1
Absolutely Continuous Functions
Def. A real valued function 𝑓 on a closed interval [𝑎, 𝑏] is said to be absolutely
continuous on [𝑎, 𝑏] if for each 𝜖 > 0, there is a 𝛿 > 0 such that for every
disjoint collection {(𝑎𝑘 , 𝑏𝑘 )}𝑛
𝑘=1 of open intervals in (𝑎, 𝑏) if
∑𝑛𝑘=1|𝑏𝑘 − 𝑎𝑘 | < 𝛿 then ∑𝑛𝑘=1|𝑓(𝑏𝑘 ) − 𝑓 (𝑎𝑘 )| < 𝜖 .
Notice that if the finite collection is a single set we get the definition for uniform
continuity. Thus absolutely continuous implies uniformly continuous (but not the
other way around).
Ex. The Cantor function 𝜑 is increasing and continuous on [0,1] (and hence
uniformly continuous), but it is not absolutely continuous.
In the 𝑛𝑡ℎ stage of construction the Cantor set is a disjoint collection
𝑛
{[𝑐𝑘 , 𝑑𝑘 ]}𝑘=2 𝑛 −𝑛
𝑘=1 of 2 subintervals of [0,1] each of length 3 .
1 2 1 2 7 8
For example 𝐴2 = [0, 9] ∪ [9 , 3] ∪ [3 , 9] ∪ [9 , 1].
𝜑 is constant on each of the intervals that comprise the complement in [0,1] of
this collection of intervals.
Since 𝜑 is increasing and 𝜑(1) − 𝜑(0) = 1;
𝑛 2 𝑛 𝑛
∑𝑘=2
𝑘=1 |𝑑𝑘 − 𝑐𝑘 | = (3) while 𝑘=2 | ( )
∑𝑘=1 𝜑 𝑑𝑘 − 𝜑(𝑐𝑘 )| = 1.
1 2 3 2𝑛 −1
(Since 𝜑 takes on the values { , , ,…, } on the 2𝑛 − 1 open
2𝑛 2𝑛 2𝑛 2𝑛
intervals).
𝑛
But if 𝜖 = 1 there is no 𝛿 > 0 where if ∑𝑘=2
𝑘=1 |𝑑𝑘 − 𝑐𝑘 | < 𝛿 then
𝑛
∑𝑘=2
𝑘=1 |𝑓 (𝑑𝑘 ) − 𝑓 (𝑐𝑘 )| < 𝜖 .
, 2
It’s not hard to show that linear combinations of absolutely continuous functions
are also absolutely continuous, however, compositions of absolutely continuous
functions need not be absolutely continuous.
Prop. If 𝑓 is Lipschitz on a closed, bounded interval [𝑎, 𝑏] then it is absolutely
continuous.
Proof: Let 𝑐 > 0 be a Lipschitz constant for 𝑓 on [𝑎, 𝑏]. So
|𝑓(𝑢) − 𝑓 (𝑣)| ≤ 𝑐|𝑢 − 𝑣| for all 𝑢, 𝑣 ∈ [𝑎, 𝑏].
𝜖
If we just take 𝛿 = then
𝑐
𝜖
∑𝑛𝑘=1|𝑏𝑘 − 𝑎𝑘 | < 𝛿 = ⟹ 𝑐 ∑𝑛𝑘=1|𝑏𝑘 − 𝑎𝑘 | < 𝜖 .
𝑐
But since 𝑓 is Lipschitz with constant 𝑐:
∑𝑛𝑘=1|𝑓(𝑏𝑘 ) − 𝑓(𝑎𝑘 )| ≤ 𝑐 ∑𝑛𝑘=1|𝑏𝑘 − 𝑎𝑘 | < 𝜖 .
Hence 𝑓 is absolutely continuous on [𝑎, 𝑏].
Note: there are functions that are absolutely continuous but are not Lipschitz. For
example 𝑓 (𝑥 ) = √𝑥 for 0 ≤ 𝑥 ≤ 1.
