non-$\sigma$-finite translation invariant measures on $\mathbb{R}$ It's known that the $\sigma$-finite translation invariant measures $\mu$ on $(\mathbb{R}, B(\mathbb{R}))$ are exactly the measures $ \mu = c \cdot \lambda$ where $c \in [0,\infty)$ and $\lambda$ is the Lebesgue measure. (An application of Fubini's theorem gives that any candidate for $\mu$ is absolutely continuous with respect to $\lambda$ and then it's not hard to check that the Radon-Nikodym derivative is a.s. constant)
I'm interested in the case where we don't assume $\mu$ is $\sigma$-finite. It's clear that if $\mu$ has atoms then it is a scalar multiple of the counting measure. This leaves the case where $\mu$ is atomless. Such measures certainly exist, for example consider $\mu$ defined by
\begin{align*}
\mu(A) = \begin{cases} 
      0 & \mbox{if } A \mbox{ is countable } \\
      \infty & \mbox{otherwise}
   \end{cases}
\end{align*}
or $\mu = \infty \cdot \lambda$ with the convention $\infty \cdot 0 = 0$. Is there a nice classification of the non-$\sigma$-finite translation invariant measures on $(\mathbb{R},B(\mathbb{R}))$? If not can we easily find many more examples than the three I give above?
 A: Not exactly an answer, but here is a fun example. Define
$B$ to the be the set of reals $x$ such the fractional part of $x$ can be written in base $100$ using only the base-$100$ digits $0,$ $1,$ $2$ and $10.$ Note that $B$ is a Borel set. Define
$$\mu_B(S)=\begin{cases}0&\text{ if $S$ can be covered by countably many translates of $B$,}\\
\infty&\text{otherwise.}\end{cases}$$
i.e. $\mu_B(S)=0$ if and only if $S$ is contained in a sumset $X+B=\{x+b\mid x\in X\text{ and }b\in B\}$ for some countable set $X$. This is countably additive on the whole of $\mathcal P(\mathbb R),$ but $\mu_B$ satisfies:


*

*$\mu_B(B)=0$

*$\mu_B(-B)=\infty$ where $-B$ means $\{-b\mid b\in B\}$

*$\mu_B(\tfrac 1 2B)=\infty$ where $\tfrac 1 2B$ means $\{\tfrac 1 2 b\mid b\in B\}$


So translation-invariance is quite a weak property for non-$\sigma$-finite measures; it does not even guarantee $\mu(S)=\mu(-S)$ and $\mu(2S)\geq \mu(S)$.
Proof of $\mu_B(-B)=\infty$
This will be a diagonalization/Baire category argument; each set $B\cap(r-B)$ is nowhere dense in $B$, with the subset topology from $\mathbb R$. Define $B_0$ to be
$$B_0=([0,3)\cup [10,11))+100\mathbb Z$$
so that $B=\bigcap_{n\geq 1}100^{-n}B_0$. For any real $r$, the set $r-B_0$ has empty intersection with at least one of the sets $[0,1),$ $[2,3),$ and $[10,11)$. To see this it suffices to verify that:


*

*if $r-B_0$ intersects $[0,1)$ then $r$ must lie in $([0,4)\cup [10,12))+100\mathbb Z$;

*if $r-B_0$ intersects $[2,3)$ then $r$ must lie in $([2,6)\cup [12,14))+100\mathbb Z$;

*if $r-B_0$ intersects $[10,11)$ then $r$ must lie in $([10,14)\cup [20,22))+100\mathbb Z$;


and these conditions cannot all be satisfied simultaneously.
Scaling by $100^{-(N+1)}$ (where $N\geq 1$) this shows that given any real $r$ and any interval $[a100^{-N},b100^{-N})\subseteq \bigcap_{n=1}^{N} 100^{-n}B_0$ (with $a,b\in\mathbb Z$) we can "add a digit" and find a subinterval $[a'100^{-N-1},b'100^{-N-1})\subset [a100^{-N},b100^{-N}) \cap \bigcap_{n=1}^{N+1} 100^{-n}B_0$ (with $a',b'\in\mathbb Z$) that has empty intersection with $r-B$. This means that $B\cap (r-B)$ is nowhere dense in $B$.
To apply this to $\mu_B(-B)$, suppose for contradiction that $-B\subseteq X+B$ with $X$ countable. Then $B\subseteq -X-B$, contradicting the fact that $B\cap(-x-B)$ is nowhere dense in $B$ for each $x$.
Proof of $\mu_B(\tfrac 1 2 B)=\infty$
Similar to above, but now we need to verify that:


*

*if $r+2B_0$ intersects $[0,1)$ then $r$ must lie in $((-6,1)\cup (-22,-19))+100\mathbb Z$;

*if $r+2B_0$ intersects $[10,11)$ then $r$ must lie in $((4,11)\cup (-12,-9))+100\mathbb Z$;


and these conditions cannot both be satisfied simultaneously.
