26.2. ABSOLUTELY CONTINUOUS FUNCTIONS 939

increasing consistent with set inclusion of intervals is also clearly true and follows directlyfrom the definition.

Now let t < V [x,y] where P0 = {x0,x1, · · · ,xn} is a partition of [x,y] . There exists apartition, P of [x,y] such that t <VP [x,y] . Without loss of generality it can be assumed that{x0,x1, · · · ,xn}⊆P since if not, you can simply add in the points of P0 and the resulting sumfor the total variation will get no smaller. Let Pi be those points of P which are containedin [xi−1,xi] . Then

t <Vp [x,y] =n

∑i=1

VPi [xi−1,xi]≤n

∑i=1

V [xi−1,xi] .

Since t <V [x,y] is arbitrary,

V [x,y]≤n

∑i=1

V [xi,xi−1] (26.2.9)

Note that 26.2.9 does not depend on f being absolutely continuous. Suppose now that f isabsolutely continuous. Let δ correspond to ε = 1. Then if [x,y] is an interval of length nolarger than δ , the definition of absolute continuity implies

V [x,y]< 1.

Then from 26.2.9

V [a,nδ ]≤n

∑i=1

V [a+(i−1)δ ,a+ iδ ]<n

∑i=1

1 = n.

Thus V is bounded on [a,b]. Now let Pi be a partition of [xi−1,xi] such that

VPi [xi−1,xi]>V [xi−1,xi]−ε

nThen letting P = ∪Pi,

−ε +n

∑i=1

V [xi−1,xi]<n

∑i=1

VPi [xi−1,xi] =VP [x,y]≤V [x,y] .

Since ε is arbitrary, 26.2.7 follows from this and 26.2.9.Now let x < y

V (y)− f (y)− (V (x)− f (x)) = V (y)−V (x)− ( f (y)− f (x))

≥ V (y)−V (x)−| f (y)− f (x)| ≥ 0.

It only remains to verify that V is absolutely continuous.Let ε > 0 be given and let δ correspond to ε/2 in the definition of absolute continuity

applied to f . Suppose ∑ni=1 |yi− xi| < δ and consider ∑

ni=1 |V (yi)−V (xi)|. By 26.2.9 this

last is no larger than ∑ni=1 V [xi,yi] . Now let Pi be a partition of [xi,yi] such that VPi [xi,yi]+

ε

2n >V [xi,yi] . Then by the definition of absolute continuity,n

∑i=1|V (yi)−V (xi)| =

n

∑i=1

V [xi,yi]

≤n

∑i=1

VPi [xi,yi]+η < ε/2+ ε/2 = ε.