Absolutely Continuous Functions
Def. A real valued function 𝑓 on a closed interval [𝑎, 𝑏] is said to be absolutely
continuous on [𝑎, 𝑏] if for each 𝜖 > 0, there is a 𝛿 > 0 such that for every
disjoint collection {(𝑎𝑘 , 𝑏𝑘 )}𝑛
𝑘=1 of open intervals in (𝑎, 𝑏) if
∑𝑛𝑘=1|𝑏𝑘 − 𝑎𝑘 | < 𝛿 then ∑𝑛𝑘=1|𝑓(𝑏𝑘 ) − 𝑓 (𝑎𝑘 )| < 𝜖 .
Notice that if the finite collection is a single set we get the definition for uniform
continuity. Thus absolutely continuous implies uniformly continuous (but not the
other way around).
Ex. The Cantor function 𝜑 is increasing and continuous on [0,1] (and hence
uniformly continuous), but it is not absolutely continuous.
In the 𝑛𝑡ℎ stage of construction the Cantor set is a disjoint collection
𝑛
{[𝑐𝑘 , 𝑑𝑘 ]}𝑘=2 𝑛 −𝑛
𝑘=1 of 2 subintervals of [0,1] each of length 3 .
1 2 1 2 7 8
For example 𝐴2 = [0, 9] ∪ [9 , 3] ∪ [3 , 9] ∪ [9 , 1].
𝜑 is constant on each of the intervals that comprise the complement in [0,1] of
this collection of intervals.
Since 𝜑 is increasing and 𝜑(1) − 𝜑(0) = 1;
𝑛 2 𝑛 𝑛
∑𝑘=2
𝑘=1 |𝑑𝑘 − 𝑐𝑘 | = (3) while 𝑘=2 | ( )
∑𝑘=1 𝜑 𝑑𝑘 − 𝜑(𝑐𝑘 )| = 1.
1 2 3 2𝑛 −1
(Since 𝜑 takes on the values { , , ,…, } on the 2𝑛 − 1 open
2𝑛 2𝑛 2𝑛 2𝑛
intervals).
𝑛
But if 𝜖 = 1 there is no 𝛿 > 0 where if ∑𝑘=2
𝑘=1 |𝑑𝑘 − 𝑐𝑘 | < 𝛿 then
𝑛
∑𝑘=2
𝑘=1 |𝑓 (𝑑𝑘 ) − 𝑓 (𝑐𝑘 )| < 𝜖 .
, 2
It’s not hard to show that linear combinations of absolutely continuous functions
are also absolutely continuous, however, compositions of absolutely continuous
functions need not be absolutely continuous.
Prop. If 𝑓 is Lipschitz on a closed, bounded interval [𝑎, 𝑏] then it is absolutely
continuous.
Proof: Let 𝑐 > 0 be a Lipschitz constant for 𝑓 on [𝑎, 𝑏]. So
|𝑓(𝑢) − 𝑓 (𝑣)| ≤ 𝑐|𝑢 − 𝑣| for all 𝑢, 𝑣 ∈ [𝑎, 𝑏].
𝜖
If we just take 𝛿 = then
𝑐
𝜖
∑𝑛𝑘=1|𝑏𝑘 − 𝑎𝑘 | < 𝛿 = ⟹ 𝑐 ∑𝑛𝑘=1|𝑏𝑘 − 𝑎𝑘 | < 𝜖 .
𝑐
But since 𝑓 is Lipschitz with constant 𝑐:
∑𝑛𝑘=1|𝑓(𝑏𝑘 ) − 𝑓(𝑎𝑘 )| ≤ 𝑐 ∑𝑛𝑘=1|𝑏𝑘 − 𝑎𝑘 | < 𝜖 .
Hence 𝑓 is absolutely continuous on [𝑎, 𝑏].
Note: there are functions that are absolutely continuous but are not Lipschitz. For
example 𝑓 (𝑥 ) = √𝑥 for 0 ≤ 𝑥 ≤ 1.
