Does $\mu(x+B)= \mu(B)$ for all balls $B$ imply that $\mu$ equals the Lebesgue measure (up to scaling)? Suppose that $\mu$ is a measure on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$ such that $\mu(K)&lt;\infty$ 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.
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 $\epsilon &gt;0$, 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 $0&lt;\epsilon&lt;1$) 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,

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

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.
[ edit- this is wrong. ]

$\newcommand{\ball}{\mathbf{B}_N} \DeclareMathOperator{\Leb}{Leb}$Actually, my comment is not 100% correct. We need to also assume that  we assign (say) $c=\mu([0,1]^d)&lt;∞$.  From this we easily get that 
$$ \mu (A) = c\Leb(A)$$
for any rectangle $A=∏_i[a_i,b_i)$ such that $b_i,a_i ∈ \mathbb{Q}$.
 Let 
\begin{align} P :=&amp; \{\text{all such rational rectangles $A$}\} \\
L_N:=&amp;\{A∈\mathcal{B}(ℝ^d) : \mu (A∩ \ball) = c\Leb(A∩ \ball) \}
\end{align} 
where $\ball$ is the ball around $\mathbf{0}$ of radius $N&lt;∞$. (The proof might be easier if you use a cube around 0 of radius $N$)
We can check that  $P$ is a π-system, $L_N$ is a $\lambda$-system and $P ⊂ L_N$. Thus by Dynkin's $\pi-\lambda$ Theorem, 
$$ \mathcal{B}(ℝ^d) = \sigma( P ) ⊂ L_N $$

So we have $\mu (A∩ \ball) = c\Leb(A∩ \ball)$ for any $N$; sending $N→∞$ gives the result.



I think this is reasonably standard, I adapted it from my unsourced written notes for the 1D case.

(response to comments)In the 1D case: we can get $$\mu[0,k) = \mu [0,1) + … + \mu[k-1,k)= c \text{Leb}[0,k)$$ by translation invariance and $\sigma$-additivity. Then $$\mu [0,k) = q\mu[0,k/q) = c \text{Leb}[0,k)$$ for any $q\in\mathbb{N}$, again by TI and $\sigma$-additivity. Now translate by a rational coordinate; $k,q$ was arbitrary, so this proves that $P⊂ L_N$ in the 1D case.

The same proof works in $d$ dimensions by first 'fixing' the last $d-1$ coordinates. the same construction yields 
$$\mu[a_1,b_1)\times[0,1)^{d-1} = c \text{Leb}[a_1,b_1)\times[0,1)^{d-1}$$
for an arbitrary rational $a_1,b_1$. Repeating once for each dimension yields the result.