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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 $[0, 1]$.

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    $\begingroup$ 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? $\endgroup$
    – peek-a-boo
    Feb 22 at 15:25
  • $\begingroup$ $f(x)$ 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 $x = 0, f(x) = 42 > 0$ $\endgroup$ Feb 22 at 15:30
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    $\begingroup$ 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). $\endgroup$
    – peek-a-boo
    Feb 22 at 15:32
  • $\begingroup$ 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? $\endgroup$ Feb 22 at 15:36
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    $\begingroup$ 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”) $\endgroup$
    – peek-a-boo
    Feb 22 at 15:37

1 Answer 1

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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.

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