Use dominated convergence theorem to show: $\lim_{\ n \to \infty} \| f \ 1_{[n,n+1]} \|_1 = 0$ Consider the Lebesgue measure $\lambda$ on $\mathbb{R}$.
Given that $f: \mathbb{R} \to \mathbb{R} $ is integrable,
Ik would like to show the following using the dominant convergence theorem:
$$\lim_{\ n \to \infty} \| f \ 1_{[n,n+1]} \|_1 = 0$$
I started as follows:
$$\begin{align}
\| f \ 1_{[n,n+1]} \|_1 &= \int_\mathbb{R} |f \ 1_{[n,n+1]}| \, d\lambda \\
                        &= \int_\mathbb{R} |f \ 1_{[n,n+1]}| \, d\lambda 
\end{align}$$
$f_n = |f \ 1_{[n,n+1]}|$ is dominated by  $|f|$, because $|f \ 1_{[n,n+1]}| \leq |f| $ for all $x \in \mathbb{R}$ and |f| is integrable, because if $f$ is integrable, then absolute value of $f$ is also. We can apply dominant convergence:
$$\begin{align}
\lim_{\ n \to \infty} \| f \ 1_{[n,n+1]} \|_1 &=\lim_{\ n \to \infty}\int_\mathbb{R} |f \ 1_{[n,n+1]}| \, d\lambda \\
                        &= \int_\mathbb{R} \lim_{\ n \to \infty}|f \ 1_{[n,n+1]}| \, d\lambda \\
                        &= \int_\mathbb{R} 0 \, d\lambda \\
                        &= 0
\end{align}$$
But I feel like I am doing something wrong.

$$
\int\limits_{\mathbb R} f\,d\lambda = \sum_{n\,\in\,\mathbb Z} \,\,\, \int\limits_{\mathbb R} f\mathbf 1_{[n,n+1]} \, d\lambda
$$
The series converges absolutely because $\|f\|_1<\infty.$ The terms of a convergent series approach $0.$
