Chapter 3: Lebesgue Measure

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1 Chapter 3: Lebesgue Measure Written by Men-Gen Tsai 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 (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( E n ) me n. [Hint: Use Proposition 1.2.] This property of a measure is called countable subadditivity. Proof: By Proposition 1.2 on page 17, since M is a σ -algebra, there is a sequence < B n > of sets in M such that B n Bm = φ for n m and B i = 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 ) = mb n me n. 3. If there is a set A in M such that ma <, then mφ = 0. Proof: Note that A = A φ and A and φ are disjoint, and thus ma = ma + mφ. Since ma <, mφ = 0 precisely. 1

2 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 l(i n ) 1. Proof 1: (due to Meng-Gen Tsai) Since 0 A, there is an open interval J 1 in {I n } such that 0 J 1. Let J 1 = (a 1, b 1 ). Note that a 1 < 0 and b 1 > 0. If b 1 1, then l(in ) l(j 1 ) = b 1 a 1 1. Suppose not. If a 1 is rational, then I can find an open interval J 2 {I n } such that a 1 J 2. If a 1 is irrational, I consider the following cases. Case 1. There is an open interval J 2 such that a 1 J 2. Case 2. There is an open interval J 2 such that a 1 is the right endpoint of J 2. Case 3. Otherwise. I claim that Case 3 is impossible. Consider the following subcollection K 1,..., K m where K i {I I n a 1 < x for all x I}. Thus I select a open interval K 0 = (x, y) nearest a 1. Hence (a 1, x) Q cannot be covered by any elements of {I n }, a contradiction. Hence we can find J 2 in Case 1 and Case 2. Continue the process to find out J 3,... Since {I n } is a finite covering of A, this process can be done in finite steps. Hence l(in ) J n = (a n+1 a n ) = a m a 1. Note that a m 1 and a n 0. Hence l(i n ) 1. 2

3 Proof 2: (due to Shin-Yi Lee) A is contained in I k, then A is contained in I k = I k. So, A = 1 I k 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 I n are disjoint; otherwise, if I m In is not empty, then we can use I = I m In to replace I m and I n. I is also an interval, and also covers A, and will make l(i n ) smaller. Hence if the sum is still 1 after our adjustment, the original one is 1 surely. (2) Now I n are disjoint, so we can suppose that I n = (a n, b n ), 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 l(i n ) = b N a 1. Also, it is easy to show that a 1 < 0 and b N > 1. Proved. 6. Given any set A and any ɛ > 0, prove that there is an open set O such that A O and m O m A + ɛ. Also, prove that there is a G G δ such that A G and m A = m G. Proof: Note that m A = inf l(in ). A I n where I n are open intervals by the definition of the outer measure. Let O = I n. O is also open. For any ɛ > 0, there exists {I n } such that m A + ɛ l(i n ). By Proposition 1 and Problem 2, l(in ) = m (I n ) m ( I n ) = m O. 3

4 Combine them and I have m O m A + ɛ. 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 + 1 n. Take G = O 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 n} covers E +y where I n = I n +y. Note that I n is also an open interval, and l(i n) = l(i n ). Hence for any ɛ > 0 there is {I n } such that m E + ɛ > l(i n ). Thus, m E + ɛ > l(i n ) = l(i n) 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. 4

5 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 (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 (E + y) = (A y) E + y; thus m (A (E + y)) = m ((A y) E + y) = m ((A y) E) (since Problem 3.7). Similarly, A (E + y) c = (A y) E 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 measurable. 5

6 Show that the condition me 1 < is necessary in Proposition 14 by giving a decreasing sequence < E n > of measurable set with φ = E n and me n = for each n. Solution: Let n E n = where A k = (k 1 4, k ). A k k=1 12. Let < E n > be a sequence of disjoint measurable sets and A any set. Then m (A E i ) = m (A E i ). Proof: By Lemma 9, m (A n n E i ) = m (A E i ). Since A E i A n E i i=1 i=1 for all n, m (A E i ) m (A n n E i ) = m (A E i ). i=1 for all n. Hence, m (A E i ) m (A E i ). Similarly, by using the fact that m is nonnegative, I have m (A E i ) m (A E i ). Therefore I got the conclusion. 6

7 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 ɛ > 0, there is an open set O E with m (O E) < ɛ. iii. Given ɛ > 0, there is a closed set F E with m (E F ) < ɛ. 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 ɛ > 0, there is a finite union U of open intervals such that m (U E) < ɛ. 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 + 2ɛ. 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) 2ɛ < ɛ. (ii) (vi): Take ɛ = 1/n for all n N, there is an open set O n E with m (O E) < 1/n. Take G = O 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 ɛ 0 > 0 such that m (O E) ɛ 0 7

8 for any open set O. Note that G is the intersection of countable open set, write G = O n. Hence n m ( O k E) ɛ 0 k=1 for all n. Hence a contradiction. m ( O k E) ɛ 0, k=