Find a measurable set A$\subset(-1,1)$ such that $f(x)=m(A\cap(-1,x))$ has $f'(0)=1/2$ Now, I have done some work on this and by using the definition of the derivative, I come to the conclusion that if such a set $A$ exists, then I should have
\begin{align*} \lim\limits_{x\rightarrow0^{+}}\frac{m(A\cap(0,x))}{x} = \frac{1}{2} = \lim\limits_{x\rightarrow0^{-}}\frac{-m(A\cap(x,0))}{x} 
\end{align*}
But now, I tried to create such an $A$ in the following way: $A = \bigcup \limits_{n=1}^{\infty} (\frac{1}{2^n}, \frac{3}{2^{n+1}})$. The idea was, that if I took an interval of measure $\frac{1}{2^{n+1}}$ in each interval $(\frac{1}{2^n}, \frac{1}{2^{n+1}})$, then eventually I could have 'half' of $(0,x)$ in each $A\cap(0,x)$ set, as the limit requires. Note that I'm working only for the part $x>0$, if I manage this I think that by symmetry it will work out in the $x<0$ part too. But I cannot prove that for this $A$ the limit is indeed $\frac{1}{2}$, so I'm not sure I'm on the right path.

Divide the interval $[2^{-k-1}, 2^{-k}]$ into $N(k)$ subintervals which are alternately included in $A$ or excluded from it.  Then the difference quotient $(f(x) - f(0))/x$ for $x$ in the first subinterval will vary between $1/2$ and $\frac{1/2 + 1/N(k)}{1 + 1/N(k)}$.  So as long as $N(k) \rightarrow \infty$ the limit will exist at $0$.
