The sum of Darboux is a Darboux function? I was thinking last days about the following problem - the sum of Darboux is a Darboux function?
Do You know a proof or counter example?
 A: No this is not true, as darboux functions are really a general class of functions, you can proof that you can write any (really any) real valued function as the sum of two Darboux functions, look here
The example of such a function from the link is, with 
$$\omega(x)=\limsup_{n\to \infty} \frac{1}{n} \sum_{i=1}^n a_i$$  where $x\in(0,1)$ has the dyadic expansion $x=0.a_1 a_2 \dots$ 
$\omega$ takes every value between $0$ and $1$ in any subinterval of $(0,1)$ and has the intermediaty value property. 
Now we denote
$$g(x)=\begin{cases}
0 & \omega (x)=x\\
\omega(x) & \text{else}\\
\end{cases}$$
This function still hase the intermediate value property but $h(x)=g(x)-x$ doesn't have the intermediate value property.
The proof is here on page 6 theorem 4.1
A: There's a simpler example. Let $g,h: \mathbb{R} \to \mathbb{R}$ be given by
$$g(x)=\begin{cases}
{\sin x^{-1}} & x \neq 0 \\
1 & x=0\\
\end{cases}$$
$$h(x)=\begin{cases}
{-\sin x^{-1}} & x \neq 0 \\
0 & x=0\\
\end{cases}$$
These are Darboux. However, 
$$g(x) + h(x) =\begin{cases}
0 & x \neq 0 \\
1 & x=0\\
\end{cases}$$
is not. 
