$\lim_{n\to \infty} \int_n^{n+1}f(x)~dx=0$ if $f\in L^2(\Bbb R)$

Question

Let $f$ be a real-valued funtion belonging to $L^2(\Bbb R)$, where $\Bbb R$ is equipped with the Lebesgue measure. I am trying to show that $$\lim_{n\to \infty} \int_n^{n+1}f(x)~dx=0$$

My attempt: Since $C_c(\Bbb R)$ is dense in $L^2(\Bbb R)$ (where $C_c(\Bbb R)$ is the real-valued collection of all continuous functions on $\Bbb R$ with compact support), there is a sequence $(f_n)$ in $C_c(\Bbb R)$ converging to $f$ in $L^2$. Thus there is a subsequence $(f_{n_k})$ such that $f_{n_k}$ converges to $f$ pointwise a.e. on $\Bbb R$. Since each $f_{n_k}$ has compact support, I tried to use this to deduce the result, but I got stuck. Any hints?

Answer 1

$|\int_n^{n+1} f(x)dx| \leq \sqrt {\int_n^{n+1} |f(x)|^{2}dx} \to 0$ by DCT.

[$I_{(n,n+1)} \to 0$ at each point. $|f|^{2}$ is your dominating integrable function].