# Discontinuous function that admits antiderivatives.

Given the function $g$ defined on $[-1, 1]$ with real values, having a plot as depicted in the image, can you prove that $g$ has antiderivatives on $[-1, 1]$? All the triangles are isosceles and are built with their bases being the intervals $[1/(n+1), 1/n]$ for each positive integer $n$, or $[-(1/n), -(1/(n+1))]$, and with height 1. Also, note that $g(0)=1/2$.

It shouldn't be hard to show that $g$ has Darboux's intermediate value property. Yet i cannot find any approach towards proving that g has indeed antiderivatives. A sum of functions that allow for antiderivatives? Some other aproach, using Fourier? The plot of the function, sketched roughly by me:

status April 7 2016
According to the comment: Define $$G(x) = \int_{-1}^x g(t)\;dt .$$ Since $g$ is bounded and measurable, it is Lebesgue integrable. From the standard texts on Lebesgue integral (or easily proved), $G'(x) = g(x)$ at any point where $g$ is continuous. So the only question remaining is:

Prove (or disprove) $$G'(0) = \frac{1}{2}$$

• Candidate for antiderivative should be $G(x) := \int_{-1}^x g(t)\,dt$. So you need to show $g$ is (imporperly) Riemann integrable (or Lebesgue integrable), and that $G'(0)=1/2$. The rest, that $G'(x)=g(x)$ at points where $g$ is continuous, should be standard. Nov 17, 2015 at 22:48
• I tried to find a definite integral of this function, since areas below the function are quite easy to express. Concluding, i reached a sort of.. G(1/k)=(1+k)/2k. What do i do when x is between 1/(k+1) and 1/k? Nov 17, 2015 at 23:03
• How can I respark interest in this question? Apr 7, 2016 at 11:51
• I edited it to show the status. That may respark interest. Apr 7, 2016 at 14:30

We claim first that $$\lim_{h \to 0+} \frac{G(h)-G(0)}{h} = \frac{1}{2} \tag{1}$$ The left-hand limit will be done in the same way, and that will show that $G'(0) = \frac{1}{2}$.

Note that $(1)$ may be written $$\lim_{h \to 0+} \frac{1}{h} \int_0^h g(t)\;dt = \frac{1}{2} ,$$ which is what we will prove.

(a) If $n$ is a positive integer, then $$\int_{1/(n+1)}^{1/n} g(t)\;dt = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+1}\right) \tag{2}$$ since it is the area of a triangle.

(b) If $n$ is a positive integer, then $$\int_0^{1/n} g(t)\;dt = \frac{1}{2}\cdot \frac{1}{n} .$$ This is proved using (2) like this: $$\int_0^{1/n} g(t)\,dt = \sum_{k=n}^\infty \int_{1/(k+1)}^{1/k} g(t)\;dt =\frac{1}{2} \sum_{k=n}^\infty \left(\frac{1}{k}-\frac{1}{k+1}\right) =\frac{1}{2}\cdot\frac{1}{n} .$$

(c) If $\frac{1}{n+1} \le x < \frac{1}{n}$, then (since $g(t) \ge 0$) $$\int_0^{1/(n+1)} g(t)\;dt \le \int_0^x g(t)\;dt \le \int_0^{1/n} g(t)\;dt \\ \frac{1}{2(n+1)}\;dt \le \int_0^x g(t)\;dt \le \frac{1}{2n} \\ \frac{1}{2(n+1)x}\;dt \le \frac{1}{x}\int_0^x g(t)\;dt \le \frac{1}{2nx} \\ \frac{n}{2(n+1)}\;dt \le \frac{1}{x}\int_0^x g(t)\;dt \le \frac{n+1}{2n}$$ and as $x \to 0+$ we have $n \to \infty$, and the limit is squeezed to $$\lim_{x \to 0+}\frac{1}{x}\int_0^x g(t)\;dt = \frac{1}{2} .$$

remark
Note that if we use parabolas instead of triangles, like this:

then we will get $G'(0) = 2/3$.

• Thanks a lot Edgar. That is complete. Apr 8, 2016 at 19:59