# Determine which of two functions are Riemann integrable

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 $$[0, 1]$$.

• btw, it’s not true that it must have no more than $|\Bbb{N}|$ many break points. It suffices, but is not necessary. Regarding $f$, is it bounded? Feb 22 at 15:25
• $f(x)$ at $x\rightarrow 0$ goes to $-\infty$, that is, this function is in no way bounded from below, but we have defined $f(x)$ at point 0, $f(0) = 42$, and actually this is the problem for me, $x \rightarrow 0, f(x) \rightarrow - \infty$, and at the point $x = 0, f(x) = 42 > 0$ Feb 22 at 15:30
• It doesn’t matter what $f(0)$ is. You’ve already correctly explained that $f$ is not bounded bleow. Since it is not bounded below, it is not bounded, so not Riemann-integrable, simply by definition. (One can ask whether it is improperly Riemann-integrable, or Lebesgue integrable, but these are completely separate questions). Feb 22 at 15:32
• thanks for the explanation, I just needed to understand about the point $x = 0$. and what about $g(x)$, I know I have not explained in detail why it is integrable according to Riemann, but in general, are my thoughts about the integrability of $g(x)$ correct or not? Feb 22 at 15:36
• yes, I would clean up the wording a little, but you have the right idea (particularly when you say “the continuity of $g(x)$ is visible”… but really you mean that “piecewise continuity of $g$ is clear”) Feb 22 at 15:37

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.