26.2. ABSOLUTELY CONTINUOUS FUNCTIONS 941

Note that up till now, no referrence has been made to the absolute continuity of f . Anyincreasing continuous function would be fine.

Now if E is a Borel set such that m(E) = 0, Then the outer regularity of m impliesthere exists an open set, V containing E such that m(V ) < δ where δ corresponds to ε inthe definition of absolute continuity of f . Then letting {Ik} be the connected componentsof V it follows E ⊆∪∞

k=1Ik with ∑k m(Ik) = m(V )< δ . Therefore, from absolute continuityof f , it follows that for Ik = (ak,bk) and each n

µ (∪nk=1Ik) =

n

∑k=1

µ (Ik) =n

∑k=1| f (bk)− f (ak)|< ε

and so letting n→ ∞,

µ (E)≤ µ (V ) =∞

∑k=1| f (bk)− f (ak)| ≤ ε.

Since ε is arbitrary, it follows µ (E) = 0. Therefore, µ≪m and so by the Radon Nikodymtheorem there exists a unique h ∈ L1 ([a,b]) such that

µ (E) =∫

Ehdm.

In particular,

µ ([a,x]) = f (x)− f (a) =∫[a,x]

hdm.

From the fundamental theorem of calculus f ′ (x) = h(x) at every Lebesgue point of h.Therefore, writing in usual notation,

f (x) = f (a)+∫ x

af ′ (t)dt

as claimed. This proves the lemma.With the above lemmas, the following is the main theorem about absolutely continuous

functions.

Theorem 26.2.5 Let f : [a,b]→R be absolutely continuous if and only if f ′ (x) exists a.e.,f ′ ∈ L1 ([a,b]) and

f (x) = f (a)+∫ x

af ′ (t)dt.

Proof: Suppose first that f is absolutely continuous. By Lemma 26.2.3 the total varia-tion function, V is absolutely continuous and f (x) = V (x)− (V (x)− f (x)) where both Vand V − f are increasing and absolutely continuous. By Lemma 26.2.4

f (x)− f (a) = V (x)−V (a)− [(V (x)− f (x))− (V (a)− f (a))]

=∫ x

aV ′ (t)dt−

∫ x

a(V − f )′ (t)dt.