Consider $f_n=n\chi_{[0,\frac{1}{n}]}$, I want to show that $f_n \to 0$ a.e. but $f_n \not\to 0$ in $L^1$.
I guess just by examining the sequence, I can see the only point where $f_n \not\to 0$ is when $x=0$. And $\{0\}$ has measure $0$. Hence, $f_n\to 0$ a.e.. But I have to write this down formally.
To do that I was thinking, for all $\delta>0$ then there exists $N(\delta)>0$ such that $m(\bigcap^k\{[0,\frac{1}{n}]\})<\delta$ for all $k>N$. Let $\epsilon>0$, then pick $N=N(\delta<\epsilon)$ then it is clear that $f_n(x)=0<\epsilon$ when $x\notin \bigcap^k\{[0,\frac{1}{n}]$ and else we have $m(\bigcap^k\{[0,\frac{1}{n}])<\epsilon$. Take epsilon to 0, and we have $x$ only diverge on a set of measure 0.
On the other hand, clearly $\int f_n = 1\not\to 0$ in $L^1$.
Is this correct? Thanks
Your proof for $f_n \to 0$ a.e. is a little overcomplicated.
Clearly, nothing of note happens outside of $[0,1]$, since $f_n \equiv 0$ there.
Consider: $x \in (0,1]$ has $f_n(x) = 0$ when? Well, it is clearly when $x \not \in [0,1/n]$, i.e. when $x > 1/n$, and hence when $n > 1/x$.
Thus, given $x \in (0,1]$, choose $N := \lceil 1/x \rceil$ in the definition of sequence convergence. The only point in $[0,1]$ you cannot contend with like this is $0$, but that is fine for a.e. convergence.