Determine which of two functions are Riemann integrable on $[0, 1]$:
$$ f(x) = \begin{cases} \ln(x), &\text{$x \in (0,1]$}\\ 42, &\text{x = 0} \end{cases} $$
$$ g(x) = \begin{cases} \left(1 + \frac{1}{n} \right)^n, &\text{$x \in \left( \frac{1}{2^n}, \frac{1}{2^{n-1}} \right] $}\\ e, &\text{x = 0} \end{cases} $$
My thinking: As far as I know, for a function to be Riemann integrable, it must be a continuous and bounded function, and it must have no more than $|\mathbb{N}|$ breakpoints.
In the function g of the expression $\frac{1}{2^n} \rightarrow 0, \frac{1}{2^{n-1}} \rightarrow 0$ when $n\rightarrow \infty$, then if $x\in \left(\frac{1}{2^n}, \frac{1}{2^{n-1}} \right]$, then $g(x)\rightarrow e$. And when $x = 0$ then $g(x) = e$, hence the boundedness of the function $g(x)$ is visible, the continuity of $g(x)$ is obvious. So $g(x)$ will be Riemann integrable?
I don't have any good idea about $f(x)$ due to some misunderstanding of the Riemann integrability of the function. I think an explanation of how to solve this problem will help me understand this topic.
I will be glad of any kind of help related to determining whether $f(x)$ and $g(x)$ are Riemann integrable on $[0, 1]$.
For $f$, it is useful to establish whether or not the function is bounded. In particular, as we know the behavior of $f$ near $0$ on $[0,1]$, i.e. $$\underset{x \rightarrow 0^+}{\lim} f = -\infty,$$ So we can conclude it cannot be Riemann integrable as it is unbounded. For $g$, you might appeal to Lebesgue's theorem for the Riemann integral. It's important to note that your idea that integrable $\rightarrow$ no more than $|\mathbb{N}|$ breakpoints is inaccurate. This is a sufficient condition, but it not necessary.
The right characterization would be that the set of discontinuous is a measure zero set, which all countable sets are. You might use this with the piecewise continuity of $g$ to show $g$ is indeed integrable.