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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.

1 Answers 1

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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.

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    @student Glad to be of help!2012-07-27