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.



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.

  • $\begingroup$ So, think of $\mu$ as a distribution of mass in space, then at a point $x$ in space, ask what is the gravitational attraction there. This is the "potential" of that vector field (or maybe physicists would do it with a minus sign and a constant factor). Certainly is is finite when you are outside the support of $\mu$ but could also be finite at some points of that support. Note: if $\mu$ is a point mass at some point, then certainly this integral is not finite at that point. $\endgroup$ – GEdgar Jul 27 '12 at 20:28
  • $\begingroup$ @Byron thanks for providing the details of the proof, I wasn't able to get access to that reference. $\endgroup$ – student Jul 27 '12 at 21:29
  • $\begingroup$ @student Glad to be of help! $\endgroup$ – user940 Jul 27 '12 at 21:37

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