Suppose $\{f_n\}$ are Lebesgue measurable functions on $[0,1]$, such that $\int_0^1 |f_n|\,d\mu=1$ for all $n$, and $f_n\to 0$ almost everywhere.
How would we show that given $\epsilon>0$, there exists a Lebesgue meausurable $E\subseteq [0,1]$ such that $\mu(E)<\epsilon$ and $$\lim_{n\to\infty}\int_E |f_n|\,d\mu=1$$?
Intuitively, this is puzzling to me, as from absolute continuity of Lebesgue Integral $\int_E |f_n|\,d\mu$ should be arbitrarily small when $\mu(E)$ becomes small, but in this case, the integral actually attains its full value.
I tried using contradiction: Suppose to the contrary there exists $\epsilon>0$ such that for all $E\subseteq [0,1]$, either $\mu(E)\geq \epsilon$ or $\lim_{n\to\infty}\int_E |f_n|\,d\mu\neq 1$. But I am stuck here.
Fatou's lemma gives $\int_0^1 |f|\,d\mu\leq\liminf\int_0^1 |f_n|\,d\mu=1$ which doesn't look too useful.
I can see that one example of $f_n$ is $f_n=n\chi_{[0,1/n]}$, where basically what happens is $f_n$ concentrates all its value in a very narrow domain.
Thanks for any help! Really stuck here.
I think you're on the right track. With Egorov you can conclude quickly, but it is a bit of overkill. You can instead try to use Fatou on the set on which the pointwise convergence occurs (or rather in a slightly smaller set)