Convolution of Measures is Bilinear

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Theorem

Let $\mu, \mu', \nu$ and $\nu'$ be measures on the Borel $\sigma$-algebra $\BB^n$ on $\R^n$.


Then for all $\lambda \in \R$:

$\paren {\lambda \mu + \mu'} * \nu = \lambda \paren {\mu * \nu} + \mu' * \nu$
$\mu * \paren {\lambda \nu + \nu'} = \lambda \paren {\mu * \nu} + \mu * \nu'$

where $*$ denotes convolution.

That is, convolution is a bilinear operation.


Proof



Sources