Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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. –  GEdgar Jul 27 '12 at 20:28
    
@Byron thanks for providing the details of the proof, I wasn't able to get access to that reference. –  student Jul 27 '12 at 21:29
    
@student Glad to be of help! –  Byron Schmuland Jul 27 '12 at 21:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.