Chapter 3: Lebesgue Measure Written by

Chapter 3: Lebesgue Measure

Written by Men-Gen Tsai email: b89902089@ntu.edu.tw 1. Let m be a countably additive measure defined for all sets in a σ-algebra

M.

If A and B are two sets in M with A ⊂ B, then mA ≤ mB. This property is called monotonicity.

Proof: B = A^S(A − B). A and A − B are disjoint. Since m is a countably additive measure, mB = mA + m(A − B). Note that m is nonnegative, and m(A − B) ≥ 0. Hence mA ≤ mB.

2. Let < E_n > be any sequence of sets in M. Then m(^SE_n) ≤ ^PmE_n. [Hint: Use Proposition 1.2.] This property of a measure is called count- able subadditivity.

Proof: By Proposition 1.2 on page 17, since M is a σ -algebra, there is a sequence < Bn > of sets in M such that BnT

Bm = φ for n 6= m

and _∞

[

i=1

B_i =

∞

[

i=1

E_i.

Since m is a countably additive measure and B_i ⊂ E_i for all i, by Problem 3.1 I have that

m(^[E_n) = m(^[B_n) = ^XmB_n≤^XmE_n. 3. If there is a set A in M such that mA < ∞, then mφ = 0.

Proof: Note that A = A^Sφ and A and φ are disjoint, and thus mA = mA + mφ.

Since mA < ∞, mφ = 0 precisely.

4.

5. Let A be the set of rational numbers between 0 and 1, and let {I_n} be a finite collection of open intervals covering A. Then ^Pl(I_n) ≥ 1.

Proof 1: (due to Meng-Gen Tsai) Since 0 ∈ A, there is an open interval J₁ in {I_n} such that 0 ∈ J₁. Let J₁ = (a₁, b₁). Note that a₁ < 0 and b₁ > 0. If b₁ ≥ 1, then

Xl(I_n) ≥ l(J₁) = b₁− a₁ ≥ 1.

Suppose not. If a₁ is rational, then I can find an open interval J₂ ∈ {I_n} such that a1 ∈ J2. If a1 is irrational, I consider the following cases.

Case 1. There is an open interval J₂ such that a₁ ∈ J₂.

Case 2. There is an open interval J₂ such that a₁ is the right endpoint of J₂.

Case 3. Otherwise.

I claim that Case 3 is impossible. Consider the following subcollection K₁, ..., K_m

where K_i ∈ {I ∈ I_n|a₁ < x for all x ∈ I}. Thus I select a open interval K0 = (x, y) nearest a1. Hence (a1, x)^TQ cannot be covered by any elements of {In}, a contradiction.

Hence we can find J₂ in Case 1 and Case 2. Continue the process to find out J₃, .... Since {I_n} is a finite covering of A, this process can be done in finite steps. Hence

Xl(In) ≥^XJn =^X(an+1− an) = am− a1. Note that a_m ≥ 1 and a_n≤ 0. Hence ^Pl(I_n) ≥ 1.

Proof 2: (due to Shin-Yi Lee) A is contained in ^SI_k, then A is con- tained in ^SI_k=^SI_k. So, |A| = 1 ≤ |^SI_k| ≤^P|I_k|.

Where |.| in Zygmund’s book is the same as m(.) is Royden’s book if I do not misunderstand.

Proof 3: (due to ljl) (1) We can suppose that Inare disjoint; otherwise, if ImT

In is not empty, then we can use I = ImT

In to replace Im and I_n. I is also an interval, and also covers A, and will make ^Pl(I_n) smaller. Hence if the sum is still ≥ 1 after our adjustment, the original one is ≥ 1 surely.

(2) Now In are disjoint, so we can suppose that In = (an, bn), n = 1, ..., N with a_i < b_i ≤ a_i+1< b_i+1 for i = 1, ..., N − 1.

(3) It is easy to show that b_i = a_i+1 (or that collection cannot cover A). Hence ^Pl(I_n) = b_N − a₁. Also, it is easy to show that a₁ < 0 and bN > 1. Proved.

6. Given any set A and any  > 0, prove that there is an open set O such that A ⊂ O and m^∗O ≤ m^∗A + . Also, prove that there is a G ∈ G_δ such that A ⊂ G and m^∗A = m^∗G.

Proof: Note that

m^∗A = inf

A⊂S

In

Xl(I_n).

where I_n are open intervals by the definition of the outer measure. Let O =^SI_n. O is also open. For any  > 0, there exists {I_n} such that

m^∗A +  ≥^Xl(I_n).

By Proposition 1 and Problem 2,

Xl(I ) = ^Xm^∗(I ) ≥ m^∗(^[I ) = m^∗O.

Combine them and I have m^∗O ≤ m^∗A + .

By previous conclusion, for any n ∈ N there is an open set O_n such that A ⊂ O_n and m^∗O_n≤ m^∗A +_n¹. Take G =^TO_n. Hence

m^∗G ≤ m^∗O_n≤ m^∗A + 1 n for all n ∈ N . Therefore

m^∗G ≤ m^∗A.

Note that A ⊂ O_n for all n, that is, A ⊂ G, that is, m^∗A ≤ m^∗G.

Hence

m^∗A = m^∗G.

