Integral of shifted measure of set

Question

Consider a measure $\mu$ on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ such that $\mu(\mathbb{R}) < \infty$. In relation to showing stability of a point process, I need to show that for any bounded set $A \in \mathcal{B}(\mathbb{R})$, $$\int_{\mathbb{R}} \mu(A - y) \,\textrm{d}y < \infty$$ where $A - y$ simply denotes the shifted set $\{x-y\mid x\in A\}$. (Or if this is not in general possible, I am looking for conditions to imply on $\mu$, such that this is the case). I can assume that $\mu$ is supported on $\mathbb{R}_+$, though I'm not convinced this changes much.

Would be grateful for any tips or thoughts!

Answer 1

This is easy if you note that what you're trying to show is finite, really is a convolution of two measures, and use Tonelli theorem. Recall that Lebesgue measure is invartiant of translations by vectors, and Lebesgue measure of a bounded set is finite, since it sits in an interval. $$ \iint_{R^2} 1_A(x+y) dx\mu(dy) = (\mu*\lambda)(A) = \int_R \lambda(A-y)\mu(dy) = \lambda(A)\mu(R) <\infty.$$

Answer 2

By Tonelli's Theorem, $$ \int_{\mathbb R}\mu(A-y)\ dy=\int_{\mathbb R}\lambda(A-y)\ d\mu(y)=\int_{\mathbb R}\lambda(A)\ d\mu(y)=\lambda(A)\mu(\mathbb R)<\infty, $$ where $\lambda$ denotes Lebesgue measure and we have used the fact that $A$ is bounded.