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)
$$
G'(0) = \frac{1}{2}
$$
 A: 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 $G'(0) = 2/3$.
