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 $\mu *\nu (A)=0$.