Pointwise a.e convergence implies boundedness Let $f_n$ be a sequence of real valued functions with
$$f_n(t) \to f(t)$$
pointwise almost every $t.$ Does it follow that $|f_n(t)| < K$ for some constant $K$? I only know this fact for pointwise convergence.
 A: What if $f_n(x)=x$ whenever $x$ is an integer multiple of n, and 0 otherwise ? This seems to contradict the result even for pointwise convergence though...
A: For extended measurable functions
Just having a limit does not say anything about boundedness. In fact we may choose $f_n(x)=n$ which converges to $f(x)=+\infty$. 
However, if we know $|f_n(x)|<K$ for all $n$ and almost all $x$, then $|f(x)|\leq K$ for almost all $x$ (to see this use that a countable union of sets of measure zero has measure zero). 
Also, if we knew $|f(x)|\leq K$ for almost all $x$, then $f_n(x)=f(x) + \infty\chi_{(n,+\infty)}(x)$ converges pointwise to $f$ - so that does not help either. 
For finite measurable functions 
Consider the function  $f(x)= \frac1x$, which is unbounded at $x=0$. Define $f_n(x)=f(x)$when $|x|>1/n$ and $f_n(x)=0$ otherwise. Then $f_n\to f$ a.e. but $f_n$ is unbounded. 
A: 
Consider $f_n(t)=\frac{1}{n}$ if $|t|\le \frac{1}{n}$ and $|t|$ if $|t|>\frac{1}{n}$, Then $\langle f_n(t)\rangle$ converges pointwise to $f(t)=|t|$ but each $f_n$ is unbounded!

