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 $L^{1}$?

Use Lebesgue's Dominated Convergence theorem for $f_n = f\cdot1_{[n,n+1]}$ Indeed $f_n\to 0$ pointwise and $|f_n(x)|\le |f(x)|$ with $\int_\mathbb{R}|f(x)|dx<\infty$, so: $$\int_n^{n+1}f(x)\,dx=\int_\mathbb{R} f_n(x)\ dx\to \int_\mathbb{R} 0\,dx=0$$
