1
$\begingroup$

Condition: $f:I\to\mathbb R$ is continuous. For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $\forall\epsilon\exists\delta$ such that $$ \sum_{k} |y_{k} - x_{k}| < \delta$$ implies $$\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon.$$

Is this condition a necessary or sufficient condition of absolution continuity? Note that the order of the logic identifiers has changed.

A function $f: I \to \mathbb{R}$ is absolutely continuous on an interval $I$ if for every $\epsilon > 0$ there is a $\delta > 0$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$ satisfies $$ \sum_{k} |y_{k} - x_{k}| < \delta$$ then $$\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon$$

$\endgroup$
1
$\begingroup$

Given any function $f:I\to\mathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $\forall\epsilon\exists\delta$ such that $$ \sum_{k} |y_{k} - x_{k}| < \delta$$ implies $$\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon$$ is trivially true.

Proof: Given any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, just choose $\delta = \frac{1}{2} \sum_{k} |y_{k} - x_{k}|$. Then the condition $ \sum_{k} |y_{k} - x_{k}| < \delta$ will be false and so the implication "$ \sum_{k} |y_{k} - x_{k}| < \delta$ implies $\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon$" will be trivially true.

So this conditionis not sufficient for absolute continuity or even continuity.

$\endgroup$
  • 1
    $\begingroup$ Yes, I also saw this after you gave the detailed explanation in the other question $\endgroup$ – High GPA Mar 28 '19 at 22:25
2
$\begingroup$

Yes, they are equivalent. Suppose you choose $\delta$ according to the usual definition of absolute continuity with $\epsilon$ repalced by $\epsilon /2$. If $(a_k.b_k)$ is a disjoint sequence of interval with total length less than $\delta$ then $\sum\limits_{k=1}^{N} |f(b_k)-f(a_k)| < \epsilon /2$ for each $N$. Let $N \to \infty$ to complete the proof.

$\endgroup$
  • $\begingroup$ Many thanks for your teaching! So you proved that the definition "$\forall\epsilon\exists\delta(\forall \text{finite subintervals we have} (\sum|y_k-x_k|<\delta \Rightarrow \sum|f(y_k)-f(x_k)|<\epsilon))$" is equivalent to "$\forall\epsilon\exists\delta(\forall \text{countable subintervals we have} (\sum|y_k-x_k|<\delta \Rightarrow \sum|f(y_k)-f(x_k)|<\epsilon))$". However, my first condition means "$\forall \text{countable subintervals}(\forall\epsilon\exists\delta \text{we have} (\sum|y_k-x_k|<\delta \Rightarrow \sum|f(y_k)-f(x_k)|<\epsilon))$". Not sure my understanding is correct, though. $\endgroup$ – High GPA Mar 25 '19 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.