Let us consider the function on $\mathbb{R}$ as
$$f(x)= \begin{cases} \frac {1}{\sqrt{x}} & \text{if } 0<x<1 \\ 0 & \text{otherwise} \end{cases}.$$
Then show that the function is integrable on $\mathbb{R}$.
Is this asking about Riemann Integrability of $f$?
$0$ is a point of infinite discontinuity of $f$, but $f$ is bounded and integrable on $[0+\delta,1]$ for all $\delta$ satisfying $0< \delta <1$. $f$ is unbounded in a nbd of $0$, so it can't be Riemann integrable. By the way the integral is improper and convergent and
$$ \int_{0}^{1} \dfrac{1}{\sqrt{x}}=2.$$.
My question is, what is the correct conclusion of the question? Please help me.