Showing that the two definitions agree almost everywhere is easy! Using the truncated transform
$$
\mathcal{H}\_\epsilon\mu(x)=\frac1\pi\int_{\lvert y-x\rvert > \epsilon}\frac{d\mu(y)}{x-y}
$$
then, by definition, $\mathcal{H}\mu(x)=\lim_{\epsilon\to0}\mathcal{H}\_\epsilon\mu(x)$ for all $x$ at which the limit exists. Convolve the identity
$$
\Re\left(\frac{1}{x+ih}\right)=\frac{x}{x^2+h^2}=\int_0^11_{\left\lbrace\lvert x\rvert > h\sqrt{t/(1-t)}\right\rbrace}\frac1x\\,dt
$$
with $\frac1\pi d\mu$ to obtain,
$$
\Re\left(\mathcal{H}\mu\right)(x+ih)=\int_0^1\mathcal{H}_{h\sqrt{t/(1-t)}}\mu(x)\\,dt.
$$
The integrand on the right hand side tends to $\mathcal{H}\mu(x)$ as $h\to0$, whenever this is defined, so bounded convergence gives $\Re(\mathcal{H}\mu)(x+ih)\to\mathcal{H}\mu(x)$.