Bourbaki

Proposition 7

Tag 09KG, permanent. Cite it as [Bourbaki, Tag 09KG].

Integration, VIII, p. 34, § 4. Convolution of measures and functions

*Let ff be a locally β\beta-integrable function on XX, ν\nu a measure on XX, and μ\mu a measure on GG. Assume that: (i) μ\mu and ff are convolvable and formula (3) of No. 1 defines locally β\beta-almost everywhere a convolution product μ∗βf\mu *_{\beta} f. (ii) χ⋅μˇ\chi \cdot \check{\mu} and ν\nu are convolvable. (iii) The function g(s,x)=f(s−1x)χ(s−1)g(s, x) = f(s^{-1}x)\chi(s^{-1}) is (μ⊗ν)(\mu \otimes \nu)*-integrable. Then ff is essentially integrable for (χ⋅μˇ)∗ν(\chi \cdot \check{\mu}) * \nu, the function μ∗βf\mu *_{\beta} f defined by (3) is ν\nu*-integrable, and*

ν(μ∗βf)=((χ⋅μˇ)∗ν)(f). \nu(\mu *_{\beta} f) = ((\chi \cdot \check{\mu}) * \nu)(f).

Since g(s,x)g(s, x) is integrable for μ⊗ν\mu \otimes \nu, the function f(sx)f(sx) is essentially integrable for (χ⋅μˇ)⊗ν(\chi \cdot \check{\mu}) \otimes \nu and ff is essentially integrable for (χ⋅μˇ)∗ν(\chi \cdot \check{\mu}) * \nu. By the Lebesgue–Fubini theorem, μ∗βf=∫g(s,x)dμ(s)\mu *_{\beta} f = \int g(s, x) d\mu(s) is ν\nu-integrable and

ν(μ∗βf)=∬f(s−1x)χ(s−1)dμ(s)dν(x)=∬f(sx)χ(s)dμˇ(s)dν(x)=((χ⋅μˇ)∗ν)(f). \begin{align*} \nu(\mu *_{\beta} f) &= \iint f(s^{-1}x)\chi(s^{-1}) d\mu(s) d\nu(x) \\ &= \iint f(sx)\chi(s) d\check{\mu}(s) d\nu(x) = ((\chi \cdot \check{\mu}) * \nu)(f). \end{align*}

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.