# 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?

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?

• Clear and helpful, thanks very much! For your question, we need $\sigma$-finite to use the Tonelli Theorem to exchange the order of the integral. Just a further question: Why do we define $\mu$ on the Borel $\sigma$-algebra? It looks fine if $\mu$ is defined on the Lebesgue $\sigma$-algebra. – user424674 Mar 13 '17 at 21:26
• @Philo: Good question. The main reason probably is that assuming $\mu$ to be defined (and a measure( on the Borel $\sigma$-algebra is a weaker assumption (after all, if $\mu$ was defined on the Lebesgue $\sigma$-algebra, we could simply restrict it to the Borel sets). Also, once we have shown that $\mu \ll m$, it follows that $\mu$ (uniquely) extends to the Lebesgue $\sigma$-algebra, by virtue of $\mu(A \cup N) = \mu(A)$ if $N$ is a subset of a Borel measurable Lebesgue null set. Finally, the Lebesgue $\sigma$-algebra is very peculiar and has some properties that are considered bad by – PhoemueX Mar 14 '17 at 9:01
• some. For example, if $f : \Bbb{R} \to \Bbb{R}$ is Lebesgue measurable and $h : \Bbb{R} \to \Bbb{R}$ is continuous, it does not follow that $f \circ h$ is Lebesgue measurable, but the analogous claim holds if "Lebesgue" is replaced by "Borel" everywhere. Also in the proof above, we used Fubini's theorem. For this, we need to know that $(x,y) \mapsto 1_{A+x}(y) = 1_A (y-x)$ is (Borel or Lebesgue) measurable. If $A$ is Borel measurable, this follows easily, but if $A$ is merely Lebesgue measurable, this is not soooo clear. – PhoemueX Mar 14 '17 at 9:03
• @Philo: For example, Barry Simon literally writes: "In summary, passing from Borel to Lebesgue measurable functions is the work of the devil." – PhoemueX Mar 14 '17 at 9:05
• That's a good reason. Thank again! – user424674 Mar 15 '17 at 15:40