# Ngô Quốc Anh

## November 5, 2009

### An equivalent criterion for absolutely continuous functions

Filed under: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205) — Ngô Quốc Anh @ 22:29

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

Definition. A finite function $f$ on a finite interval $[a,b]$ is said to be absolute continuous if and only if for given $\varepsilon > 0$, there exists $\delta > 0$ such that

$\displaystyle\sum_k |f(b_k) - f(a_k)| < \varepsilon$

for any collection (finite or not) $\{[a_k, b_k]\}$ of non-overlapping subintervals of $[a, b]$ with $\sum (b_k - a_k) < \delta$.

We now prove the following result.

Statement. Show that $f$ is absolutely continuous on $[a, b]$ if and only if given $\varepsilon > 0$, there exists $\delta > 0$ such that

$\displaystyle \Big|\sum_k (f(b_k) - f(a_k)) \Big| < \varepsilon$

for any finite collection $\{[a_k, b_k]\}$ of non-overlapping subintervals of $[a, b]$ with $\sum (b_k - a_k) < \delta$.

Proof. If $f$ is absolutely continuous on $[a, b]$, then the result is easily obtained by using the definition and the fact that $|x + y| \leq |x| + |y|$ for every $x,y \in \mathbb R$.

Now we prove that

for given $\varepsilon > 0$, there exists $\delta > 0$ such that $\sum |f(b_k) - f(a_k)| < \varepsilon$ for any finite collection $\{[a_k, b_k]\}$ of non-overlapping subintervals of $[a, b]$ with $\sum (b_k - a_k) < \delta$.

Indeed, we split the collection $\{[a_k, b_k]\}$ into two types:

• type $A$ are all $k$ such that $f(b_k) - f(a_k) \geq 0$ and
• type $B$ are all $k$ such that $f(b_k) - f(a_k) < 0$.

For given $\varepsilon > 0$, there exists $\delta > 0$ such that $| \sum (f(b_k) - f(a_k))| < \frac{\varepsilon}{3}$ for any finite collection $\{[a_k, b_k]\}$ with $\sum (b_k - a_k) < \delta$. Then for $k \in A$ we also have

$\displaystyle\sum_{k \in A} \left( f(b_k) - f(a_k) \right) = \Big| \sum_{k \in A} (f(b_k) - f(a_k))\Big| < \frac{\varepsilon}{3}$.

Similarly,

$\displaystyle\sum_{k \in B} \left( f(a_k) - f(b_k) \right) = \Big| \sum_{k \in B} (f(b_k) - f(a_k))\Big| < \frac{\varepsilon}{3}$.

From the following inequality $a + b \leq |a-b| + b + b$ with $a,b \geq 0$ we deduce that

$\displaystyle\sum\limits_{k = 1}^n {\left| {f\left( {{b_k}} \right) - f\left( {{a_k}} \right)} \right|} = \underbrace {\sum\limits_{k \in A} {f\left( {{b_k}} \right) - f\left( {{a_k}} \right)} }_a + \underbrace {\sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)} }_b$

with the fact that

$\displaystyle a+b \leqslant \left| {\sum\limits_{k \in A} {f\left( {{b_k}} \right) - f\left( {{a_k}} \right)} - \sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)} } \right| + \sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)} + \sum\limits_{k \in B} {f\left( {{a_k}} \right) - f\left( {{b_k}} \right)}.$

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

$\displaystyle\frac{\varepsilon }{3} + \frac{\varepsilon }{3} + \frac{\varepsilon }{3} = \varepsilon.$

Thus, we have proved that for given $\varepsilon >0$, there exists $\delta>0$ such that

$\displaystyle\sum\limits_{k = 1}^n {\left| {f\left( {b_k } \right) - f\left( {a_k } \right)} \right|} < \varepsilon$

for any finite collection $\{[a_k, b_k]\}$ of non-overlapping subintervals of $[a, b]$ with $\sum (b_k - a_k) < \delta$. Letting $n \to \infty$ we can claim that $f$ is absolutely continuous.