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Suppose $\mu$ is a finite Borel measure on $\mathbb{R}$ and $m$ is lebesgue measure. Then, if $\mu(A+x)$ is continuous in $x$, then $\mu \ll m$.

My approach was simply use the definition of continuity, Let $A\in \mathcal{B}(\mathbb{R})$ with $m(A) = 0$, then $$ m(A+(y-y_0))\le m(A) + |y-y_0| = |y-y_0|<\delta\Rightarrow |\mu(A+y)-\mu(A+y_0)|<\epsilon $$

I think I am totally doing wrong. I hope that anyone could give me direction. Thanks.

I could not understand A note on translation continuity of probability measures by S. Zabell. and looked at $μ≪m$ finite Borel implies $x↦μ(A+x)$ is continuous

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    $\begingroup$ For all $A$? You didn't say so. $\endgroup$ – GEdgar Nov 13 '17 at 22:29
  • $\begingroup$ @GEdgar sorry yes for all $A$ in Borel set. $\endgroup$ – user1292919 Nov 14 '17 at 0:39
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Here is Zabell's argument (on $\Bbb R$): If $\mu$ fails to be absolutely continuous with respect to $m$, then there is a Borel set $B$ with $\mu(B)>0$ but $m(B)=0$. By translation invariance of Lebesgue measure, you then have $m(B+t)=0$ for every $t\in\Bbb R$. (Here $B+t:=\{x+t:x\in B\}$ is the translate of $B$ by $t$.) By Fubini, $$ \int_{\Bbb R} \mu(B+s)\,ds = \int_{\Bbb R} m(B-x)\,\mu(dx)=0. $$ It follows that $\mu(B+s)=0$ for Lebesgue-a.e. $s\in\Bbb R$. By the assumed continuity of $s\mapsto\mu(B+s)$, you even have $\mu(B+s)=0$ for all $s$. In particular, take $s=0$ to see that $\mu(B)=0$, a contradiction.

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