This is question involves a problem from one of my previous questions:
Show that the function $\dfrac{1}{\sqrt{x}}$ is Riemann integrable on $[2^{-2n} , 2^{-2n+2}]$ for any positive integer $n$.
My Work
$\int^{2^{-2n+2}}_{x = 2^{-2n}} \dfrac{1}{\sqrt{x}} = 2(2^{-n})$
Therefore, for any $n \ge 1$, the function is bounded and continuous on the interval $[2^{-2n} , 2^{-2n+2}]$. Therefore, the function is Riemann integrable on this interval.
When finding the area under the curve, we get $\sum^{\infty}_{n=1} 2(2^{-n}) = 2$. Therefore, the area $= 2$ is defined.
However, although the area is defined, apparently the function is not Riemann integrable.
I am really struggling to understand this. We found that the function $\dfrac{1}{\sqrt{x}}$ is Riemann integrable on the interval $[2^{-2n} , 2^{-2n+2}]$ for any $n \ge 1$. We also found that the area is defined for any $n \ge 1$. But apparently the function is not Riemann integrable? This seems contradictory.
I would greatly appreciate it if people could please take the time to help me understand what is going on here.
You have to distinguish between different things:
1) For any $n$, the function $x\mapsto 1/\sqrt{x}$ is Riemann integrable on $[2^{-2n},2^{-2n+2}]$ since the function is continuous on that compact interval. (That is a standard theorem in the theory of Riemann integrals.
2) On $(0,1)$ the function $x\mapsto 1/\sqrt{x}$ is not bounded, and thus, according to a theorem not Riemann integrable on that interval.
3) If $\lim_{\epsilon \to 0+}\int_\epsilon^1 f(x)\,dx$ exists then the generalized Riemann integral $\int_0^1 f(x)\,dx$ exists (and is defined to be that limit). Your function satisfies this property.