pointwise convergence in $\sigma$-algebra

Solution 1:

The other answers using $\inf$, $\sup$, $\limsup$ and $\liminf$ are nice and efficient, but your idea can also be made into an argument:

Observe that a sequence of real numbers converges to a real number if and only if it is a Cauchy sequence. Thus $$ \begin{align*} \{t \in T\,&:\,\lim_{n\to\infty} f_n(t) \text{ exists and is in }\mathbb{R}\} \\ & = \{t \in T\,:\,(\forall k \in \mathbb{N})\,(\exists N \in \mathbb{N})\,(\forall m,n \geq N) \; |f_n(t) - f_m(t)| \lt 1/k\} \\ & = \bigcap_{k = 1}^\infty \bigcup_{N = 1}^\infty \bigcap_{n,m \geq N} \{t \in T\,:\,\lvert f_n(t)-f_m(t)\rvert \lt 1/k\} \in \mathcal{A} \end{align*} $$ because $\mathcal{A}$ is a $\sigma$-algebra and $\lvert f_n(t) - f_m(t)\rvert$ is a measurable function, so that $\{t \in T\,:\,\lvert f_n(t)-f_m(t)\rvert \lt 1/k\} \in \mathcal{A}$.

Solution 2:

Hint: Given a $t\in T$, $\lim_{n\to\infty}f_n(t)$ exists if and only if $$\limsup_{n\to\infty}f_n(t)=\liminf_{n\to\infty}f_n(t),$$ and this quantity is finite. Show that the functions $$g(t)=\limsup_{n\to\infty}f_n(t),\quad\quad h(t)=\liminf_{n\to\infty}f_n(t)$$ are $\mathcal{A}$-measurable, hence the set where they are equal is measurable, hence the set where they are equal and finite is measurable.

Solution 3:

Show the function $x \mapsto \sup_n f_n(x)$ is measurable. It follows from this that $x \mapsto \inf_n f_n(x)$ is measurable. From these, it follows that $\overline{f}(x) = \limsup_n f_n(x)$ and $\underline{f}(x) = \liminf_n f_n(x)$ are measurable. Then define the measurable function $\phi(x) = \overline{f}(x) - \underline{f}(x)$. Then consider the set $\phi^{-1} \{0\} \cap \smash{\overline{f}}^{-1} \mathbb{R} \cap \underline{f}^{-1} \mathbb{R}$, which is measurable.