$\mu * \nu$ a finite Borel measure in $\mathbb{R}$? Let $\mu$ and $\nu$ be two finite Borel measures on $\mathbb{R}$. For any Borel set $A \subset \mathbb{R}$, define$$\mu * \nu(A) = \mu \times \nu(\{(x, y) \in \mathbb{R}^2 : x + y \in A\}).$$Is $\mu * \nu$ necessarily a finite Borel measure in $\mathbb{R}$?
Thoughts. I know that the set $\{(x, y) \in \mathbb{R}^2 : x + y \in A\}$ is Borel when $A$ is Borel.
 A: Let $\left(\Omega_{i},\mathcal{A}_{i}\right)$ be measurable spaces
for $i=1,2$ and let $\rho$ be a measure on $\mathcal{A}_{1}$. 
Every measurable function $f:\Omega_{1}\to\Omega_{2}$ induces a measure
on $\mathcal{A}_{2}$ by the prescription $A\mapsto\rho\left(f^{-1}\left(A\right)\right)$.
This measure is denoted as $\rho f^{-1}$.
Observe that $\rho f^{-1}\left(\Omega_{2}\right)=\rho\left(f^{-1}\left(\Omega_{2}\right)\right)=\rho\left(\Omega_{1}\right)$
showing that every $\rho f^{-1}$ is a finite measure if $\rho$ is
a finite measure.
Special case: $\Omega_{1}=\mathbb{R}^{2}$, $\Omega_{2}=\mathbb{R}$
and the $\mathcal{A}_{i}$ are the Borel $\sigma$-algebras on these
sets. 
Let $\rho$ be the product measure $\mu\times\nu$ where $\mu,\nu$
are measures on $\left(\Omega_{2}=\mathbb{R},\mathcal{A}_{2}\right)$.
Function $f:\mathbb{R}^{2}\to\mathbb{R}$ prescribed by $\left\langle x,y\right\rangle \mapsto x+y$
is measurable so $\rho f^{-1}$ is a well defined measure.
For $A\in\mathcal{A}_{2}$ observe that: $$\rho f^{-1}\left(A\right)=\rho\left(f^{-1}\left(A\right)\right)=\mu\times\nu\left(\left\{ \left\langle x,y\right\rangle \mid x+y\in A\right\} \right)=\mu\star\nu\left(A\right)$$
That means exactly that: $$\rho f^{-1}=\mu\star\nu$$ We conclude that $\mu\star\nu$
is a finite measure if $\rho$ is a finite measure, which is evidently the case if $\mu,\nu$ are finite measures.
A: Yes, as @Math1000 said, you have that: $(\mu * \nu)(A) \leq (\mu \times \nu)(\mathbb{R}^2) = \mu(\mathbb{R})\nu(\mathbb{R}) < \infty$.
However, to give you a little more intuition on your question, you can use Fubini's theorem:
$(\mu \times \nu)(A) = \int \mu(dx) \int \nu(dy) 1\{(x, y) \in \mathbb{R}^2 : x + y \in A\}$. Letting $A_x = \{ y : x + y \in A\}$ for almost every $x$, you can see this integral equals $\int \mu(dx) \nu(A_x)$.
Additionally, you can visualize the set $\{(x, y) \in \mathbb{R}^2 : x + y \in A\}$ by placing $A$ on the $x$-axis, i.e. $\{(x,0):x\in A\}$, and drawing a line with slope $-1$ through every point of $\{(x,0):x\in A\}$.
