Boundedness of the integral

Question

Let $f$ be a real-valued function that is Riemann-integrable in every closed, bounded subinterval of $(0, \infty)$. Suppose that the improper integral $\int_0^{\infty}f(x)dx$ exists. Then how can I prove that there exists a real number $ M > 0$ such that for all $ 0\leq a < b$, $|\int_a^{b}f(x)dx| < M$?

Answer 1

The function $F(t)=\int_0^t f(x)\,dx$ is continuous by the hypothesis that $f$ is Riemann-integrable in every closed, bounded subinterval of $(0,\infty)$. Indeed, consider some $t_0>0$ and fix $\delta>0$ such that $$I_{t_0,\delta}=[t_0-\delta,t_0+\delta]\subseteq (0,\infty)$$ Since, $f$ is Riemann-integrable on $I_{t_0,\delta}$ by definition it should be bounded also in that interval. Hence, there is some $M_1>0$ (depending on $t_0,\delta$) such that $|f(x)|\leq M_1$ for every $x\in I_{t_0,\delta}$. Then, if $|t-t_0|<\delta$ we have \begin{align} |F(t)-F(t_0)|&=\biggl| \int_t^{t_0} f(x)\,dx \biggr|\\ &\leq \int_{t}^{t_0}|f(x)|\,dx\\ &\leq M_1|t_0-t| \end{align} Hence, $\lim_{t\to t_0}|F(t)-F(t_0)|=0$ which shows that $F$ is continuous at $t_0$. Since, $F$ is continuous on $[0,\infty)$ and $\lim_{t\to \infty}F(t)$ exists you can can conclude that there exists some $M>0$ such that $$\biggl|\int_{\alpha}^\beta f(x)\,dx\biggr|\leq M$$ for every $\alpha<\beta$ (check this step).