Annalee Gomm Math 714: Assignment #2

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1 Annalee Gomm Math 714: Assignment # Verify that if A M, λ(a = 0, and B A, then B M and λ(b = 0. Suppose that A M with λ(a = 0, and let B be any subset of A. By the nonnegativity and monotonicity of Lebesgue outer measure, we have 0 λ (B λ (A = 0, and so λ (B = 0. Using Proposition 3.4 from our text, we conclude that B M. Moreover, we have λ(b = λ (B = 0, as claimed A set is called a G δ -set if it is the intersection of a countable number of open sets; and a set is called an F σ -set if it is the union of a countable number of closed sets. Note that G δ -sets and F σ -sets are Borel sets. Now suppose that M. (a Show that there is a G δ -set, G, and an F σ -set, F, such that F G and λ( \ F = λ(g \ = 0. (b Referring to part (a, deduce that λ(f = λ( = λ(g. We begin by proving the following claim: for each ɛ > 0, there is an open set, O, with O and λ(o \ < ɛ. First suppose that λ( <. We have { } λ( = λ ( = inf l(i n : {I n } open intervals, I n, and so λ(+ɛ is not a lower bound for { l(i n : {I n } open intervals, I n}. This means that there is some sequence {J n } of open intervals such that J n and l(j n < λ( + ɛ. If we put O = J n, then O is open and O. We also have λ(o λ(j n = l(j n < λ( + ɛ, and so, since λ( is finite, λ(o \ = λ(o λ( < ɛ. This proves the claim for M with finite measure. Now let M be arbitrary, and fix ɛ > 0. Put n = {x : x < n} = n ( n, n for each natural number n, so that λ( n 2n < and = n. By what we showed Math 714: Assignment #2 1

2 above, for each n we can find an open set O n such that n O n and λ(o n \ n < ɛ/2 n. Let O = O n. Then O is open, O, and ( λ(o \ = λ O n \ n completing the proof of the claim. ( = λ (O n \ n < = ɛ, λ(o n \ n ɛ/2 n For each n N, let G n be an open set containing such that λ(g n \ < 1/n. The existence of such a set is guaranteed by the claim proved above. Put G = G n. Then G is a G δ -set and G. We have for all n, so we conclude that λ(g \ = 0. λ(g \ λ(g n \ < 1/n Note that c M, so the claim above allows us to find, for each n N, an open set O n such that c O n and λ(o n \ c < 1/n. Let F n = On. c Then F n is closed (since O n is open, and c O n implies that F n. Since \ F n = O n = O n = O n \ c, we have λ( \ F n = λ(o n \ c < 1/n. Now let F = F n, so that F is an F σ -set with F. We have λ( \ F λ( \ F n < 1/n for each n, and so λ( \ F = 0. Finally, observe that since F, we have Similarly, since G, we have Hence λ(f = λ( = λ(g. λ( = λ(f + λ( \ F = λ(f + 0 = λ(f. λ(g = λ( + λ(g \ = λ( + 0 = λ(. Math 714: Assignment #2 2

3 4.16. Prove that the measure space, (R, M, λ, is the completion of the measure space, (R, B, λ B. Use the following steps: (a Verify that B M by employing xercise (b Show that B M by applying xercise (c Prove that λ = λ B. Let B B. That is, suppose we can write B = A C, where A B and C is a subset of some Borel set with Lebesgue measure 0. By xercise 3.32, C M. Since A B M and C M, the union B = A C must be in M as well, M being closed under unions. This shows that B M. Now let M. Choose F B and G B as in xercise 3.44 so that F G, λ( \ F = λ(g \ = 0, and λ(f = λ( = λ(g. Note that G \ F = G F c B since B is closed under complements and unions, and that λ(g \ F = λ ( (G \ ( \ F = λ( \ F + λ(g \ = 0. This means that \ F is a subset of a Borel set with measure 0. We have = F ( \ F, and so is the union of a Borel set and a subset of a Borel set with measure 0. This means that B, and consequently we conclude that B M. Having proved that M = B, it remains to show that λ = λ B. To this end, suppose M = B. We can write = A C, where A B and C is a subset of some Borel set with Lebesgue measure 0 (so λ(c = 0. We have A, hence λ(a λ(, and λ( = λ(a C λ(a + λ(c = λ(a + 0. It follows that λ( = λ(a. Since λ B ( = λ(a by definition, we have λ( = λ B (, completing the proof Suppose that f L 1 (, A, µ. Show that for each ɛ > 0, there is a δ > 0 such that µ( < δ implies f dµ < ɛ. Define a nondecreasing sequence {f n } of nonnegative functions by { f(x if f(x n; f n (x = n otherwise for each n N. This sequence converges pointwise to f, so by the Monotone Convergence Theorem, lim f n dµ = f dµ. n Math 714: Assignment #2 3

