Use dominated convergence theorem to show: $\lim_{\ n \to \infty} \| f \ 1_{[n,n+1]} \|_1 = 0$

Question

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.

Answer 1

Solution is correct but looks overcomplicated. Here is how I would write it.

You must show $$\int |f|I_{[n,n+1]}d \lambda \to 0$$

But you have $|f|I_{[n,n+1]} \to 0$ pointwise, $|f|I_{[n,n+1]} \leq |f|$ for all $n \geq 1$ and $\int |f| d \lambda < \infty$, so by the dominated convergence theorem we can conclude.

Answer 2

Your solution is correct and you mentioned all the important arguments, I see no problem here.

Answer 3

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