Show that for any integrable function... Show that for any integrable function $f$ on $\mathbb{R}$ we have   $$\lim_{n\to\infty}\int_{n}^{n+1}f(x) dx = 0$$
What’s the way to solve it. Can I use Lebesgue dominated convergence theorem? Or I have to use the idea that the step functions are dense in $L^{1}$?

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*}
