Solutions of Homework #4 1.

Solutions of Homework #4

1. Claim1 : λ(E + s) = λ(E).

λ^∗(E + s) = inf {Σλ^∗(I_n)|[

I_n⊇ E + s, I_n: open}

= inf {Σλ^∗(I_n− s)|[

(I_n− s) ⊇ E, I_n: open}

= inf {Σλ^∗(In)|[

(In− s) ⊇ E, In: open}

= inf {Σλ^∗(I_n)|[

I_n⊇ E, I_n: open} = λ^∗(E) Therefore, λ(E + s) = λ(E).

Claim2 : E ∈ L ⇒ E + s ∈ L.

Let A ⊆ R and B = A − s.

λ^∗(A ∩ (E + s)) + λ^∗(A \ (E + s))

= λ^∗((B + s) ∩ (E + s)) + λ^∗((B + s) \ (E + s))

= λ^∗((B ∩ E) + s) + λ^∗((B \ E) + s)

= λ^∗(B ∩ E) + λ^∗(B \ E) = λ^∗(B) = λ^∗(A).

Therefore, E + s ∈ L.

Claim3 : λ(rE) = |r|λ(E), r ∈ R.

λ^∗(rE) = inf {Σλ^∗(In)|[

In ⊇ rE, In: open}

= inf {Σλ^∗(rI_n)|[

I_n⊇ rE, I_n: open}

= |r|inf {Σλ^∗(I_n)|[

I_n ⊇ E, I_n: open} = |r|λ(E).

Therefore, λ(rE) = |r|λ(E).

Claim4 : E ∈ L ⇒ rE ∈ L.

If r = 0, then rE = {0} ∈ L.

For r 6= 0, let A ⊆ R and B = ¹_rA.

λ^∗(A ∩ (rE)) + λ^∗(A \ (rE))

= λ^∗((rB) ∩ (rE)) + λ^∗((rB) \ (rE))

= λ^∗(r(B ∩ E)) + λ(r(B \ E))

= |r|λ^∗(B ∩ E) + |r|λ(B \ E)

= |r|λ^∗(B) = λ^∗(A) Therefore rE ∈ L.

2. Let E be Lebesgue measurable with λ(E) < ∞. Then λ(E) = λ^∗(E).

By the definition of λ^∗, ∀ε > 0, ∃ open intervals {A_n} such that S∞

n=1An⊃ E and λ(

∞

[

n=1

An) ≤

∞

X

n=1

λ(An) < λ^∗(E) + ε

2 = λ(E) + ε

2 (1)

Since λ(E) < ∞, P∞

n=1λ(A_n) converges.

Therefore with respect to this ε, ∃n₀ ∈ N s.t.

∞

X

n=n0+1

λ(A_n) < ε

2. (2)

Let G =Sn0

n=1A_n, then because A⁰_ns are open intervals , we can write G = SK0

k=1Ik, where Ik is a finite sequence of mutually disjoit open intervals. Thus, we have

G M E ⊂ (G \ E) ∪ (

∞

[

n=n0+1

A_n) ⊂ ((

∞

[

n=1

A_n) \ E) ∪ (

∞

[

n=n0+1

A_n)

Since E is Lebesgue measurable and by (1) (2), we have λ(G M E) < ε

2+ ε 2 = ε 3. Let B1 = [0.5, 0.6) and

B_n= {0.a₁a₂· · · a_n−15|a_i ∈ I, ∀i = 1, . . . , n−1}, where I = 0, 1, 2, . . . , 9.

Let A₁ = ([0, 1] \ B₁), A₂ = (A₁\ B₂), . . ., A_n= (A_n−1\ B_n).

Therefore,

A₁ ⊇ A₂ ⊇ · · · ⊇ A_n⊇ · · · .

⇒ A =T∞ n=1A_n.

Since λ(A_n) = (₁₀⁹)ⁿ, ⇒ λ(A₁) = ₁₀⁹ < ∞. Thus we have

λ(A) = λ(

∞

\

n=1

A_n) = lim

n→∞λ(A_n) = lim

n→∞( 9

10)ⁿ = 0 4. There are two examples! Given 0 < ε < 1.

1^o The first stage of the construction is to subdivide [0, 1] into three parts. The length of the middle part is ₃^ε, and the remaining two parts have equal length. We denote the middle part by A₁. At the second stage, we subdivide each of the remaining two parts at the first stage into three parts. The length of each middle part is

ε

9. Then let A₂ = A₁∪(the two middle parts at the second stage).

Use similar algorithm to construct {A_k} for all k ∈ N and observe that A_k % and each A_k is measurable.

Set E =S∞ n=1A_n.

λE = lim

n→∞λA_n= ε

3+ 2 ∗ ε

9+ 4 ∗ ε 3³ + · · ·

= ε 3(1 + 2

3 + (2

3)²+ · · · )

= ε 3∗ 1

1 − ²₃ = ε Therefore λ(E) = ε.

Claim : E is dense in [0, 1].

Pf :∀x ∈ [0, 1]

case< 1 >: x ∈ E ⇒ ∀r > 0B(x, r) ∩ E 6= ∅.

case< 2 >: x ∈ ([0, 1] \ E) ⇒ x ∈ (T∞

n=1(A_n)^c) ∩ [0, 1].

⇒ x ∈ A^c_n ∩ [0, 1], ∀n. We can estimate dist(x, An). For example,

dist(x, A_n) ≤ 1 − λ(An) 2ⁿ < 1

2ⁿ. Thus x is a limit point ofS A_n = E.

2^o Let E⁰ = (0, ε), and E = E⁰∪ (Q ∩ [0, 1]).

Since Q is dense in [0, 1], E is also dense in [0, 1]. Moreover, ε = λ(0, ε) ≤ λ(E) ≤ λ(E⁰) + λ(Q) = ε + 0 = ε Therefore λ(E) = ε.