Does $\int_0^0 |f(x)|\mathrm{d}(x)=0$ hold for this improper integral?

Question

Let $f$ be Riemann integrable on $[\epsilon,b)$ for each $0<\epsilon<b$, and such that $\int_0^{b} |f(x)|\mathrm{d}x<\infty$. Does it follow that $\int_0^{b}|f(x)|\mathrm{d}(x)\to 0$ as $b\to 0$? It seems obvious for several reasons, one of which is because the length of the interval over which we integrate tends to $0$, but I'm not sure how to deal with the possibility that $f$ blows up near 0?

Answer 1

Let $g(t)=\int_t^{c} |f(x)|dx$ where $c$ an fixed positive number. It is given that $L=\lim_{t \to 0+} g(t)$ exists and is finite. From this it follows that $g(t)-g(s) \to 0$ as $t, s \to 0+$. This also implies that $g(t)-L \to 0$ as $ t \to 0+$. But $\int_0^{b}|f(x)|dx$ is nothing but $g(b)-L$. Hence this integral tends to $0$.