- Suppose you have where ℱ⊇ℬand also μ< ∞ for all r > 0. Let
Show that

cannot be uncountable. Explain why there exists a strictly increasing sequence r

_{n}→∞ such that μ= 0. In other words, the skin of the ball has measure zero except for possibly countably many values of the radius r. - In constructing Lebesgue Stieltjes measure on ℝ, we defined the outer measure
as
It was also shown that F

− F= μ. Show that this implies that μ is outer regular on P. That is, for any setIn particular, this holds for all A ∈ℱ the σ algebra of measurable sets. Now show that if

is a measure space such that X is a complete separable metric space (Polish space) and if μ is outer regular on ℱ⊇ℬand finite on every ball, then μ must be inner regular on each set of ℱ. That is,Hint: Let

be the increasing sequence of Problem 1. Also letB

_{n}≡ B. Thus A_{n}⊆ B_{n}. Let A ∈ℱ and A ⊆ A_{n}⊆A_{n}. Then show that there exists an open set V_{n}⊇A_{n}∖ A, V_{n}⊆ B_{n}such thatThen explain why V

_{n}^{C}∩A_{n}⊆ A and μ< ε. It might help to draw a picture on this last part. Thus there is a closed set H contained in A such that μ< ε. Now recall the interesting result about regularity in Polish space. Thus there is K compact such that μ< ε. Of course μ is not finite but μ restricted to B_{n}is. Now let F be arbitrary. Then let l < μand argue that l < μfor some n. Then use what was just shown. - Suppose you have any measure space . The problem is that it might not be a complete measure space. That is, you might have μ= 0 and G ⊆ F but Gℱ. Define the followingon P.
Show first that

is an outer measure. Next show that it agrees with μ on ℱ and that for every E ∈ℱ,From the Caratheodory procedure for constructing a measure space, there exists a σ algebra

which contains ℱ on whichis a complete measure. This is called the completion of the measure space. - Lebesgue measure was discussed. Recall that m= b − a and it is defined on a σ algebra which contains the Borel sets, more generally on P. Also recall that m is translation invariant. Let x ∼ y if and only if x − y ∈ ℚ. Show this is an equivalence relation. Now let W be a set of positive measure which is contained in. For x ∈ W, letdenote those y ∈ W such that x ∼ y. Thus the equivalence classes partition W. Use axiom of choice to obtain a set S ⊆ W such that S consists of exactly one element from each equivalence class. Let T denote the rational numbers in. Consider T + S ⊆. Explain why T + S ⊇ W. For T ≡, explain why the sets
_{j}are disjoint. Now suppose S is measurable. Then show that you have a contradiction if m= 0 since m> 0 and you also have a contradiction if m> 0 because T + S consists of countably many disjoint sets. Explain why S cannot be measurable. Thus there exists T ⊆ ℝ such thatIs there an open interval

such that if T =, then the above inequality holds? - Consider the following nested sequence of compact sets, {P
_{n}}.Let P_{1}=, P_{2}=∪, etc. To go from P_{n}to P_{n+1}, delete the open interval which is the middle third of each closed interval in P_{n}. Let P = ∩_{n=1}^{∞}P_{n}. By the finite intersection property of compact sets, P≠∅. Show m(P) = 0. If you feel ambitious also show there is a one to one onto mapping of [0,1] to P. The set P is called the Cantor set. Thus, although P has measure zero, it has the same number of points in it asin the sense that there is a one to one and onto mapping from one to the other. Hint: There are various ways of doing this last part but the most enlightenment is obtained by exploiting the topological properties of the Cantor set rather than some silly representation in terms of sums of powers of two and three. All you need to do is use the Schroder Bernstein theorem and show there is an onto map from the Cantor set to. If you do this right and remember the theorems about characterizations of compact metric spaces, Proposition 2.5.5 on Page 65, you may get a pretty good idea why every compact metric space is the continuous image of the Cantor set. - Consider the sequence of functions defined in the following way. Let f
_{1}= x on [0,1]. To get from f_{n}to f_{n+1}, let f_{n+1}= f_{n}on all intervals where f_{n}is constant. If f_{n}is nonconstant on [a,b], let f_{n+1}(a) = f_{n}(a),f_{n+1}(b) = f_{n}(b),f_{n+1}is piecewise linear and equal to( f_{n}(a) + f_{n}(b)) on the middle third of [a,b]. Sketch a few of these and you will see the pattern. The process of modifying a nonconstant section of the graph of this function is illustrated in the following picture.Show {f

