This question was proposed as part of a test for PhD applicants but considered too hard and rejected. I tried unsucessfully to solve it for quite some time. For anyone wishing to try his luck..
Let $f:[0,1]\rightarrow \Bbb{R}$ be a measurable, almost everywhere strictly positive function. Also let $(A_n)$ be a sequence of countable subsets of $[0,1]$ such that for $n\rightarrow\infty$ we have $$\int_{A_n}f(x)dx\rightarrow0 $$
Show that for $n\rightarrow\infty$ we also have $$\lambda(A_n)\rightarrow0$$
Fix $\epsilon > 0$. For each $k \in \mathbb N$ define $E_k = \{x : f(x) > 1/k\}$. Then $E_1 \subset E_2 \subset \cdots$ and the union of the $E_k$'s is the whole interval $[0,1]$. Thus $\lambda(E_k) \to \lambda([0,1])$ so there exists an index $k$ satisfying $\lambda(E_k) > 1 - \epsilon$ and thus $\lambda([0,1] \setminus E_k) < \epsilon$.
For any index $n$ you get $$\lambda (A_n) = \lambda(A_n \cap E_k) + \lambda(A_n \setminus E_k)$$ where $$\lambda(A_n \cap E_k) = \int_{A_n \cap E_k} \, dx \le k \int_{A_n} f(x) \, dx$$ and $$ \lambda(A_n \setminus E_k) \le \lambda([0,1] \setminus E_k) < \epsilon.$$ Since $\displaystyle \int_{A_n} f(x) \, dx \to 0$ you get $\limsup \lambda(A_n) \le \epsilon.$ Since $\epsilon > 0$ is arbitrary you get $\lambda(A_n) \to 0.$