# convergence of integrand in infinite integral

If infinite integral $\int_a^{+\infty}|f(x)|dx$ converges, i.e. $\int_a^{+\infty}f(x)dx$ absolutely converges, then $f$ is bounded on $[a,+\infty)$?

Thanks a lot.

Consider $$f(x) = \begin{cases} \frac{1}{2\sqrt{x}} & x \in (0,1) \\ 0 & \text{else} \end{cases}$$
Then $$\int_0^\infty f(x) dx = \int_0^1 \frac{1}{2\sqrt{x}} dx = 1$$ but $$\lim_{x\to 0} f(x) = +\infty$$
so $f$ is not bounded on $[0,\infty)$
• If $f$ is continuous on $[a,+\infty)$? – James Chan Feb 14 '17 at 17:10
• Then your statement holds due to the necessary condition $\lim_{x\to\infty} |f(x)| = 0$ hence $|f| \le \varepsilon$ for each $\varepsilon > 0$ and an arbitrage large $M>0$. And $f$ is bounded on the compact interval $[a,M]$ because it's continuous, so it holds $$|f| \le \max\{\max_{x\in [a,M]} |f(x)|,\varepsilon\}$$ – Gono Feb 15 '17 at 8:16