# Convolution of two measures is absolutely steady

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! :)

If $$\lambda (A)=0$$ then $$\lambda (A-y)=0$$ for all $$y$$ so $$\mu (A-y)=0$$ for all $$y$$ and $$\mu *\nu (A)=0$$.
• 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 (?) – pcalc Oct 29 '18 at 7:49
• @pcalc Absolutely right. You only need absolute continuity of $\mu$. – Kavi Rama Murthy Oct 29 '18 at 7:59