Prove that ${\lim\limits_{n \to \infty}{\left(\int\limits_{n}^\infty |f(x)|dx + \int\limits_{-\infty}^{-n} |f(x)|dx \right)}= 0}$

Question

Let $f \in L^1(\mathbb{R})$.

Prove that $\displaystyle{\lim_{n \to \infty}\displaystyle{\left(\int_{n}^{+\infty} |f(x)|dx + \int_{-\infty}^{-n} |f(x)|dx \right)}= 0}$

My Attempt:

For the first integral:

Let $f_n = 1_{[n,\infty)}f$ be a sequence of measurable functions dominated by $f$.

Then we say $f_n$ converges pointwise to $0$, for every $x$ and by applying Dominated Convergence Theorem we have:

$\displaystyle\lim_{n \to \infty}\left(\int_{n}^{+\infty} |f(x)| dx\right) = \displaystyle\lim_{n \to \infty}\left(\int_{\mathbb{R}}|f_n(x)| dx\right) = \displaystyle\lim_{n \to \infty}\left(\int_{\mathbb{R}}|f_n(x) - 0(x)| dx\right) = 0$

Now for the second integral, I'm quite stuck on how to prove it since the bounds of the integration are negative, should I use the same sequence of measurable functions on the first integral to prove the second integral?

Exactly the same argument with $f_n$ replaced by $f1_{(-\infty,-n]}$ — Kavi Rama Murthy, Jul 07 '21 at 05:59
This question seems related: $\int_{\mathbb R} |f(x)| dx < \infty \implies \int^{\infty}_{n} |f(x)| dx \to 0$ as $n \to \infty$? — Martin Sleziak, Jul 07 '21 at 06:01
@KaviRamaMurthy thanks, it kept for a while if I'm gonna use similar arguments since it might be wrong but thank you for further clarification. — Dencio, Jul 07 '21 at 06:13
Change $f_n$ to $1_{B_n}$ where $B_n = (-\infty,n]\cup [n,\infty)$, your proof works in that case without any changes. — Sam, Jul 08 '21 at 03:40

Prove that ${\lim\limits_{n \to \infty}{\left(\int\limits_{n}^\infty |f(x)|dx + \int\limits_{-\infty}^{-n} |f(x)|dx \right)}= 0}$

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