Suppose that is a sequence of absolutely continuous functions defined on such that for every and

for every . Prove:
- the series
converges for each pointwise to a function 
- the function
is absolutely continuous on ![{\displaystyle [0,1]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b26fc5fee30dada6f15cc080115cae582606d7f3)
![{\displaystyle f^{\prime }(x)=\sum _{n=1}^{\infty }f_{n}^{\prime }(x)\quad a.e.\,\,x\in [0,1]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6418d51c14ff959c2890ab33358a7a35eabc9835)
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Absolutely Continuous <==> Indefinite Integral[edit | edit source]
is absolutely continuous if and only if
can be written as an indefinite integral i.e. for all
Apply Inequalities,Sum over n, and Use Hypothesis[edit | edit source]
Let
be given. Then,
Hence
Summing both sides of the inequality over
and applying the hypothesis yields pointwise convergence of the series
,
Absolutely continuous <==> Indefinite Integral[edit | edit source]
Let
.
We want to show:
Rewrite f(x) and Apply Lebesgue Dominated Convergence Theorem[edit | edit source]
Justification for Lebesgue Dominated Convergence Theorem[edit | edit source]
Therefore
is integrable
The above inequality also implies
a.e on
. Therefore,
a.e on
to a finite value.
Since
, by the Fundamental Theorem of Calculus
a.e. ![{\displaystyle x\in [0,1]\!\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5820a535671944261c3c9edd23e9a4419e10bd8)
Check Criteria for Lebesgue Dominated Convergence Theorem[edit | edit source]
Define
,
.
Since
is positive, then so is
, i.e.,
and
. Hence,
Let
. Since
, then
, i.e.,
.
integral of g_n converges to integral of g =[edit | edit source]
Hence,

hat{f_n} converges to hat{f} a.e.[edit | edit source]
Note that
is equivalent to

i.e.

Since the criteria of the LDCT are fulfilled, we have that
, i.e.,
Show that if is absolutely continuous on and , then is absolutely continuous on
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Show that g(x)=|x|^p is Lipschitz[edit | edit source]
Consider some interval
and let
and
be two points in the interval
.
Also let
for all
Therefore
is Lipschitz in the interval
Apply definitions to g(f(x))[edit | edit source]
Since
is absolutely continuous on
, given
, there exists
such that if
is a finite collection of nonoverlapping intervals of
such that
then
Consider
. Since
is Lipschitz
Therefore
is absolutely continuous.
f(x)= x^4sin^2(\frac{1}{x^2}) is Lipschitz (and then AC)[edit | edit source]
Consider
. The derivate of f is given by
.
The derivative is bounded (in fact, on any finite interval), so
is Lipschitz.
Hence, f is AC
|f|^{1/2} is not of bounded variation (and then is not AC)[edit | edit source]
Consider the partition
. Then,
Then, T(f) goes to
as
goes to
.
Then,
is not of bounded variation and then is not AC