Bourbaki

Exercise 11, § 2

Integration, VII, p. 89

Bourbaki marks this one as harder than the rest.

Let G be a locally compact group, H a closed subgroup of G, λ\lambda a left Haar measure on G; assume that there exists on G/H a measure μ\mu invariant under G such that λ=μ#\lambda = \mu^\# and μ(G/H)<+∞\mu(G/H) < +\infty. Let ν\nu be a nonzero positive measure on G/H such that ν(G/H)<+∞\nu(G/H) < +\infty. Finally, let hh be a function on GG having the properties of No. 4, Prop. 8.

a) Let AA be a Borel subset of G/HG/H. Show that

∫Gh(s)ν(s−1A) dλ(s)=ν(G/H)μ(A) \int_G h(s)\nu(s^{-1}A)\,d\lambda(s) = \nu(G/H)\mu(A)

(make use of Prop. 9 a) of No. 4).

b) Let AiA_i (1≤i≤n1 \leq i \leq n) be Borel subsets of G/HG/H and, for every ii, let bi=sup⁡s∈Gν(s−1Ai)b_i = \sup_{s \in G} \nu(s^{-1}A_i). Show that there exist two distinct indices i,ji, j such that

∫Gh(s)ν(s−1Ai)ν(s−1Aj) dλ(s)≥(ν(G/H))2μ(G/H)((∑k=1nμ(Ak))2−μ(G/H)ν(G/H)∑k=1nbkμ(Ak)) \int_G h(s)\nu(s^{-1}A_i)\nu(s^{-1}A_j)\,d\lambda(s) \geq \frac{(\nu(G/H))^2}{\mu(G/H)} \left( \left( \sum_{k=1}^n \mu(A_k) \right)^2 - \frac{\mu(G/H)}{\nu(G/H)} \sum_{k=1}^n b_k \mu(A_k) \right)

(argue as in §1, Exer. 28 a)).

Solution

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Transcribed from the printed volume. Not translated.