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

Solution 1:

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.

Solution 2:

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 $$ \lim_{n\to\infty}\int_n^\infty|f(x)|dx=0.$$