$ where $A_{\pm\infty}$ are the sets of $x$ such that the limit is equal to $\pm\infty$ respectively, and $A$ is the set of $x$ such that the limit exists and is real.
Now $\lim_{n\to\infty}f_n(x)=\infty$ if for all $M\geq 1$ there exists $N$ such that $f_n(x)\geq M$ for all $n\geq N$. Therefore $$ A_{\infty}=\bigcap_{M=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{n=N}^{\infty}\{x:f_n(x)\geq M\}$$ hence is measurable. A similar argument works for $A_{-\infty}$.
Finally, $\{f_n(x)\}$ converges to a real number if and only if $f_n(x)$ is real for all but finitely many $n$ and $\{f_n(x)\}$ is a Cauchy sequence, hence $$ A=\bigcap_{k=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{m,n=N}^{\infty}\bigcup_{r\in\mathbb{Q}}\{x:r-k^{-1}<f_n(x),f_n(x)<r\}$$ which is measurable.