I am self studying "Representation Theorems in Hardy Spaces" by Javad Mashreghi. Here's a claim [in Page 72 of the textbook] that I am unable to prove:
If $\sigma$ is a singular measure of $\mathbb T = \{ z \in \mathbb C : |z|=1 \}$with respect to the Lebesgue measure, we have \begin{align*} \lim_{t \to 0} \frac{\sigma\left( \left\{ e^{is} : s \in (\theta - t, \theta + t \right\} \right)}{2t} = 0 \end{align*} for almost all $e^{i\theta} \in \mathbb T$.
I don't see how to prove this claim. Singularity tells me that if $\lambda$ is the Lebesgue measure on $\mathbb T$ then there exists sets Borel sets $A,B$ such that $\mathbb T = A \cup B$, $A \cap B = \emptyset$ and $\sigma (A)=\lambda (B)=0$. Also, we have the Lebesgue density theorem which talks about something similar. I tried using Lebesgue density theorem using sets $A$ and $B$ but it did not get me anywhere. Is there something that possibly needs to be done?
Hints are appreciated!