8.8. EXERCISES 211

there could be points of J other than a,b on the vertical sides of C.) In the first case wherew is above l, go from c to q along L and then along Jb to p then to z and along η to thetop line of the box. (Note that neither q nor p can be in {a,b} so this cannot intersect Jt .)This curve does not intersect Jt in contradiction to Proposition 8.7.20 since Jt has pointswith first components a,b on the extreme left and right and this new curve through z hassome second components equal to c,d. If w is below l, use the curve along L from d to lto z and then along η to the bottom line of C. This fails to intersect Jb and so contradictsProposition 8.7.20 because this curve has some second components equal to c,d and thecurve Jb has some first components equal to a,b. Thus the component which contains z isa bounded component. Hence JC has at least two components.

Suppose V is another bounded component and that U is the bounded component justdescribed containing z. Thus V is contained in the inside of the box C. Moving up on L letr be the last point of J encountered. Thus by definition, r ∈ Jt . Moving down on L from rlet q be the last point of J encountered. It is in Jb as explained earlier. Now go from r to lon Jt . Neither of r, l is an endpoint of Jt . Then go from l to p along the segment in U andfrom p to q on Jb avoiding the end points a,b. Including a ray from r pointing up and aray from q pointing down, this set of points B contains no points of V because the segmentbetween l,p is in U . Also a and b are in different components of BC. Now for δ smallenough, B(a,δ ) and B(b,δ ) contain no points of B and by Lemma 8.7.17, there is a pointa1 of V in B(a,δ ) and a point b1 of V in B(b,δ ) and so V fails to be connected after all. ■

8.8 Exercises1. Let (X ,F ,µ) be a regular measure space. For example, it could be R with Lebesgue

measure. Why do we care about a measure space being regular? This problem willshow why. Suppose that closures of balls are compact as in the case of R.

(a) Let µ (E)<∞. By regularity, there exists K ⊆E ⊆V where K is compact and Vis open such that µ (V \K)< ε . Show there exists W open such that K ⊆ W̄ ⊆Vand W̄ is compact. Now show there exists a function h such that h has values in[0,1] ,h(x) = 1 for x ∈ K, and h(x) equals 0 off W . Hint: You might considerProblem 12 on Page 154.

(b) Show that∫|XE −h|dµ < ε

(c) Next suppose s = ∑ni=1 ciXEi is a nonnegative simple function where each

µ (Ei) < ∞. Show there exists a continuous nonnegative function h whichequals zero off some compact set such that

∫|s−h|dµ < ε

(d) Now suppose f ≥ 0 and f ∈ L1 (Ω) . Show that there exists h ≥ 0 which iscontinuous and equals zero off a compact set such that∫

| f −h|dµ < ε

(e) If f ∈ L1 (Ω) with complex values, show the conclusion in the above part ofthis problem is the same. That is, Cc (Rp) is dense in L1 (Rp).

2. Let F be an increasing function defined on R. For f a continuous function havingcompact support in R, consider the functional L f ≡

∫f dF where here the integral

signifies∫ b

a f dF where spt( f )⊆ [a,b] and the integral is the ordinary Rieman Stielt-jes integral. For a discussion of these, see my single variable advanced calculus