Suppose that $\mu$ is a signed finite Borel measure which is singular with respect to the Lebesgue measure in $[0,1]$. Is it true that there always exist a point $x\in [0,1]$ such that \begin{equation*} \lim_{\delta \to 0^+} \frac{\mu([x-\delta,x+\delta])}{\delta } = + \infty \quad \text{or} \quad -\infty \quad ? \end{equation*}
In the case that $\mu$ is positive then it is known (See Rudin "Real and Complex Analysis" Theorem 8.9) that for $\mu - a.e. x$ the above limit is equal to $+\infty$.
Yes, this works. Write $d\mu = s\, d|\mu|$, with $|\mu|$ denoting the total variation of $\mu$ and $s(x)=\pm 1$. We can recover $s(x)$ for $|\mu|$-a.e. $x$ as the derivative $s(x)=\lim_{|I|\to 0} \mu(I)/|\mu|(I)$. (Here and below, I write $I$ to denote an interval centered at $x$.)
Since a positive singular measure $\nu$ is supported by the set $\{ \nu(I)/|I| \to\infty\}$, we have $|I|/|\mu|(I)\to 0$ for $|\mu|$-a.e. $x$. Thus if we had $\liminf \mu(I)/|I|<\infty$ for all $x$, then $s(x)$ can never be $1$, so $\mu$ has no positive part.
(The argument in fact shows that $\mu(I)/|I|\to\infty$ almost everywhere with respect to $\mu_+$, the positive part of $\mu$.)