7. Prove that m^∗ is translation invariant.

Proof: Let E be a set. Consider the countable collections {I_n} of open intervals that cover E. Then {I_n⁰} covers E + y where I_n⁰ = In+ y. Note that I_n⁰ is also an open interval, and

l(I_n⁰) = l(I_n).

Hence for any  > 0 there is {I_n} such that m^∗E +  > ^Pl(I_n). Thus, m^∗E +  >^Xl(I_n) =^Xl(I_n⁰) ≥ m^∗(E + y),

that is,

m^∗E ≥ m^∗(E + y).

Similarly (regard E as (E + y) − y),

m^∗E ≤ m^∗(E + y).

Hence m^∗E = m^∗(E + y), that is, m^∗ is translation invariant.

8.

9. Show that if E is a measurable set, then each translate E + y of E is also measurable.

Proof: Since E is a measurable set, for each set A I have m^∗A = m^∗(A^\E) + m^∗(A^\E^c).

Suppose x ∈ A^T(E + y), then

x ∈ A^\(E + y) ⇔ x ∈ A and x ∈ (E + y)

⇔ x ∈ A and x − y ∈ E

⇔ x − y ∈ (A − y) and x − y ∈ E

⇔ x − y ∈ (A − y)^\E

⇔ x ∈ (A − y)^\E + y.

Hence A^T(E + y) = (A − y)^TE + y; thus

m^∗(A^\(E + y)) = m^∗((A − y)^\E + y)

= m^∗((A − y)^\E)

(since Problem 3.7). Similarly, A^T(E + y)^c = (A − y)^TE^c+ y; thus m^∗(A^\(E + y)^c) = m^∗((A − y)^\E^c+ y)

= m^∗((A − y)^\E^c) Hence

m^∗(A^\(E + y)) + m^∗(A^\(E + y)^c)

= m^∗((A − y)^\E) + m^∗((A − y)^\E^c)

= m^∗(A − y)

= m^∗A

for each set A (since E is measurable). Therefore, E + y is also mea- surable.

10.

11. Show that the condition mE₁ < ∞ is necessary in Proposition 14 by giving a decreasing sequence < E_n > of measurable set with φ =^TE_n and mE_n= ∞ for each n.

Solution: Let

E_n=

n

[

k=1

A_k where A_k = (k − ¹₄, k + ¹₄).

12. Let < En > be a sequence of disjoint measurable sets and A any set.

Then

m^∗(A^\

∞

[

i=1

E_i) =

∞

X

i=1

m^∗(A^\E_i).

Proof: By Lemma 9, m^∗(A^\

n

[

i=1

E_i) =

n

X

i=1

m^∗(A^\E_i).

Since

A^\

∞

[

i=1

Ei ⊂ A^\

n

[

i=1

Ei

for all n,

m^∗(A^\

∞

[

i=1

Ei) ≥ m^∗(A^\

n

[

i=1

Ei) =

n

X

i=1

m^∗(A^\Ei).

for all n. Hence,

m^∗(A^\

∞

[

i=1

E_i) ≥

∞

X

i=1

m^∗(A^\E_i).

Similarly, by using the fact that m^∗ is nonnegative, I have m^∗(A^\

∞

[

i=1

E_i) ≤

∞

X

i=1

m^∗(A^\E_i).

Therefore I got the conclusion.

13. Prove Proposition 15. [Hints: a. Show that for m^∗E < ∞, (i) ⇒ (ii)

⇔ (vi) (cf. Proposition 5).

b. Use (a) to show that for arbitrary sets E, (i) ⇒ (ii) ⇒ (iv) ⇒ (i).

c. Use (b) to show that (i) ⇒ (iii) ⇒ (v) ⇒ (i).]

Proposition 15: Let E be a given set. Then the following five statements are equivalent:

i. E is measurable.

ii. Given  > 0, there is an open set O ⊃ E with m^∗(O − E) < .

iii. Given  > 0, there is a closed set F ⊂ E with m^∗(E − F ) < .

iv. There is a G in G_δ with E ⊂ G, m^∗(G − E) = 0.

v. There is an F in F_σ with F ⊂ E, m ∗ (E − F ) = 0.

If m^∗E is finite, the above statements are equivalent to:

vi. Given  > 0, there is a finite union U of open intervals such that m^∗(U 4 E) < .

Proof of (a):

(i) ⇒ (ii): Since E is measurable, by Proposition 5 there is an open set O such that E ⊂ O and m^∗O ≤ m^∗E + 2. By Lemma 9

m^∗O = m^∗(O − E) + m^∗(O^\E)

= m^∗(O − E) + m^∗E

since E ⊂ O). Since m^∗E is finite, m^∗(O − E) ≤ 2 < .

(ii) ⇒ (vi): Take  = 1/n for all n ∈ N , there is an open set O_n ⊂ E with m^∗(O − E) < 1/n. Take G =^TO_n; G is open and G ∈ G_δ. And

m ∗ (G − E) ≤ m ∗ (O_n− E) < 1/n for all n ∈ N . Hence m^∗(G − E) = 0.

(vi) ⇒ (ii): If not, there is a real ₀ > 0 such that

∗

for any open set O. Note that G is the intersection of countable open set, write G =^TO_n. Hence

m^∗(

n

\

k=1

O_k− E) ≥ ₀ for all n. Hence

m^∗(

∞

\

k=1

Ok− E) ≥ 0, a contradiction.

14.