Your function $h$ is the Newtonian potential of the measure $\mu$. When the measure $\mu$ is finite, this function is superharmonic ([1], Theorem 6.3), and thus finite almost everywhere with respect to Lebesgue measure ([1], Theorem 4.10).
In addition, $h$ is Lebesgue integrable over every compact subset of $\mathbb{R}^3$.
Reference: [1] Introduction to Potential Theory by L.L. Helms (Wiley, 1969)
Added: That last statement gives a clue on how to find a direct proof. Let $K$ be a compact subset of $\mathbb{R}^3$ and integrate $h$ over $K$: $\int_Kh(x)\,dx=\int_K \int_{\mathbb{R}^3} {1\over\|x-y\|}\,\mu(dy)\,dx = \int_{\mathbb{R}^3} \int_K {1\over\|x-y\|}\,dx \,\mu(dy).$
Let's show that $g(y):= \int_K {1\over\|x-y\|}\,dx$ is a bounded function.
For $y$ with distance greater than 1 from $K$ we have $g(y)\leq \lambda(K)$.
On the other hand, there is a fixed radius $R$ so that for all other $y$, the set $K$ is contained in the ball $B(y,R)$.
So, for such $y$,
$g(y)\leq \int_{B(y,R)} {1\over\|x-y\|}\,dx = \int_{B(0,R)} {1\over\|x\|}\,dx = 4\pi \int_0^R {1\over r}\, r^2\,dr = 2\pi R^2.$
Combining these bounds shows that $g$ is a bounded function, and since $\mu$ is finite, this means that the integral of $h$ over $K$ is finite. This implies that $h$ is finite Lebesgue almost everywhere.