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I have some troubles solving the following problem and hope some of you can help me;

Consider on $( \mathbb{R}, \mathcal{B})$ the absolutely steady probability measure $\mu$, $\mu \ll \lambda$ and the discrete probability measure $\nu \ll \xi|_{\mathbb{N}}$.

Now I have to show, that the convolution measure is absolutely steady too, $\mu * \nu \ll \lambda$.

We defined the convolution only for the corresponding densities $f,g \in \mathcal{M}^{+}(\mathbb{R},\mathcal{B})$ as $f*g(s) := \int_{\mathbb{R}} f(s-y) g(y) d\lambda(y)$.

Well I might find densities for my given measures via Radon-Nikodym's theorem, but due to the measures aren't given explicitly I don't see a chance in going this way. So I guess I have to use some kind of convolution of the measures it self, which made me looking up for that in a textbook.

There I found (for sigma-finite measures) $\mu * \nu(A) := \int \mu(A-y)\nu(dy)\hspace{3mm} \forall A \in \mathcal{B} $.

This even made me more confused, because I haven't seen the Notation $A-y$ yet.

Does anyone have an idea how I can solve this problem?

Thank you very much! :)

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If $\lambda (A)=0$ then $\lambda (A-y)=0$ for all $y$ so $\mu (A-y)=0$ for all $y$ and $\mu *\nu (A)=0$.

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  • $\begingroup$ Hi! Thanks for your respond! That (and your comment above) makes many Things clearer. So that means for my special example, that I don't need the information, that $\nu = \xi|_{\mathbb{N}}$ ? Because due to Lebesgue measure's translation invariance we are already finished, as you concluded in your answer (?) $\endgroup$ – pcalc Oct 29 '18 at 7:49
  • $\begingroup$ @pcalc Absolutely right. You only need absolute continuity of $\mu$. $\endgroup$ – Kavi Rama Murthy Oct 29 '18 at 7:59
  • $\begingroup$ Ah - great! Thanks again! $\endgroup$ – pcalc Oct 29 '18 at 8:34

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