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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.

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    $\begingroup$ What is your definition for $\mu \times \nu$ : is it en.wikipedia.org/wiki/Product_measure ? And is $\mu \ast \nu$ additive (is it a measure en.wikipedia.org/wiki/Sigma_additivity ) ? $\endgroup$ – reuns Jun 4 '16 at 1:28
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    $\begingroup$ Since $\mu$ and $\nu$ are finite, $$\mu*\nu(A) = \mu\times\nu(\{(x,y): x+y\in A\})\leqslant \mu\times\nu(\mathbb R^2)=\mu(\mathbb R)\nu(\mathbb R)<\infty, $$ so trivially $\mu*\nu$ is finite. $\endgroup$ – Math1000 Jun 4 '16 at 2:49
  • $\begingroup$ Lebesgue measure is finite now? $\endgroup$ – Math1000 Jun 4 '16 at 3:13
  • $\begingroup$ @Math1000 : just saw his set is weird, $E_A = \{(x,y) \ \mid \ x+y \in A\}$, so $\mu \ast \nu([a,b]) = \mu \times \nu(E_{[a,b]}) $ where the set $E_{[a,b]}$ is a strip of width $b-a$ in the $\mathbb{R}^2$ plane, delimited between the lines $x+y = b$ and $x+y = a$. since $\mu \times \nu$ is a finite measure on $\mathbb{R}^2$ it is indeed $\sigma$-additive, and finite $\endgroup$ – reuns Jun 4 '16 at 3:18
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    $\begingroup$ The set isn't "weird"; $\mu*\nu$ is the convolution of the two measures. $\endgroup$ – Math1000 Jun 4 '16 at 3:59
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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 $\{(x,0):x\in A\}$.

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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.

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