How can I show that $f$ is continuous in $x_0$ iff $\mu(C_{x_0})=0$?

Question

The problem is the following:

Let $\mu$ be a finite Borel measure on $\mathbb{R}^2$. For fixed $r>0$, let $C_x = \{y : |y − x| = r\}$ and define $f : \mathbb{R}^2 \to\mathbb{R}$ by $f(x)= μ[C_x]$. Prove that $f$ is continuous at $x_0$ if and only if $\mu[C_{x_0}] = 0$.

I couldn't even start. What can I do with "$\mu(C_{x})-\mu(C_{x_0})$"? The sets $C_x$ are disjoint, equal or they either have one or two points in common. But I don't know how this Borel measure works, it's an arbitrary one. I don't know if the measure of a single point is zero or not, I know nothing.

Any hint will be really appreciated?

Answer 1

Lemma 1. There is at most countable set of points of positive measure.

Proof. Let $A$ be a set of all points with positive measure and assume $A$ is uncountable. For a number $M\in\mathbb{R}$ define

$$A_M=\{x\in A\ |\ \mu(x) > M\}$$

If $A$ is uncountable then there exists $M$ such that $A_M$ is infinite. Indeed, otherwise

$$\bigcup_{n\in\mathbb{N}}A_{\frac{1}{n}} = A$$ would be countable as a countable sum of finite sets. Let $M$ be such that $A_M$ is infinite. Then pick a sequence of distinct points from $(a_n)\subset A_M$. You can do that since $A_M$ is infinte. Then

$$\mu(\bigcup\{a_n\})=\sum\mu(\{a_n\})\geq\sum M=\infty$$

and thus $\bigcup\{a_n\}$ has an infinite measure. Contradiction with finitness of $\mu$. $\Box$

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"$\Rightarrow$" Assume that $\mu(C_x)\neq 0$ for some $x\in\mathbb{R}^2$. Consider a sequence $x_n$ converging to $x$. Pick that sequence in such a way that $x_n\neq x_m$ for $n\neq m$ (it's quite simply to construct it).

Since $f$ is continous then $f(x_n)$ converges to $f(x)=\mu(C_x)\neq 0$. In particular for almost all $n$ we have $f(x_n)\neq 0$. Without loss of generality assume that $f(x_n)\neq 0$ for all $n$.

As you've noted $C_x$ and $C_y$ have at most two common points if $x\neq y$. We can refine our sequence $x_n$ even more. Pick $x_n$ in such a way that $C_{x_n}\cap C_{x_m}$ is of measure $0$ (whenever $n\neq m$). This can be done since $C_x\cap C_y$ has at most 2 points and there is at most countable number of points (Lemma 1) with positive measure (opposed to uncountable number of points in $\mathbb{R}^2$). Put

$$P=\bigcup_{n\neq m} C_x\cap C_y$$

Since $P$ is a countable union of finite sets of measure $0$ then it is countable and of measure $0$. Define

$$B_n=C_{x_n}-P$$

(the difference of sets). Note that each $B_n$ is actually $C_{x_n}$ minus a countable set of measure $0$. So it is Borel and its measure doesn't change:

$$\mu(B_n)=\mu(C_{x_n})=f(x_n)\neq 0$$

Also by definition $B_n\cap B_m=\emptyset$ for $n\neq m$ (that's because we've removed all common points). In particular countable additivity of $\mu$ applies and so

$$\mu(\bigcup B_n)=\sum\mu(B_n)=\infty$$

The last equality comes from the fact that $\mu(B_n)=f(x_n)$ converges to something positive (and thus the series is divergent to infinity).

Contradiction since $\mu$ is finite. $\Box$

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"$\Leftarrow$" Assume that $\mu(C_x)=0$ for some $x$ and assume that $f$ is not continous at $x$. Then there exists a sequence $(x_n)$ convergent to $x$ such that $f(x_n)$ is not convergent to $f(x)=0$. Since $f(x_n)\geq 0$ is not convergent to $0$ then there exists a positive constant $M>0$ and a subsequence $f(x_{n_k})$ such that $f(x_{n_k})>M$. Without a loss of generality we may assume that $f(x_n)$ is bounded from below by a positive constant $M$.

Now consider the set

$$P=\bigcup_{n\neq m} C_x\cap C_y$$

again. Now put

$$B_n=C_{x_n} - P$$

Obviously $B_n$ are pairwise disjoint and

$$\mu(B_n) \geq \mu(C_{x_n}) - \mu(P) \geq M-\mu(P)$$

Let $M'=M-\mu(P)$. Note that $M'$ is constant and does not depend on $n$. So due to countable additivity we get

$$\mu(\bigcup B_n)=\sum\mu(B_n)\geq \sum M'= \infty$$

EDIT: Actually the last equality $\sum M'=\infty$ does not have to hold, since $M'$ might be $0$. Uhhh... TODO.