Funky Sets

Funky Sets Groupwork - Measure Theory Stefan Maier, Jean Moraes and Hyunjung Ra Exercise 20 Show that there exist closed sets A and B s.t. m(A) = m(B) = 0 but m(A + B) > 0

Proof Let A be the cantor set, we already now that m(A) = 0. Let B = 12 A = x { 2 , x ∈ A}, then m(B) = 21 m(A) = 12 .0 = 0, which means that A and B have measure zero. It is not hard to see that A and B are closed sets. Let L the union of all open intervals that are excluded from [0, 1] in the process of construction T of the Cantor set, then A = Lc [0, 1]. Since Lc is closed, then A is closed, which implies that B is closed also. Now let A + B = {z = x + y, x ∈ A, y ∈ B}, we have to show that m(A + B) > 0.Observe that A + B is closed and the it is measurable. To show that m(A+B) > 0, we will prove that [0, 1] ⊂ A+B, which implies that m (A + B) ≥ m ([0, 1]) ≥ 1. Given z ∈ [0, 1], we can write the ternary expansion of z. z = 0.a1 a2 a3 a4 ..., where ai ∈ {0, 1, 2}∀i. Let x = 0.b1 b2 b3 b4 ..., where bi = 0 if ai = 1 and bi = ai if ai 6= 1 and y = 0.c1 c2 c3 c4 ..., where ci = 1 if ai = 1 and ci = 0 if ai 6= 1, then by construction z = x + y. But observe that the ternary expansion of x has just digits 0’s and 2’s, which implies that x ∈ A and the ternary expansion of y has just digits 0’s and 1’s, which implies that 2y ∈ A, which imply y ∈ B. Therefore, z ∈ A + B ⇒ [0, 1] ⊂ A + B ⇒ m (A + B) ≥ 1. Exercise 32a Let N denote the nonmeasurable subset of I = [0, 1] constructed in section 1.3 of Stein and Shakarchi. If E ⊂ N , E ∈ M, then m(E) = 0. Proof Consider the union of the rational translates Ek = E + rk of E, where ∞ {rk }k=1 = Q ∩ [−1, 1]:

1

∞ [

Ek ⊂ [−1, 2]

k=1

Now take the exterior measure on both sides: Ã∞ ! [ m∗ Ek ≤ m∗ ([−1, 2]) k=1

Since E ∈ M and therefore Ek ∈ M, since Ek ∩ El = ∅ ∀k 6= l and since the measure is translational invariant, we have Ã∞ ! Ã∞ ! ∞ ∞ [ [ X X m∗ Ek = m Ek = m (Ek ) = m (E) ≤ 3 k=1

k=1

k=1

k=1

⇒ m (E) = 0

Exercise 32b For all G s.t G ⊂ R,

m∗ (G) > 0 there exist E ⊂ G, such that E 6∈ M.

Proof We are going to split up this claim into two. We are going to prove a ”light version” of our claim, i.e. assume G to be bounded. But if we show that this assumption is made without loss of generality, we are done. This is the w.l.o.g.-claim: ³ ´ ˜ ⊆ G, 0 < m∗ G ˜ < ∞, G ˜ bounded. ∀G ⊂ R, m∗ (G) > 0 ∃ G We write G as a countable union of bounded sets: G=

+∞ [

(G ∩ [j, j + 1)) :=

j=−∞

+∞ [

Gj ,

j=−∞

since ∪+∞ j=−∞ [j, j + 1) = R. Taking the countable subadditivity of the exterior measure into account, this leads to: m∗ (G) ≤

+∞ X

m∗ (Gj )

(1)

j=−∞

All Gj are bounded subsets of G. So we only have to prove that one of them has exterior measure > 0. We are going to do this by assuming the contrary: Assume: m∗ (Gj ) = 0

∀j.

Because of (1) this yields m∗ (G) = 0. So we have reached contradiction and therefore at least one of the Gj must have m∗ (Gj ) > 0. 2

Now we claim the ”light version” of 32 b). This is always without loss of ˜ ⊂ G if G itsself is not bounded. generality, since we always can find a bounded G ˜ then it must be contained in G, too. If a nonmeasurable set is contained in G, ³ ´ ˜ ⊂ R, bounded, m∗ G ˜ > 0 ∃ E ⊂ G, ˜ such that E 6∈ M. For every G Consider the equivalence classes defined in setion 1.3 of Stein, Shakarchi. According to the axiom of choice, for a collection of nonempty subsets {Eα } ˜ ∃ a choice function fch : α → xα , where xα ∈ Eα . Let the subsets be of G n o ˜ x − α ∈ Q and define E = {xα }. Eα = x : x ∈ G, ˜ is bounded, ∃ I = [a, b], such that G ˜ ⊂ I. Consider now the Since G ∞ rational translates Ek = E + rk of E, where {rk }k=1 = Q ∩ [−(b − a), b − a]. So now we have: ˜⊂ G

∞ [

Ej ⊂ [2a − b, 2b − a]

j=1

Now we take the exterior measure:   ∞ ³ ´ [ ˜ ≤ m∗  0 < m∗ G Ej  ≤ 3 (b − a) j=1

Now assume E ∈ M ⇒ Ej ∈ M. Since the measure is translation invariant and the Ej are disjoint, we get 0
0 and the latter m(E) = 0. Therefore, E 6∈ M. Exercise 33 Show that N c = I − N satisfies m∗ (N c ) = 1 and ³ [ ´ m∗ (N c ) + m∗ (N ) 6= m∗ N c N

Proof Assume m∗ (N c ) < 1 then given ² > 0 ∃ an open set U s.t. U ⊂ I, N c ⊂ U and m∗ (U ) < 1 − ². If N c ⊂ U then U c ⊂ N and since U is measurable then U c is also measurable. by the previous exercise, m (U c ) = 0. S So, c However, I = U U ⇒ 1 = m (I) = m (U )+m (U c ) < 1, absurd. Therefore m∗ (N c ) = 1. Since N is nonmeasurable the m∗ (N ) > 0. Then ´ ³ [ 1 = m∗ (I) = m∗ {N } {N }c 6= m∗ ({N }c ) + m∗ (N ) > 1 + 0 = 1. 3