Proposition 8

Let $f$ and $g$ be two locally $\beta$-integrable functions on $X$ and let $\mu \in \mathscr{M}(G)$. Assume that:

(i) $\mu$ and $f$ are convolvable and the formula (3) of No. 1 defines locally $\beta$-almost everywhere a convolution product $\mu *_{\beta} f$.

(ii) $\chi \cdot \check{\mu}$ and $g$ are convolvable and the formula (3) of No. 1 (with $\mu$ replaced by $\chi \cdot \check{\mu}$ and $f$ by $g$) defines locally $\beta$-almost everywhere a convolution product $(\chi \cdot \check{\mu}) *_{\beta} g$.

(iii) There exists a function $\psi$ on $G$, equal locally $\mu$-almost everywhere to 1, such that the function

is $(\mu \otimes \beta)$-integrable.

Then the functions $g(x)((\mu *_{\beta} f)(x))$ and $f(x)(((\chi \cdot \check{\mu}) *_{\beta} g)(x))$ are essentially $\beta$-integrable, and

For, by (iii) and the Lebesgue–Fubini theorem, the function

is $\beta$-integrable, and

But $\psi \cdot \mu = \mu$, consequently

locally $\beta$-almost everywhere. This shows that the function

is essentially $\beta$-integrable and that

On the other hand, Lemma 1 shows that the function

is integrable for $(\chi \cdot \mu) \otimes \beta$. Therefore the function $(s, x) \mapsto g(s^{-1}x)f(x)\psi(s^{-1})$ is integrable for $\check{\mu} \otimes \check{\beta}$, and

But $\check{\psi} \cdot \check{\mu} = \check{\mu}$ and so $\int g(s^{-1}x)\psi(s^{-1})\, d\check{\mu}(s) = ((\chi \cdot \check{\mu}) *^\beta g)(x)$ locally $\beta$-almost everywhere. This shows that the function

is essentially $\beta$-integrable and that

This proves the proposition.