Show existence on subsequence converging to $0$

Question

Let $f_n : [0,1] \to [0,\infty)$ be a sequence of functions such that $\int f_n \to 0$ as $n\to \infty$. Then there exists a subsequence such that $f_{n_k}(x) \to 0$ a.e as $n\to \infty$ for $0\leq x\leq 1$.

I don't know how to go about this question, any help would be appreciated.

Answer 1

Hints: Denote by $\lambda$ the Lebesgue measure on $[0,1]$.

1. Show that for any $\epsilon>0$, $$\lambda(\{x \in [0,1]; f_n(x) \geq \epsilon\}) \leq \frac{1}{\epsilon} \int f_n \, d\lambda.$$ Deduce that $$\lim_{n \to \infty} \lambda(\{x \in [0,1]; f_n(x) \geq \epsilon\})=0. \tag{1}$$
2. Choose $\epsilon = 2^{-k}$. By $(1)$, there exists $n_k$ such that $$\lambda(\{x; f_{n_k}(x) \geq 2^{-k}\}) \leq 2^{-k}.$$ Without loss of generality, $n_{k+1} \geq n_k$.
3. Show that $$\lambda(\{x; f_{n_k}(x) \geq 2^{-k} \, \text{for infinitely many $k$}\}) \leq \lim_{K \to \infty} \lambda \left( \bigcup_{k \geq K} \{x; f_{n_k}(x) \geq 2^{-k}\} \right).$$ Use the subadditivity of $\lambda$ and the choice of $n_k$ to show that the limit on the right-hand side equals $0$.
4. By step 3, $$\lambda(\{x; |f_{n_k}(x)| \leq 2^{-k} \, \, \text{for $k$ sufficiently large}\}) = 1.$$ Conclude.