In mathematics, absolute continuity is a smoothness property which is stricter than continuity and uniform continuity.

**Definition**. *A finite function on a finite interval is said to be absolute continuous if and only if for given , there exists such that *

* *

*for any collection **(finite or not) ** of non-overlapping subintervals of with *.

We now prove the following result.

**Statement**. Show that is absolutely continuous on if and only if given , there exists such that

for any finite collection of non-overlapping subintervals of with .

*Proof*. If is absolutely continuous on , then the result is easily obtained by using the definition and the fact that for every .

Now we prove that

for given , there exists such that for any finite collection of non-overlapping subintervals of with .

Indeed, we split the collection into two types:

- type are all such that and
- type are all such that .

For given , there exists such that for any finite collection with . Then for we also have

.

Similarly,

.

From the following inequality with we deduce that

with the fact that

Note that the right hand side of the above inequality is bounded from above by

Thus, we have proved that for given , there exists such that

for any finite collection of non-overlapping subintervals of with . Letting we can claim that is absolutely continuous.

### Like this:

Like Loading...

*Related*

I think you are excellent. Your web is very useful for people who studies analysis, study = learn and do research.

Much achievement! Congratulations

Comment by ULF — November 12, 2009 @ 5:10

Thank you.

Comment by Ngô Quốc Anh — November 12, 2009 @ 17:05