940 CHAPTER 26. INTEGRALS AND DERIVATIVES

and shows V is absolutely continuous as claimed.

Lemma 26.2.4 Suppose f : [a,b]→ R is absolutely continuous and increasing. Then f ′

exists a.e., is in L1 ([a,b]) , and

f (x) = f (a)+∫ x

af ′ (t)dt.

Proof: Define L, a positive linear functional on C ([a,b]) by

Lg≡∫ b

agd f

where this integral is the Riemann Stieltjes integral with respect to the integrating function,f . By the Riesz representation theorem for positive linear functionals, there exists a uniqueRadon measure, µ such that Lg =

∫gdµ. Now consider the following picture for gn ∈

C ([a,b]) in which gn equals 1 for x between x+1/n and y.

x y+1/nx+1/n y

Then gn (t)→X(x,y] (t) pointwise. Therefore, by the dominated convergence theorem,

µ ((x,y]) = limn→∞

∫gndµ.

However, (f (y)− f

(x+

1n

))≤

∫gndµ =

∫ b

agnd f ≤

(f(

y+1n

)− f (y)

)+

(f (y)− f

(x+

1n

))+

(f(

x+1n

)− f (x)

)and so as n→ ∞ the continuity of f implies

µ ((x,y]) = f (y)− f (x) .

Similarly, µ (x,y) = f (y)− f (y) and µ ([x,y]) = f (y)− f (x) , the argument used to estab-lish this being very similar to the above. It follows in particular that

f (x)− f (a) =∫[a,x]

dµ.