Prove that $\lim_{n\to\infty} \int_{E_n}f =0$ Let $f\in L^1(E)$ and $(E_n)$ be a sequence of measurable subsets of $E$ such that $\lim_{n\to\infty} m(E_n)=0$. Prove that $\lim_{n\to\infty} \int_{E_n}f =0$. 
I know this can be done using an $\varepsilon - \delta$ lemma, but Im looking for an answer using convergence theorems or something. If you've got something, please let me know.
 A: I would be tempting to apply the dominated convergence theorem, but the sequence $\left(f\mathbf 1_{E_n}\right)_{n\geqslant 1}$ may not converge to $0$ almost everywhere. For example, if $2^N\leqslant n\lt 2^{N+1}$, define $E_n=\left[\left(n-2^N\right)  2^{-N},\left(n+1-2^N\right)  2^{-N}   \right)  $. Its Lebesgue measure is $2^{-n}$ but $\left| f\right| \sup_{2^N\leqslant n\lt 2^{N+1}} \mathbf 1_{E_n}=\left| f\right|$. 
If the series $\sum_{n=1}^{+\infty}m\left(E_n\right)$ was convergent, then we would have the almost sure convergence to $0$. This issue can be solved using subsequences.              
Suppose that $\left(\int_{E_n}   f\mathbf 1_{E_n}\mathrm dm\right)_{n\geqslant 1}$ does not converge to $0$: there exists a positive $\delta$ and an increasing sequence of positive integers $\left(n_j\right)_{j=1}^{+\infty}$ such that for each $j\geqslant 1$, $\left|\int_{E_{n_j}} f\mathrm dm\right|\gt\delta$. Now, there exists an increasing sequence of integers $\left(l_i\right)_{i\geqslant 1}$ such that for each $i$,   $m\left(E_{l_i}\right)\leqslant i^{-2}$ and $\left\{l_i,i\geqslant 1\right\}\subset \left\{n_j,j\geqslant 1\right\}$. Indeed, define $l_1$ such that $l_1=n_{j_0}$ for some $j_0$ and $m\left(E_{n_{j_0}}\right)\lt 1$ and
$$l_{i+1} :=\inf\left\{n_j\mid n_j\gt l_i\mbox{ and }m\left(E_{l_{ i+1} }\right)\leqslant \left(i+1\right)^{-2}    \right\}.$$
Let $A_i:=E_{l_i}$. Since   $\left\{l_i,i\geqslant 1\right\}\subset \left\{n_j,j\geqslant 1\right\}$, we have for each $i\geqslant 1$, 
$$\left|\int_{A_i} f\mathrm dm\right|\gt\delta\mbox{ and }m\left(A_i\right)\lt i^{-2}.$$
The sequence $\left(f\mathbf 1_{A_i }\right)_{i\geqslant 1}$ converges to $0$ pointwise and the dominated convergence theorem yields a contradiction.      
A: Suppose $f=I_A$ where $A$ is a measurable set. Then it is trivial that
$$0\leq\int_{E_n} f = m(A\cap E_n)\leq m(E_n)$$
so $\lim_n \int_{E_n} f =0$.
Now extend this to simple functions, and next to every $f\geq 0$ using the monotonous convergence theorem:
Let $f=\lim_n f_n$, where $f_n$ are simple functions. We know that, for each $n\in \mathbb N$, $\lim_m \int_{E_m} f_n=0$. Now, using the monotonous convergence theorem:
$$\int_{E_m} f=\int_{E_m} \lim_n f_n = \lim_n \int_{E_m} f_n$$
and now $$\lim_m \int_{E_m} f = \lim_n \lim_m \int_{E_m} f_n=0$$
Last, for $f\in L^1$, write $f=f^+-f^-$.
