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.

Proof

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Sources

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