Bourbaki

Exercise 23, § 1

Integration, VII, p. 83

Bourbaki marks this one as harder than the rest.

Let GG be a locally compact abelian group, written additively, μ\mu a Haar measure on GG, and A,BA, B two integrable subsets of GG.

⁴Suppressed from the second edition, the lemma asserts that if f,gf, g are numerical functions ≥0\geq 0 with gg measurable, then ∫∗fgdμ=inf⁡∫∗φgdμ\int^* fg d\mu = \inf \int^* \varphi g d\mu as φ\varphi runs over the set of measurable functions ≥f\geq f.

a) For every s∈Gs \in G, let

A′=σs(A,B)=A∪(B+s),B′=τs(A,B)=(A−s)∩B. A' = \sigma_s(A, B) = A \cup (B + s), \quad B' = \tau_s(A, B) = (A - s) \cap B.

Show that μ(A′)+μ(B′)=μ(A)+μ(B)\mu(A') + \mu(B') = \mu(A) + \mu(B) and A′+B′⊂A+BA' + B' \subset A + B.

b) Assume that 0 belongs to A∩BA \cap B. A pair (A′,B′)(A', B') of integrable subsets of GG is said to be derived from (A,B)(A, B) if there exist a sequence (sk)1≤k≤n(s_k)_{1 \leq k \leq n} of elements of GG and two sequences (Ak)0≤k≤n,(Bk)0≤k≤n(A_k)_{0 \leq k \leq n}, (B_k)_{0 \leq k \leq n} of subsets of GG such that A0=A,B0=B,Ak=σsk(Ak−1,Bk−1),Bk=τsk(Ak−1,Bk−1)A_0 = A, B_0 = B, A_k = \sigma_{s_k}(A_{k-1}, B_{k-1}), B_k = \tau_{s_k}(A_{k-1}, B_{k-1}) for 1≤k≤n,sk∈Ak−11 \leq k \leq n, s_k \in A_{k-1} for 1≤k≤n1 \leq k \leq n, and A′=An,B′=BnA' = A_n, B' = B_n. Show that there exists a sequence of pairs (En,Fn)(E_n, F_n) such that E0=A,F0=B,(En+1,Fn+1)E_0 = A, F_0 = B, (E_{n+1}, F_{n+1}) is derived from (En,Fn)(E_n, F_n), and μ((En−s)∩Fn)≥μ(Fn+1)−2−n\mu((E_n - s) \cap F_n) \geq \mu(F_{n+1}) - 2^{-n} for every nn and every s∈Ens \in E_n. Set E∞=⋃nEn,F∞=⋂nFnE_\infty = \bigcup_n E_n, F_\infty = \bigcap_n F_n. Show that for every s∈E∞s \in E_\infty,

μ((E∞−s)∩F∞)=μ(F∞). \mu((E_\infty - s) \cap F_\infty) = \mu(F_\infty).

c) Assume that μ(F∞)>0\mu(F_\infty) > 0. Show that the function

f(s)=μ((E∞−s)∩F∞) f(s) = \mu((E_\infty - s) \cap F_\infty)

can only take on the values 0 and μ(F∞)\mu(F_\infty), and that the set CC of s∈Gs \in G such that f(s)=μ(F∞)f(s) = \mu(F_\infty) is open and closed, is such that μ(C)=μ(E∞)\mu(C) = \mu(E_\infty), and is the closure of E∞E_\infty (use Exer. 20 a) and b)). On the other hand let DD be the set of s∈F∞s \in F_\infty such that the intersection of F∞F_\infty with every neighborhood of ss has measure >0> 0. Show that μ(D)=μ(F∞)\mu(D) = \mu(F_\infty) and that

E∞+D⊂C; E_\infty + D \subset C;

from this, deduce that DD is contained in the subgroup H(C)H(C) defined in Exer. 22, and that H(C)H(C) is compact and open in GG. Finally, show that C+H(C)=CC + H(C) = C, that μ(C)≥μ(A)+μ(B)−μ(H(C))\mu(C) \geq \mu(A) + \mu(B) - \mu(H(C)), and that C⊂A+BC \subset A + B (consider the measure of E∞∩(c−F∞)E_\infty \cap (c - F_\infty) for every c∈Cc \in C).

d) Deduce from c) that, for two integrable subsets A,BA, B of GG: either μ∗(A+B)≥μ(A)+μ(B)\mu_*(A + B) \geq \mu(A) + \mu(B); or else there exists a compact open subgroup HH of GG such that A+BA + B contains a coset of HH, in which case μ∗(A+B)≥μ(A)+μ(B)−μ(H)\mu_*(A + B) \geq \mu(A) + \mu(B) - \mu(H). The case that GG is connected.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.