- Let ℕ be the positive integers and let ℱ denote the set of all subsets of ℕ. Explain why ℕ
is a σ algebra. You could let μbe the number of elements of S. This is called counting measure. Explain why μ is a measure.
- Show f : Ω → ℝ is measurable if and only if f
^{−1}is measurable whenever U is an open set. Hint: This is pretty easy if you recall that every open set is the disjoint union of countably many connected components. - The smallest σ algebra on ℝ which contains the open intervals, denoted by ℬ is called the Borel sets. Show that ℬ contains all open sets and is also the smallest σ algebra which contains all open sets. Show that all continuous functions g : ℝ → ℝ are ℬ measurable. A word of advice pertaining to Borel sets: Don’t try to describe a typical Borel set. Instead, use the definition that it is a set in the smallest σ algebra containing the open sets.
- Show that f : Ω → ℝ is measurable if and only if f
^{−1}is measurable for every Borel ℬ. Recall ℬ is the smallest σ algebra which contains the open sets. Hint: Let G be those sets B such that f^{−1}is measurable. Argue it is a σ algebra. - Now suppose f : Ω → ℝ where is a measureable space. Suppose g : ℝ → ℝ is ℬ measurable. Explain why g ∘ f is ℱ measurable.
- The open sets in ℝ
^{n}are defined to be all sets U which are unions of open rectangles of the formShow that all open sets in ℝ

^{n}are a countable union of such open rectangles. If a pi system K consists of products of open intervals like the above, show that σis ℬ the Borel sets. Hint: There are countably many open rectangles of the formShow that an arbitrary open rectangle is the union of open rectangles of this sort having rational end points.

- ↑Show that a set of the form ∏
_{i=1}^{n}B_{i}is a Borel set in ℝ^{n}if each B_{i}is a Borel set in ℝ. The Borel sets in ℝ^{n}are the smallest σ algebra which contains the open sets. Hint: You might let f_{i}: ℝ^{n}→ ℝ be the projection map. Explain why f_{i}^{−1}is a Borel set when B is a Borel set in ℝ. You know f_{i}is continuous and that it follows that it is Borel measurable. Now consider intersections of sets like this. - Lebesgue measure was discussed. Recall that m= b−a and it is defined on a σ algebra which contains the Borel sets. It comes from an outer measure defined 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. - 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 9. 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 [94]. Let
g(x) = x + f(x) where f is the strange function of Problem 10. Let P be the Cantor set of Problem 9.
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 8 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.

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