Integration, VII, p. 83
Bourbaki marks this one as harder than the rest.
Let G be a locally compact abelian group, written additively, μ a Haar measure on G, and A,B two integrable subsets of G.
⁴Suppressed from the second edition, the lemma asserts that if f,g are numerical functions ≥0 with g measurable, then ∫∗fgdμ=inf∫∗φgdμ as φ runs over the set of measurable functions ≥f.
a) For every s∈G, let
A′=σs(A,B)=A∪(B+s),B′=τs(A,B)=(A−s)∩B.
Show that μ(A′)+μ(B′)=μ(A)+μ(B) and A′+B′⊂A+B.
b) Assume that 0 belongs to A∩B. A pair (A′,B′) of integrable subsets of G is said to be derived from (A,B) if there exist a sequence (sk)1≤k≤n of elements of G and two sequences (Ak)0≤k≤n,(Bk)0≤k≤n of subsets of G such that A0=A,B0=B,Ak=σsk(Ak−1,Bk−1),Bk=τsk(Ak−1,Bk−1) for 1≤k≤n,sk∈Ak−1 for 1≤k≤n, and A′=An,B′=Bn. Show that there exists a sequence of pairs (En,Fn) such that E0=A,F0=B,(En+1,Fn+1) is derived from (En,Fn), and μ((En−s)∩Fn)≥μ(Fn+1)−2−n for every n and every s∈En. Set E∞=⋃nEn,F∞=⋂nFn. Show that for every s∈E∞,
μ((E∞−s)∩F∞)=μ(F∞).
c) Assume that μ(F∞)>0. Show that the function
f(s)=μ((E∞−s)∩F∞)
can only take on the values 0 and μ(F∞), and that the set C of s∈G such that f(s)=μ(F∞) is open and closed, is such that μ(C)=μ(E∞), and is the closure of E∞ (use Exer. 20 a) and b)). On the other hand let D be the set of s∈F∞ such that the intersection of F∞ with every neighborhood of s has measure >0. Show that μ(D)=μ(F∞) and that
E∞+D⊂C;
from this, deduce that D is contained in the subgroup H(C) defined in Exer. 22, and that H(C) is compact and open in G. Finally, show that C+H(C)=C, that μ(C)≥μ(A)+μ(B)−μ(H(C)), and that C⊂A+B (consider the measure of E∞∩(c−F∞) for every c∈C).
d) Deduce from c) that, for two integrable subsets A,B of G: either μ∗(A+B)≥μ(A)+μ(B); or else there exists a compact open subgroup H of G such that A+B contains a coset of H, in which case μ∗(A+B)≥μ(A)+μ(B)−μ(H). The case that G is connected.