14.14. EXERCISES 351
measurable.) Show there exist Lebesgue measurable sets which are not Borel mea-surable. Hint: The function, φ is Lebesgue measurable. Now recall that Borel ◦measurable = measurable.
18. Suppose you have a function f which is of bounded variation on every closed in-terval. Explain why this function has a derivative off some Borel set of measurezero.
19. Suppose f : R→ R is Lipschitz. This means | f (x)− f (y)| ≤ K |x− y| . Then letg(x) ≡ (K +1)x− f (x) . Explain why g is monotone. Note f (x) = (K +1)x−g(x) . Explain why f ′ (x) exists off a Borel set of measure zero. This is a case ofRademacher’s theorem which says that a Lipschitz map is differentiable almost ev-erywhere. In the general case, the function is vector valued and defined on Rp ratherthan R.
20. Suppose f ≥ 0, is Lebesgue measurable, see Problem 14, or if you didn’t do that one,let it be Borel measurable. Also suppose f X[a,b] has a finite integral for any finiteinterval [a,b]. Let F (x)≡
∫ x0 f (t)dm. Show that this function of x is continuous ev-
erywhere on [a,b] and then explain why this function has a derivative off some Borelset of measure zero. Hint: To show continuous, note that for x < b,X[x,x+h] f (x)→ 0as h→ 0. Consider using the dominated convergence theorem.
21. Let S ̸= /0 and define dist(x,S)≡ inf{|x− y| : y ∈ S} . Show x→ dist(x,S) is contin-uous. In fact, show that
|dist(x,S)−dist(y,S)| ≤ |x− y|
22. ↑From regularity theorems for measures on B (R), you know that if µ (E)< ∞, thenthere is a compact K and open V such that K ⊆ E ⊆ V and µ (V \K) < δ . Here µ
is a Borel measure which is finite on bounded sets. Show that for such E, there isa continuous function f which is 1 on K, 0 off V , and has values in [0,1] . Use thisto show that there exists a continuous function h such that
∫|XE −h|dµ < ε. Hint:
You might get such compact sets Kn and open sets Vn such that µ (Vn \Kn) < 1/2n
and let hn be as just shown. Then do some sort of argument involving the dominatedconvergence theorem.
23. ↑Let µ be a measure on B (R) which is finite on bounded intervals and∫| f |dµ < ∞.
Show there exists a simple function s such that |s(x)| ≤ | f (x)| for all x and∫| f (x)− s(x)|dµ < ε/2
Next show there is a continuous function h such that∫|s(x)−h(x)|dµ < ε/2
Conclude that if f is real valued and measurable with∫| f |dµ < ∞, then there is h
continuous such that∫| f −h|dµ < ε .
24. ↑Let µ be a measure on B (R) which is finite on bounded intervals and∫| f |dµ < ∞.
Show that if fn ≡ f X[−n,n], then if n is large enough,∫| f − fn|dµ < ε