measures defined by shifting is absolutely continuous with respect to Lebesgue measure Let $\mu$ be a non-zero positive $\sigma$-finite measure on the Borel sigma-algebra $\mathscr{B}(\mathbb{R})$. For $x \in \mathbb{R}$ define a measure $\mu_x$ by
$\mu_x(A) = \mu(A + x)$.
Assume that for any $x,y\in \mathbb{R}$, $\mu_x \ll \mu_y$($\mu_x$ is absolutely continuous with respect to $\mu_y$). Prove that $\mu \ll m$, where $m$ is the Lebesgue measure.
This is a homework problem, I have no idea where to start. Can anyone give a hint or sketch?
 A: Let $A \subset \Bbb{R}$ be Borel measurable with $m(A) = 0$. We want to show $\mu(A) = 0$.
Since $\mu_x \ll \mu_y$ for all $x,y$, it suffices to show that there is at least one $x \in \Bbb{R}$ with $\mu_x(A) = 0$ (why exactly?)
Now, in measure theory, it is sometimes easier to show that a set is very big instead of just showing that it is nonempty. In our case, if we knew that that set $G := \{x \,:\, \mu(A + x) = 0\}$ of "good points" had full measure (i.e., $G^c$ is a Lebesgue null-set), then we would be done.
To see that this is indeed true, consider
$$
\int \mu (A+x) d m (x)
$$
and try to write this in different ways.
If you don't find it yourself, uncover the spoiler below (but first try it, otherwise you rob yourself of a nice experience).

 $$\int \mu(A+x) dm(x) = \int \int 1_{A+x}(y) d\mu(y) dm(x) = \int \int 1_{A+x}(y)dm(x) d\mu(y) = \int m(y - A) d\mu(y)  = \int m(A) d\mu(y) = 0.$$

 Since the integral vanishes, the set $\{x \,:\, \mu(A+x) > 0\} = G^c$ must be a Lebesgue null set. As seen above, this completes the proof.

Side remark: For the actual computation, we did not use that $\mu_x \ll \mu_y$. This was only used at the very beginning of the proof.
Question: (Where) did we use that $\mu$ is $\sigma$-finite?
