Let $\mu$ be a finite Borel measure on $\Bbb R$, which is absolutely continuous with respect to the Lebesgue measure $m$. Prove that $x \mapsto \mu(A+x)$ is continuous for every Borel set $A \subseteq \Bbb R$.

Thoughts: $\mid \mu(A+x) -\mu(A-x) \mid = \mid \int_{-\infty}^{A+x} d\mu - \int_{-\infty}^{A+x_0} d\mu \mid = \mid \int_{A+x_0}^{A+x} d\mu \mid =\mid x-x_0 \mid$, hence continuous. I'm not sure whether this is a correct arguement.

2nd thought: sorry, I regarded $\mu$ as a distribution function in probability. Another question, why can we use Radon-Nikodym to write $\mu$ as a convolution?

  • $\begingroup$ How do you conclude $$\left| \int_{A+x_0}^{A+x} d\mu \right| = |x-x_0|$$ ...? (And: What do you mean by the left-hand side? This notation is hardly rigorous.) $\endgroup$ – saz Sep 1 '15 at 7:53
  • 1
    $\begingroup$ What does the symbol $\int_{-\infty}^{A+x}$ mean? $\endgroup$ – Prahlad Vaidyanathan Sep 1 '15 at 8:29
  • 2
    $\begingroup$ Hint: By Radon-Nikodym, $\mu(A+x)$ can be written as the convolution of $\chi_A$ with an $L^1$ function - such a convolution product is continuous. $\endgroup$ – Prahlad Vaidyanathan Sep 1 '15 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.