Prove that, if $\mu(\Omega)<\infty$, $(f_n)\subset L^1(\Omega,\mu)$ and $\int_{\Omega}|f_n|d\mu\leq M<\infty$, then $$\left[(\forall\,\varepsilon>0)(\exists\,\delta>0):(\mu(E)<\delta\rightarrow \sup_n\int_Ef_n\,d\mu<\varepsilon)\right] \Leftrightarrow \left[\lim_{k\to\infty}\sup_n\mu(x:|f_n(x)|\geq k) = 0\right]$$
I already prove $(\Rightarrow)$, but for the other implication I get stuck. I used that if $F = \{x:|f_n(x)|\geq k\}$ $$\int_Ef_n\,d\mu = \int_{E\cap F}f_n\,d\mu + \int_{E\cap F^c}f_n\,d\mu$$ the second integral on the right is bounded by $\mu(E)k$, so we can work it as $\mu(E)<\delta$. But I can't bound the other integral, because the hypothesis works for $\mu(F)$ and the $f_n$ on the integral is giving me problems. How can I work out that integral?
Consider the unit interval with Lebesgue measure and $f_n(x):=n\chi_{(0,n^{—1})}$. We have $\lVert f_n\rVert_{L^1}=1$ and $$\mu\{x:|f_n(x)|\geqslant k\}=\begin{cases}1/n &\mbox{ if }k\leqslant n;\\ 0&\mbox{ otherwise};\end{cases}$$ so that $\sup_n\mu\{x:|f_n(x)|\geqslant k\}=\frac 1k\to 0$. But the other property is not satisfied because for $\varepsilon\lt 1/2$, there is no $\delta$ such that $\sup_n\int_{(0,\delta)}|f_n|\mathrm d\mu\lt \varepsilon$ (consider $n\geqslant \frac 1{\delta}$).