Bourbaki

Exercise 8, § 4

Integration, IV, p. 122

Bourbaki marks this one as harder than the rest.

If the torus T=R/ZT = \mathbf{R}/\mathbf{Z} is canonically identified with the interval [0,1[[0, 1[ of R\mathbf{R}, the Lebesgue measure μ\mu on this interval is a measure on TT; for every set A⊂TA \subset T and every z∈Tz \in T, μ∗(A+z)=μ∗(A)\mu^*(A + z) = \mu^*(A).

a) Show that there exists a subgroup H0H_0 of TT such that the sets Hn=rn+H0H_n = r_n + H_0, where rnr_n runs over the set of rational numbers contained in [0,1[[0, 1[, form a partition of TT (regarding R\mathbf{R} as a vector space over Q\mathbf{Q}, note that in R\mathbf{R} the subspace Q\mathbf{Q} admits a supplement).

b) Let HH be a set that is the union of any finite number of sets HnH_n; show that HH is not integrable and that μ∗(H)=0\mu_*(H) = 0 (note that every subgroup of the additive group Q/Z\mathbf{Q}/\mathbf{Z} generated by a finite number of elements has infinite index in Q/Z\mathbf{Q}/\mathbf{Z}; from this, deduce that there exists a partition of TT into a countable infinity of sets PnP_n, each of which contains a set of the form z+Hz + H).

c) Deduce from b) an example of a decreasing sequence (An)(A_n) of subsets of TT, whose intersection is empty, and is such that μ∗(An)=1\mu^*(A_n) = 1 for all nn.

d) Let AA be an integrable set such that H0⊂AH_0 \subset A and μ(A)=μ∗(H0)>0\mu(A) = \mu^*(H_0) > 0; show that, for every integrable set B⊂AB \subset A, the sets B1=B∩H0B_1 = B \cap H_0 and B2=B∩CH0B_2 = B \cap \mathbf{C}H_0 form a partition of BB such that μ∗(B1)=μ∗(B2)=μ(B)\mu^*(B_1) = \mu^*(B_2) = \mu(B) and μ∗(B1)=μ∗(B2)=0\mu_*(B_1) = \mu_*(B_2) = 0 (make use of Exer. 7 b)).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.