Bourbaki

Exercise 26, § 1

Integration, VII, p. 85

Bourbaki marks this one as harder than the rest.

Let G be a locally compact group, μ\mu a left Haar measure on G, kk an integer ≥1\geq 1, and A an integrable set. Show that for every ε>0\varepsilon > 0, there exists a neighborhood U of e in G having the following property: for every finite subset S of kk elements in U, the set of s∈Gs \in G such that sS⊂AsS \subset A has measure ≥(1−ε)μ(A)\geq (1 - \varepsilon) \mu(A). (Reduce to the case that A is compact and take U to be such that

μ(AU)≤(1−εk−1)μ(A). \mu(AU) \leq \left( 1 - \frac{\varepsilon}{k-1} \right) \mu(A) .

For every finite subset H of G, set p(H)=Card(H)p(H) = \mathrm{Card}(H) and q(H)=1q(H) = 1 if H≠∅H \neq \varnothing, q(H)=0q(H) = 0 if H=∅H = \varnothing; by evaluating the integrals

∫Gp(A∩sS) dsand∫Gq(A∩sS) ds, \int_G p(A \cap sS) \, ds \quad \text{and} \quad \int_G q(A \cap sS) \, ds ,

show that if, for h≤kh \leq k, MhM_h denotes the set of s∈Gs \in G for which Card(A∩sS)=h\mathrm{Card}(A \cap sS) = h, then

∑h=1khμ(Mh)=k⋅μ(A)and∑h=1kμ(Mh)≤μ(AU). \sum_{h=1}^k h \mu(M_h) = k \cdot \mu(A) \quad \text{and} \quad \sum_{h=1}^k \mu(M_h) \leq \mu(AU) .

From this, conclude that ∑h=1k−1hμ(Mh)≤kε⋅μ(A)\sum_{h=1}^{k-1} h \mu(M_h) \leq k \varepsilon \cdot \mu(A).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.