Does $\mu(x+B)= \mu(B)$ for all balls $B$ imply that $\mu$ equals the Lebesgue measure (up to scaling)?

Question

Suppose that $\mu$ is a measure on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$ such that $\mu(K)<\infty$ for any compact set $K$ and

$$\mu(x+B) = \mu(B) \tag{1}$$

for all $x \in \mathbb{R}^d$ and open balls $B$. Does this imply that $\mu$ equals up to a constant the Lebesgue measure $\lambda^d$, i.e. does it hold that $\mu = c \lambda^d$ for some constant $c \in [0,\infty]$?

My attempts:

It is well-known that if $(1)$ holds for all ($d$-dimensional) rectangles, then the claim holds true. So if we knew that any rectangle can be covered by disjoint balls, then we would be done. However, as far as I know, this is actually not possible. (See this question)
Define $$\mathcal{D} := \{A \in \mathcal{B}(\mathbb{R}^d); \forall x: \mu(x+A) = \mu(A)\}$$ and show that $\mathcal{D}$ is a Dynkin system. If we knew this, then the fact that the balls are contained in $\mathcal{D}$ would imply $\mathcal{D} = \mathcal{B}(\mathbb{R}^d)$. My problem: Since $\mu$ is (in general) not a finite measure, I don't see how to prove the implication $A \in \mathcal{D} \implies A^c \in \mathcal{D}$. I also considered defining $$\mathcal{D}_R := \{A \in \mathcal{B}(B(0,R)); \forall x: \mu(x+A) = \mu(A)\};$$ then it is easy to show that $\mathcal{D}_R$ is a Dynkin system, but, unfortunately, we cannot conclude that the balls are contained in $\mathcal{D}_R$ (since the intersection $B \cap B(0,R)$ is in general not a ball).

Any ideas, counterexamples,...?

Edit: @NateEldredge suggested as a counterexample the measure $$\mu(A) := \sum_{q \in \mathbb{Q}} \delta_q(A)$$ in case that $\mu$ does not need to be finite on compact sets.

Counterexample: let $\mu(A) := 2\text{Leb}(A)$. (It is true up to scaling) — Calvin Khor, May 29 '15 at 16:19
@CalvinKhor Well, it's not a counterexample, because I ask whether $\mu = c \lambda^d$ for some constant $c$. (But I see, I should add "up to a scaling" to the title of my question.) — saz, May 29 '15 at 16:24
You will need an assumption like finiteness on compact sets to rule out something like counting measure. — Nate Eldredge, May 29 '15 at 16:41
@NateEldredge Yeah, I'm fine with assuming $\sigma$-finiteness or finiteness on compact sets. — saz, May 29 '15 at 16:44
I edited: $\sigma$-finiteness is not sufficient, since you can take counting measure on the rationals. — Nate Eldredge, May 29 '15 at 16:45
@NateEldredge Well, actually, $c$ might be infinite, so counting measures aren't a counterexample, right? — saz, May 29 '15 at 16:46
But counting measure (e.g. on rationals) satisfies the hypothesis yet does not equal infinity times Lebesgue measure. — Nate Eldredge, May 29 '15 at 16:47
@NateEldredge Ah, I see; you are right, thank you very much... — saz, May 29 '15 at 16:50

score 4 · Accepted Answer · answered May 29 '15 at 22:40

Yes, this is true. The following is based on the proof of uniqueness of Haar measure in Hewitt and Ross (if I remember correctly):

The general idea is to write an arbitrary $C_c$ function as an infinite linear combination of characteristic functions of balls. Rigorously, this amounts to approximation by convolution.

I denote the usual Lebeague measure by $\lambda$, and integration against this measure is written $dx$.

For $\epsilon >0$, let $f_\epsilon = \chi_{B_\epsilon}/\lambda(B_\epsilon)$, where $\chi_{\dots}$ denotes the indicator function of the set in the index and $B_\epsilon$ is the ball with radius $\epsilon$ around the origin.

Let $h\in C_c$ be arbitrary. It is a classical fact that the convolution $f_\epsilon \ast h$ converges locally uniformly to $h$, has support in a fixed compact set (for $0<\epsilon<1$) and is uniformly bounded. Using dominated convergence, we get $f_\epsilon \ast h \to h$ in $L^1(\nu)$ or every locally finite measure $\nu$.

Hence,

$$ \begin{eqnarray*} \int h d\mu &=&\lim_\epsilon \int f_\epsilon \ast h d\mu\\ &=& \lim_\epsilon \int\int f_\epsilon (y) h(x-y)\, dy \, d\mu (x)\\ &\overset{z=x-y}{=} &\lim_\epsilon \int \int f_\epsilon (x-z) h(z)\,dz \, d\mu (x)\\ &\overset{Fubini}{=}& \int h(z)\int f_\epsilon (x-z)\,d\mu (x)\, dz\\ &=& \lim_\epsilon \int h(z)\, dz \cdot \frac{\mu (B_\epsilon )}{\lambda (B_\epsilon )}. \end{eqnarray*} $$

Fix some $h_0\in C_c$ with $h\geq 0$ and $h\not\equiv 0$. Then the above calculation shows

$$ c:=\frac{\int h_0 \,d\mu}{\int h_0dx}=\lim_\epsilon \frac{\mu (B_\epsilon)}{\lambda(B_\epsilon)} $$ (in particular, the limit exists and we take the limit $\epsilon \downarrow 0$), as well as

$$ \int h\,d\mu =c \int h \,dx $$ for all $h \in C_c$. This easily implies the claim.

Thank you very much for this nice proof. :) – saz May 30 '15 at 04:54 — saz, May 30 '15 at 04:54

Does $\mu(x+B)= \mu(B)$ for all balls $B$ imply that $\mu$ equals the Lebesgue measure (up to scaling)?

1 Answers1