Finiteness of an integral w.r.t. a finite Borel measure. Suppose $\mu$ is a finite Borel measure on $\mathbb{R}^3$. Define $h : \mathbb{R}^3 \rightarrow \mathbb{R}$ by
$$h(x) = \int_{\mathbb{R}^3} \dfrac{d\mu(y)}{\|x - y\|}.$$
Question 1: Must $h(x)$ be finite for almost every $x$, w.r.t. Lebesgue measure?
Question 2: Suppose $h(x)$ is indeed finite for (Lebesgue) almost every $x$. Does this imply that $\mu$ and $m^3$ (the Lebesgue measure on $\mathbb{R}^3$) are mutually singular?
I'm also interested in any comments about interesting conditions that could imply a.e. finiteness for $h$. Since interesting is somewhat vague, I'm not stating this as a formal question.
Thanks.
 A: 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. 
