Comment on Lemma 8.2.

Let  v(E)=¡Ò_E f(t)dt,  E€B. We have seen that v(I)=0 for all open intervals I, hence v(O)=0 for all open sets O (as such O can be written as a countable union of open pairwise disjoint  intervals).

Suppose f is real valued (the case of complex f  follows easily from this).  We let 

P= { x € (a,b): f(x)¡Ý 0}, N={x € (a,b) : f(x)<0}. 

Then P, N defines a Hahn decomposition for v and

v⁺(E)=¡Ò_E X_P f dt,  v^-(E)= -¡Ò_E X_N f dt,  E € B  (X_P= the characteritic function of P, etc.) defines the Jordan decomposition of v. 

 Since Lebesgue measure is regular, there is a decreasing sequence O_n of open sets containing P with intersection G  such that G \ P has Lebesgue measure zero. Now this implies v(G \ P)=0. Since v(G)=lim_n v(O_n) =0, it follows that v(P)=0. Similarly v(N)=0. Hence f⁺= f X_P = 0 ae and f^- =-f X_N = 0 ae. Therefore, f=0 ae as claimed ?