Suppose $\{f_n\} \subset L^2(\mathbb{R})$ is a sequence of functions converging to $0$ in the $L^2$ norm. I.e.,
$$\bigg(\int_{\mathbb{R}} \vert f_n(x) \vert^2\bigg)^{\frac{1}{2}} \rightarrow 0; \space \text{as $n$ tends to $\infty$}$$
Prove there exists a convergent subsequence converging to $0$ a.e.
My thoughts, since $L^2$ is a metric space, use the triangle inequality on $0,f_n,f_{n_k}$? But how to contstruct the $f_{n_k}$? Sorry if this is a duplicate, I couldn't find it anywhere.
We recall a few facts:
We say $f_n \xrightarrow[]{\text{measure}} f$ if $$\lambda(|f_n - f| > \epsilon) \xrightarrow[]{n \rightarrow \infty}0,$$ where $\lambda$ is Lebesgue measure.
Convergence in measure implies convergence almost everywhere of a subsequence
The goal is to show that $f_n \xrightarrow[]{\|\cdot\|_2} 0$ implies that $f_n \xrightarrow[]{\text{measure}}0$, where $$\|f\|_2 := \left( \int_\mathbb{R} |f|^2 d\lambda \right)^{1/2}.$$ The result follows immediately once we have this.
Notice that $f_n \xrightarrow[]{\|\cdot\|_2} 0$ implies that $$\lim_{n \rightarrow \infty} \int_{\mathbb{R}} |f_n|^2 d\lambda = 0.$$ Let $A(n,\epsilon) := \{x \in \mathbb{R} \ | \ |f_n(x)| > \epsilon\}$. Then we have
$$ \int_{\mathbb{R}} |f_n|^2 d\lambda \geq \int_{A(n,\epsilon)} |f_n|^2 d\lambda > \int_{A(n,\epsilon)} \epsilon^2 d\lambda = \epsilon^2 \lambda(A(n,\epsilon)) = \epsilon^2 \lambda( |f_n| > \epsilon),$$
so
$$\lambda(|f_n| > \epsilon) < \frac{\|f_n\|_2^2}{\epsilon^2}.$$
Now take $n \rightarrow \infty$ and we have convergence in measure.
