Suppose that $\mu$ is a signed finite Borel probability 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 $+\infty$.

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 $+\infty$.

Suppose that $\mu$ is a signed finite Borel probability 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*}

Suppose that $\mu$ is a signed finite Borel probability 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 $+\infty$.