I have recently stumbled upon the following function $f [0,1] \rightarrow \mathbb{R}$:

$$ f(x) = \begin{cases} e, & \text{if } x = 0 \\ \sum_{n=0}^y \frac{1}{n!}, & \text{if } x>0 \end{cases} $$ where $y = [\frac{1}{x}]$ is the integer part of $\frac{1}{x}$. The claim is that it is Riemann integrable but has infinitely many discontinuities. I clearly can see that it is Riemann Integrable since it is a decreasing function on the given interval $[0,1]$. However, I am not sure about the proof of being infinitely discontinuous. Here is the proof that came with this example:

Take $n \in \mathbb{N}$ and set $x_0 = \frac{1}{n}$. Then, $\lim_{x\to x_0^-} f(x) - \lim_{x\to x_0^+} f(x) = \frac{1}{n!}$

Here I understand that discontinuity follows, however, I am not sure how they obtained that the difference is $\frac{1}{n!}$.

Thank you in advance!

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    $\begingroup$ @Starfall I might have explained it not very well but what I mean here is that it is a monotonic function, i.e. strictly decreasing and so Riemann Integrable. $\endgroup$ – IVU Apr 28 '17 at 21:36
  • $\begingroup$ @IVU: Neat example! $\endgroup$ – dantopa Apr 28 '17 at 21:41
  • $\begingroup$ that should be $f(x).$ $\endgroup$ – zhw. Apr 29 '17 at 19:41
  • $\begingroup$ @zhw. Indeed, it should be, thank you $\endgroup$ – IVU Apr 30 '17 at 13:48

We are seeing the step discontinuity.

Let $x_{0}=\frac{1}{n}=\frac{1}{3}$. As we approach from the left, $$\color{blue}{2.9,\ 2.99,\ 2.999},\dots$$ The function value is $$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!}$$ As we approach from the right, $$\color{red}{ 3.1,\ 3.01,\ 3.001}$$ The function value is $$\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \boxed{\frac{1}{3!}}$$

The difference is $\frac{1}{n!} = \frac{1}{3!}$


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    $\begingroup$ Oh, this makes sense now, thank you! $\endgroup$ – IVU Apr 28 '17 at 22:15
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    $\begingroup$ You need to take $x_0 = 1/3$ and take the inverses of the other numbers as well. Other than this, +1. $\endgroup$ – Michał Miśkiewicz Apr 28 '17 at 22:34
  • $\begingroup$ @Michał Miśkiewicz: You careful reading and comment helps me and many other users. Thanks. $\endgroup$ – dantopa Apr 29 '17 at 3:58

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