Are the functions $f: (0, 1) \rightarrow \mathbb{R}$ with $f(x) = \frac{1}{x^2}$ for all $x \in (0, 1)$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ with $g(x) = 1$ for all $x \in \mathbb{R}$ Riemann integrable?
I think $g$ is trivially true, since $g(x) = 1$ for any $x$, and thus the left sum will always be equal to the right sum.
But, I don't quite know how I should approach the case for $f$. Since $(0, 1)$ is not closed, it means it's not compact. From what I understand, Riemann integrals are defined only for compact sets. Does this mean $f$ is not integrable?
I don't think I'm quite understanding Riemann integration well, so any help would be appreciated. Thank you.
Apparently the text is addressing about improper Riemann integrals.
$f$ is not improper Riemann integral, as can be seen by $\lim_{n\rightarrow\infty}\displaystyle\int_{1/n}^{1/2}f(x)dx=\infty$.
$g$ is not improper Riemann integral, as can be seen by $\lim_{n\rightarrow\infty}\displaystyle\int_{0}^{n}g(x)dx=\infty$.