$L^1$ functions at $\infty$ to $\infty$

Question

I am having trouble with a practice prelim question:

If $f \in L^1(\mathbb{R})$ then $\lim_{n \rightarrow \infty} \int_n^{\infty} f(x)dx = 0$

I know that $f$ is bounded, but I am not if I should add and subtract by something convenient or if I should try something else.

Answer 1

Well the part $\int_{-\infty}^nf(x)dx$ must converge to $\int_{-\infty}^{\infty}f(x)dx$ so the part $\int_n^{-\infty}f(x)dx$ must go to zero. Just write the whole integral as a sum of those two parts and take the limit.

Answer 2

Actually, $f$ is not necessarily bounded. $f(x)=\frac{\chi_{(0,1]}(x)}{\sqrt{x}}$ is an unbounded (Lebesgue) integrable function (its integral is $2$).

Hint: try to argue that $\int_{-\infty}^n f(x) dx$ converges to $\int_{-\infty}^\infty f(x) dx$ by using the dominated convergence theorem. Then notice that $\int_{-\infty}^\infty f(x) dx - \int_{-\infty}^n f(x) dx = \int_n^\infty f(x) dx.$

An additional hint: a nice trick in problems involving the application of some integral convergence theorem is to make the domain of all the integrals be the entire space, by multiplying by the characteristic function of the set over which you want to integrate. So for instance $\int_n^\infty f(x) dx = \int_{-\infty}^\infty f(x) \chi_{[n,\infty)}(x) dx$.