Bourbaki

Exercise 19, § 3

Integration, VIII

Bourbaki marks this one as harder than the rest.

The notations and conventions are those of Exercise 15.

a) Now prove that if a bounded positive measure μ\mu on GG is such that for every function f∈Lp(G,β)f \in L^p(G, \beta) (pp given, 1≤p<+∞1 \leq p < +\infty) the relation μ∗f≥0\mu * f \geq 0 locally almost everywhere implies f≥0f \geq 0 locally almost everywhere, then one can conclude that μ\mu is a point measure in each of the following cases: α)\alpha) GG is compact ; β)\beta) GG is discrete ; γ)\gamma) GG is commutative (use Exercise 18) (*). For p=+∞p = +\infty, the same conclusion holds without any additional assumption on GG.

b) Let μ\mu be a bounded positive measure on GG such that: 1o1^o γ(μ)\gamma(\mu) is a surjective endomorphism of L1(G,β)L^1(G, \beta) ; 2o2^o for every function f∈L1(G,β)f \in L^1(G, \beta) the relation μ∗f≥0\mu * f \geq 0 locally almost everywhere implies

(*) For an example where μ\mu is non-point and such that μ∗f≥0\mu * f \geq 0 implies f≥0f \geq 0, see Williamson, Proc. Edimb. Math. Soc., 1957, pp. 71-77.

f⩾0f \geqslant 0 locally almost everywhere. Now prove that μ\mu is a point measure. (Observe that for every function g∈L∞(G,β)g \in L^\infty(G, \beta), the relation μ~∗g⩾0\tilde{\mu} * g \geqslant 0 locally almost everywhere implies g⩾0g \geqslant 0 locally almost everywhere).

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.