Some results in measure theory

Proposition. Let $\mu$ be a positive Borel measure on $\mathbb{R}$ and $\epsilon > 0$. Then for almost every $x$, we have $$\int_\mathbb{R} \frac{d\mu(t)}{|x - t|^{1 + \epsilon}} = +\infty.$$
Proof: For each positive integer $N$, define $E_N = \{x \in \mathbb{R}: \int_\mathbb{R} d\mu(t) / |x - t|^{1 + \epsilon} \leq N\}$. Let $I$ be any bounded closed interval with $E_N \cap I$ nonempty. Pick $x \in E_N \cap I$ and write $I = [x - \delta_1, x + \delta_2]$. We have $$N \geq \int_\mathbb{R} \frac{d\mu(t)}{|x - t|^{1 + \epsilon}} \geq \int_I \frac{d\mu(t)}{|x - t|^{1 + \epsilon}} \geq \frac{\mu(I)}{\max\{\delta_1, \delta_2\}^{1 + \epsilon}} \geq \frac{\mu(E_N \cap I)}{(\delta_1 + \delta_2)^{1 + \epsilon}},$$ i.e. $\mu(E_N \cap I) \leq Nm(I)^{1 + \epsilon}$ where $m(I)$ is the length of the interval $I$. Now let $[a, b] \subseteq \mathbb{R}$ and $n \in \mathbb{N}$. Decompose $[a, b]$ into a union of $n$ closed subintervals of length $(b - a)/n$, then $$\mu(E_N \cap [a, b]) \leq nN\left(\frac{b - a}{n}\right)^{1 + \epsilon} \rightarrow 0$$ as $n \rightarrow \infty$. Hence $\mu(E_N \cap [a, b]) = 0$. It follows that $\mu(E_N) = 0$ for all $N \in \mathbb{N}$ and our result follows.

(To be continued)