Show that for any integrable function...

Since $f \in L^1$, $$ \begin{align*} \int_{\mathbb{R}} |f(x)| \ dx &= \sum_{n \in \mathbb{Z}}\int_n^{n+1}|f(x)| \ dx < \infty \end{align*} $$ Since the series converges, we must have $$ \lim_{n\to\infty} \int_n^{n+1}|f(x)| = 0 $$

Note that $\int_{0}^{\infty}|f(x)|dx<\infty$ and by monotone convergence theorem, $\int_{0}^{n}|f(x)|dx\uparrow\int_{0}^{\infty}|f(x)|dx$. Now we use the simple Cauchy criterion:

\begin{align*} \left|\int_{0}^{m}|f(x)|dx-\int_{0}^{n}|f(x)|dx\right|<\varepsilon \end{align*} for $m,n\geq N$. Now pick $m=n+1$ and we have \begin{align*} \left|\int_{n}^{n+1}f(x)dx\right|\leq\int_{n}^{n+1}|f(x)|dx<\varepsilon,\quad n\geq N. \end{align*}