# pointwise convergence in $\sigma$-algebra

Let $(T,\mathcal{A})$ be a $\sigma$-algebra, and $(f_{n})$ a sequence of $\mathcal{A}$-measurable functions from $T$ to $\mathbb{R}$. Show that the set $$\{ t \in T: \lim_{n\rightarrow \infty}f_{n}(t)\text{ exists and is finite}\}$$ is in $\mathcal{A}$.

I try with the definition of pointwise convergence, that is, $\forall \epsilon >0$, $\exists n_{0}$ such that $|f_{n}(t)-f(t)|<\epsilon$, now $f_{n}(t) \in (f(t)-\epsilon,f(t)+\epsilon)$ and like $f_{n}$ is measurable the set is in $\mathcal{A}$. Can I use this to resolve the problem, or does there exist another way?

Any help is appreciated.

-
I've edited your post a bit, hopefully making it easier to understand, but if I accidentally changed what you wanted to say, I apologize, and please do feel free to fix it. –  Zev Chonoles May 16 '12 at 2:03

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 $\{t \in T\,:\,\lvert f_n(t)-f_m(t)\rvert \lt 1/k\} \in \mathcal{A}$.

-

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.

-
You've used this statement that "each subset of any measurable set is measurable" in the last line. But is it true? –  Fardad Pouran Mar 12 at 20:56
I never said that statement. What I used in the last line is that if a subset $X\subseteq[-\infty,\infty]$ of the extended real line is measurable, then so is $X\cap(-\infty,\infty)$, i.e. the finite part of $X$. –  Zev Chonoles Mar 12 at 21:50

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.

-
Thanks for the edit! –  copper.hat May 16 '12 at 2:24
No problem; another way of doing it is to insert a {} before the exponent, so that it doesn't line it up with the top of the box containing the overlined $f$; here's four versions: \overline{f}^{-1}, {\overline{f}}^{-1}, \overline{f}{}^{-1}, and \smash{\overline{f}}^{-1}: $\overline{f}^{-1}$, ${\overline{f}}^{-1}$, $\overline{f}{}^{-1}$, $\smash{\overline{f}}^{-1}$. –  Arturo Magidin May 16 '12 at 2:48
Thanks again. A bit sad that I have used Latex for over 25 years and still can't quite get it right! –  copper.hat May 16 '12 at 2:50
It wasn't wrong, it just looks better the other way. (-; –  Arturo Magidin May 16 '12 at 2:51