Question

If a function is absolutely continuous on [0,n) for every n then it is absolutely continuous

Answer 1

4. Absolute Continuity
A real-valued function f defined on [a, b] is said to be
absolutely continuous on [a, b] if
∀ > 0 ∃δ > 0 ∀n ∈ N ∀{(xi
, x
0
i
)}
n
i=1 of nonoverlapping intervals with
Xn
i=1
|x
0
i − xi
| < δ =⇒
Xn
i=1
|f(x
0
i
) − f(xi)| <
Note that (i) every absolutely continuous function is uniformly continuous,
(ii) every indefinite integral F(x) = Z x
a
f(t)dt for all x ∈ [a, b]
where f is integrable on [a, b] is absolutely continuous.
Proof. Let > 0.
By Proposition 4.14; there is a δ > 0 such that ∀{(xi
, x
0
i
)}
n
i=1 of
nonoverlapping intervals in [a, b] with m(
[n
i=1
(xi
, x
0
i
)) = Xn
i=1
|x
0
i − xi
| < δ
we have
Xn
i=1
|F(x
0
i
) − F(xi)| =
Xn
i=1
|
Z x
0
i
xi
f(t)dt|
≤
Xn
i=1
Z x
0
i
xi
|f(t)|dt
=
Z
Sn
i=1(xi,x
0
i
)
|f(t)|dt
=
Z
Sn
i=1(xi,x
0
i
)
|f|
< .
Hence F(x) is absolutely continuous.
(iii) the sum and difference of two absolutely continuous functions
is absolutely continuous.
Proof. Let f, g be absolutely continuous functions
Let > 0; there is a δ > 0 such that ∀{(xi
, x
0
i
)}
n
i=1 of nonoverlapping inter-
vals in [a, b] with Xn
i=1
|x
0
i − xi
| < δ we have
Xn
i=1
|(f + g)(x
0
i
) − (f + g)(xi)| =
Xn
i=1
|f(x
0
i
) − f(xi) + g(x
0
i
) − g(xi)|
≤
Xn
i=1
|f(x
0
i
) − f(xi)| +
Xn
i=1
|g(x
0
i
) − g(xi)|
<

2
+

2
=
and
Xn
i=1
|(f − g)(x
0
i
) − (f − g)(xi)| =
Xn
i=1
|f(x
0
i
) − f(xi) + g(xi) − g(x
0
i
)|
≤
Xn
i=1
|f(x
0
i
) − f(xi)| +
Xn
i=1
|g(x
0
i
) − g(xi)|
<

2
+

2
=
Hence the sum and difference of two absolutely continuous functions
is absolutely continuous.