$\int_{\mathbb R} |f(x)| dx < \infty \implies \int^{\infty}_{n} |f(x)| dx \to 0$ as $n \to \infty$?

Question

Suppose that $\int_{\mathbb R} |f(x)| dx < \infty$.

Can we say that $\int^{\infty}_{n} |f(x)| dx \to 0$ as $n \to \infty$ If yes, how to justify?

Answer 1

Yes.

Let $f_n=1_{[n;\infty)}f$.

Then $f_n$ is a sequence of measurable functions dominated by $f$.

The pointwise limit of $f_n(x)$ is $0$ for every $x$ and thus, by the dominated convergence theorem

$\lim_{n\to\infty}\int_n^\infty|f(x)|dx=\lim_{n\to\infty}\int_\mathbb{R} |f_n(x)|dx=\lim_{n\to\infty}\int_\mathbb{R} |f_n(x)-0(x)|dx=0$.

QED.

Answer 2

Yes.

$$\int_n^\infty \lvert f\rvert\,dx=\int_{\Bbb R} 1_{[n,\infty)}\lvert f\rvert\,dx$$

Since $1_{[n.\infty)}\lvert f\rvert\le\lvert f\rvert$ and $1_{[n,\infty)}\lvert f\rvert\stackrel{n\to\infty}\longrightarrow 0\ $ pointwise, you can use dominated convergence theorem.

Answer 3

Note $$ \int^{\infty}_0|f(x)|dx\le\int_{-\infty}^0|f(x)|dx+\int^{\infty}_0|f(x)|dx=\int_{\mathbb{R}}|f(x)|dx<\infty$$ and hence for $\forall \varepsilon>0$, there is $N\in\mathbb{N}$ such that, for $\forall m, n\ge N$ ($m>n$), $$ \int_n^m|f(x)|dx<\varepsilon$$ Letting $m\to\infty$ gives $$ \int_n^\infty|f(x)|dx\le\varepsilon$$ namely $$ \lim_{n\to\infty}\int_n^\infty|f(x)|dx=0.$$