Convolution of Measurable Function and Measure is Bilinear

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

Let $f, f': \R^n \to \R$ be $\BB^n$-measurable functions.

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


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

provided the convolutions in these expressions exist.

That is, convolution $*$ is a bilinear operation.