# Convolution of Measurable Function and Measure is Bilinear

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## 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.

## Proof

## Sources

- 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $14.5 \ \text{(i)}$