8.9. EXERCISES 199

of Problem 16. This function is called the Cantor function.It is a very importantexample to remember. Note it has derivative equal to zero a.e. and yet it succeedsin climbing from 0 to 1. Explain why this interesting function cannot be recoveredby integrating its derivative. (It is not absolutely continuous, explained later.) Hint:This isn’t too hard if you focus on getting a careful estimate on the difference betweentwo successive functions in the list considering only a typical small interval in whichthe change takes place. The above picture should be helpful.

18. ↑ This problem gives a very interesting example found in the book by McShane[33]. Let g(x) = x+ f (x) where f is the strange function of Problem 17. Let P be theCantor set of Problem 16. Let [0,1]\P=∪∞

j=1I j where I j is open and I j∩Ik = /0 if j ̸=k. These intervals are the connected components of the complement of the Cantor set.Show m(g(I j)) = m(I j) so m(g(∪∞

j=1I j)) = ∑∞j=1 m(g(I j)) = ∑

∞j=1 m(I j) = 1. Thus

m(g(P)) = 1 because g([0,1]) = [0,2]. By Problem 15 there exists a set, A ⊆ g(P)which is non measurable. Define φ(x) =XA(g(x)). Thus φ(x) = 0 unless x∈ P. Tellwhy φ is measurable. (Recall m(P) = 0 and Lebesgue measure is complete.) Nowshow that XA(y) = φ(g−1(y)) for y ∈ [0,2]. Tell why g is strictly increasing andg−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 arenot Borel measurable. Hint: The function, φ is Lebesgue measurable. Now recallthat Borel ◦ measurable = measurable.

19. Show that every countable set of real numbers is of Lebesgue measure zero.

20. The Cantor set is obtained by starting with [0,1], delete the middle third, the open set(1/3,2/3). Now do the same for the two remaining closed intervals. This results ina nested sequence of compact sets. The intersection of all of these is the Cantor set.Show that you can take out open intervals in the middle which are not necessarilymiddle 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→ Jwhere J is the usual Cantor set which also has measure 0.

21. Suppose you have a π system K of sets of Ω and suppose G ⊇K and that G isclosed with respect to complements and that whenever {Fk} is a decreasing sequenceof sets of G it follows that ∩kFk ∈ G . Show that then G contains σ (K ). This is analternative formulation of Dynkin’s lemma. It was shown after the Dynkin lemmathat closure with respect to countable intersections is equivalent.

22. Let (Ω,F ,µ) be a measure space and let s(ω) = ∑ni=0 ciXEi (ω) where the Ei are

distinct measurable sets but the ci might not be. Thus the ci are the finitely manyvalues of s. Say each ci ≥ 0 and c0 = 0. Define

∫sdµ as ∑i ciµ (Ei). Show that this is

well defined and that if you have s(ω) = ∑ni=1 ciXEi (ω) , t (ω) = ∑

mj=1 d jXFj (ω) ,

then for a,b nonnegative numbers, as(ω)+ bt (ω) can be written also in this formand that

∫(as+bt)dµ = a

∫sdµ + b

∫tdµ . Hint: s(ω) = ∑i ∑ j ciXEi∩Fj (ω) =

∑ j ∑i ciXEi∩Fj (ω) and (as+bt)(ω) = ∑ j ∑i (aci +bd j)XEi∩Fj (ω).

23. ↑Having defined the integral of nonnegative simple functions in the above problem,letting f be nonnegative and measurable. Define∫

f dµ ≡ sup{∫

sdµ : 0≤ s≤ f ,s simple}.