Lebesgue Integral that does not go to zero even when domain of integration is arbitrarily small

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.

• You are dealing with a sequence of $f_n$ that converges to a $\delta$ function which is not absolutely continuous. Commented Oct 27, 2016 at 7:20
• Thanks. Another line of thought I am working on now is using convergence in measure. Commented Oct 27, 2016 at 7:25

• @yoyostein You're welcome. If you don't want to use Egorov, you can consider the negligible set $E$ outside of which the pointwise convergence occurs, take a set $\tilde E\supset E$ of measure $\epsilon$ and apply Fatou on $\tilde E^c$.