_{n}} converges uniformly on [0,1]. If f(x) = lim_{n→∞}f_{n}(x), show that f(0) = 0,f(1) = 1,f is continuous, and f^{′}(x) = 0 for all xP where P is the Cantor set of Problem 5. This function is called the Cantor function.It is a very important example to remember. Note it has derivative equal to zero a.e. and yet it succeeds in climbing from 0 to 1. Explain why this interesting function is not absolutely continuous although it is continuous. Hint: This isn’t too hard if you focus on getting a careful estimate on the difference between two successive functions in the list considering only a typical small interval in which the change takes place. The above picture should be helpful. - ↑ This problem gives a very interesting example found in the book by McShane [22].
Let g(x) = x + f(x) where f is the strange function of Problem 6. Let P be the Cantor
set of Problem 5. Let [0,1] ∖ P = ∪
_{j=1}^{∞}I_{j}where I_{j}is open and I_{j}∩ I_{k}= ∅ if j≠k. These intervals are the connected components of the complement of the Cantor set. Show m(g(I_{j})) = m(I_{j}) soThus m(g(P)) = 1 because g([0,1]) = [0,2]. By Problem 4 there exists a set, A ⊆ g

which is non measurable. Define ϕ(x) = X_{A}(g(x)). Thus ϕ(x) = 0 unless x ∈ P. Tell why ϕ is measurable. (Recall m(P) = 0 and Lebesgue measure is complete.) Now show that X_{A}(y) = ϕ(g^{−1}(y)) for y ∈ [0,2]. Tell why g^{−1}is continuous but ϕ ∘ g^{−1}is not measurable. (This is an example of measurable ∘ continuous ≠ measurable.) Show there exist Lebesgue measurable sets which are not Borel measurable. Hint: The function, ϕ is Lebesgue measurable. Now recall that Borel ∘ measurable = measurable. - Let the rational numbers in be
_{k=1}^{∞}and defineShow that lim

_{n→∞}f_{n}= fwhere f is one on the rational numbers and 0 on the irrational numbers. Explain why each f_{n}is Riemann integrable but f is not. However, each f_{n}is actually a simple function and its Lebesgue and Riemann integral is equal to 0. Apply the monotone convergence theorem to conclude that f is Lebesgue integrable and in fact, ∫ fdm = 0. - Show that every countable set of real numbers is of measure zero.
- Review the Cantor set in Problem 12 on Page 232. You deleted middle third open intervals. Show that you can take out open intervals in the middle which are not necessarily middle thirds, and end up with a set C which has Lebesgue measure equal to 1 − ε. Also show if you can that there exists a continuous and one to one map f : C → J where J is the usual Cantor set of Problem 12 which also has measure 0.
- Give an example of a sequence of functions ,f
_{n}≥ 0 and a function f ≥ 0 such that f= liminf_{n→∞}f_{n}but ∫ fdm < liminf_{n→∞}∫ f_{n}dm so you get strict inequality in Fatou’s lemma. - Let f be a nonnegative Riemann integrable function defined on . Thus there is a unique number between all the upper sums and lower sums. First explain why, if a
_{i}≥ 0,Explain why there exists an increasing sequence of Borel measurable functions

converging to a Borel measurable function g, and a decreasing sequence of functionswhich are also Borel measurable converging to a Borel measurable function h such that g_{n}≤ f ≤ h_{n},dm = 0. Explain whyis a set of measure zero. Then explain why f is measurable and ∫_{a}^{b}fdx = ∫ fdm so that the Riemann integral gives the same answer as the Lebesgue integral.

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