Let $[0,1]=\cup A_n$ and assume that $\lambda(A_i\cap E_j) \to 0$ as $j\to \infty$ for all $i$. Show that $\lambda(E_j)\to 0$ ($E_j\subseteq [0,1]$) Let $[0,1]=\cup A_n$ and assume that $\lambda(A_i\cap E_j) \to 0$ as $j\to \infty$ for all $i$. Show that $\lambda(E_j)\to 0$ ($E_j\subseteq [0,1]$)
My proof: Assume towards contradiction that is not the case. Then there is a subsequence s.t $\lambda(E_{n_k})>\epsilon_0$ for some $\epsilon_0>0$. Now we can select a sequence s.t $\lambda(A_i\cap E_i)<\frac{\epsilon_0}{2^i}$ for each $i$ where $E_i$ is an element of our subsequence. This implies that $\lambda(\cup (A_i \cap E_i))<\sum\frac{\epsilon_0}{2^i}=\epsilon$ so $\lambda(\cup E_i)<\epsilon_0$ which is a contradiction.
Is this correct? Is there a better cleaner proof?
 A: The step "we can select a sequence s.t $\lambda(A_i\cap E_i)<\frac{\epsilon_0}{2^i}$ for each $i$" has to be made more precise. It is not clear a priori that $\lambda(A_{n_k}\cap E_{n_k})$ goes to zero. However, we know that for all fixed $i$, $\lambda(A_{i}\cap E_{n_k})\to 0$ as $k$ goes to infinity. We can choose $k_i$ such that $\lambda(A_{i}\cap E_{n_{k_i}})\leqslant 2^{-i-1}\epsilon_0$. Then 
$$
\epsilon_0\leqslant \lambda\left(\bigcup_{i\geqslant 1}E_{n_{k_i}}\right)
\leqslant \lambda\left(\bigcup_{i\geqslant 1}\left(E_{n_{k_i}}\cap A_i\right)\right)
+\lambda\left(\left(\bigcup_{i\geqslant 1}A_i\right)^c\right)\leqslant 2^{-1}\epsilon_0,$$
a contradiction. 
Alternatively, suppose without loss of generality that $(A_i)$ is pairwise disjoint (replace $A_i$ by $A_i\setminus \bigcap_{k=1}^{i-1}A_k^c$ if necessary). Let $c_{i,j}:=\lambda\left(A_i\cap E_j\right)$. We know that for all $i$, $c_{i,j}\to 0$ and $\sup_{j\geqslant 1}\lvert c_{i,j}\rvert\leqslant \lambda(A_i)$ hence $\sum_{i\geqslant 1}\sup_{j\geqslant 1}\lvert c_{i,j}\rvert$ is summable. Apply the dominated convergence theorem to get that $\sum_{i\geqslant 1}c_{i,j}\to 0$, which translates (since the $A_i$ are disjoint) into $\lambda(E_j)\to 0$.