4 Since f n dµ f dµ < for all n, this means that we can choose an n large enough to ensure f dµ f n dµ = ( f f n dµ < ɛ/2. (1 Next, note that f n n, and so f n dµ n dµ = nµ( for any A. Put δ = ɛ/2n. Then µ( < δ implies f n dµ nµ( < ɛ/2. (2 It follows that if µ( < δ, then f dµ = ( f f n dµ + f n dµ ( f f n dµ + f n dµ } {{ } }{{ } < ɛ, which is what we wanted to show. < ɛ/2 by (1 < ɛ/2 by ( Denote by B 2 the smallest σ-algebra of subsets of R 2 that contains all open sets of R 2. Members of B 2 are called two-dimensional Borel sets. (a Show that B 2 = B B. (b A measure on B 2 is called a two-dimensional Borel measure. Suppose that µ and ν are finite two-dimensional Borel measures such that µ(a B = ν(a B for all A, B B. Prove that µ = ν. Let us first prove that B 2 is contained in B B. To do this we will use the fact that the set, T, of open rectangles of the form (a, b (c, d with a, b, c, d Q, forms a countable basis for the topology on R 2. If U is an open subset of R 2, then we can write U as a countable union of elements of T. But each element of T is of the form (a, b (c, d, where (a, b B and (c, d B, so clearly elements of T are contained in B B. Hence U is a countable union of elements of B B, and therefore U B B. This shows that B B is a σ-algebra containing all open sets of R 2. Since B 2 is the smallest σ-algebra containing the open sets, it follows that B 2 B B. To show that B B B 2, we will show that every measurable rectangle is in B 2. Consider the set B x = {A R : A B} of subsets of R 2. Note that (A c R = (A R c and ( A n R = (A n R. Math 714: Assignment #2 4

5 Therefore, since each A B can be obtained by taking complements and countable unions of open subsets of R, each A R B x can be obtained by taking complements and countable unions of sets of the form (U R, where U R is open. But each (U R is open in R 2, and therefore is an member of B 2, as are complements and countable unions of sets of this form. It follows that B x B 2. Similar reasoning shows that B y = {R B : B B} B 2. Now suppose that A B and B B. Then A B = (A R (R B is the intersection of two members of B 2, and so A B B 2 since B 2 is closed under unions. This shows that B 2 is a σ-algebra containing the set U = {A B : A B and B B} of measurable rectangles. As B B is the smallest σ-algebra containing U, it follows that B B B 2. Having also shown the reverse containment, we now conclude that B 2 = B B. Now suppose that µ and ν are finite two-dimensional Borel measures such that µ(a B = ν(a B for all A, B B. This means that µ U = ν U, where U = {A B : A B and B B} is the semialgebra of measurable rectangles of R 2. The σ-algebra generated by U is B B, which by what we showed above is equal to B 2. Since µ and ν are finite measures, Corollary 4.7 one of the corollaries to the uniqueness portion of the xtension Theorem implies that µ = ν on the measurable space (R 2, B 2.. Math 714: Assignment #2 5