Convolution of Measures is Bilinear

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Theorem

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


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

$\left({\lambda \mu + \mu'}\right) * \nu = \lambda \left({\mu * \nu}\right) + \mu' * \nu$
$\mu * \left({\lambda \nu + \nu'}\right) = \lambda \left({\mu * \nu}\right) + \mu * \nu'$

where $*$ denotes convolution.

That is, convolution is a bilinear operation.


Proof


Sources