Convolution of Measurable Function and Measure is Bilinear

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

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

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


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

provided the convolutions in these expressions exist.

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