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Different definition of continuity



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The measure of the image of a set of measure zeroAbsolutely continuous function admits weak derivativeEquivalent definitions of absolutely continuous functionsDoes absolute continuity of $f$ on $[epsilon,1]$ and continuity at $f=0$ imply absolute continuity on $[0,1]$?Proof that continuous function respects sequential continuityIs $sqrtx, xin [0,1]$ absolutely continuous?Absolutely Continuous function using sums.A question about absolute continuityAbsolute continuity of increasing functions on an intervalEquivalence condition of Absolute Continuity










1












$begingroup$


Condition: $f:Itomathbb R$ is continuous. For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ 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 $f: I to mathbbR$ 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
$$sum_k |f(y_k) - f(x_k)| < epsilon$$











share|cite|improve this question











$endgroup$
















    1












    $begingroup$


    Condition: $f:Itomathbb R$ is continuous. For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ 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 $f: I to mathbbR$ 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
    $$sum_k |f(y_k) - f(x_k)| < epsilon$$











    share|cite|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      Condition: $f:Itomathbb R$ is continuous. For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ 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 $f: I to mathbbR$ 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
      $$sum_k |f(y_k) - f(x_k)| < epsilon$$











      share|cite|improve this question











      $endgroup$




      Condition: $f:Itomathbb R$ is continuous. For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ 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 $f: I to mathbbR$ 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
      $$sum_k |f(y_k) - f(x_k)| < epsilon$$








      real-analysis calculus limits analysis continuity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 26 at 1:47







      High GPA

















      asked Mar 25 at 8:19









      High GPAHigh GPA

      914422




      914422




















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Given any function $f:Itomathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ 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 $delta = frac12 sum_k |y_k - x_k|$. 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.






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Yes, I also saw this after you gave the detailed explanation in the other question
            $endgroup$
            – High GPA
            Mar 28 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 $sumlimits_k=1^N |f(b_k)-f(a_k)| < epsilon /2$ for each $N$. Let $N to infty$ to complete the proof.






          share|cite|improve this answer









          $endgroup$












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












          Your Answer








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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Given any function $f:Itomathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ 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 $delta = frac12 sum_k |y_k - x_k|$. 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.






          share|cite|improve this answer









          $endgroup$








          • 1




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















          1












          $begingroup$

          Given any function $f:Itomathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ 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 $delta = frac12 sum_k |y_k - x_k|$. 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.






          share|cite|improve this answer









          $endgroup$








          • 1




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













          1












          1








          1





          $begingroup$

          Given any function $f:Itomathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ 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 $delta = frac12 sum_k |y_k - x_k|$. 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.






          share|cite|improve this answer









          $endgroup$



          Given any function $f:Itomathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ 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 $delta = frac12 sum_k |y_k - x_k|$. 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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 28 at 10:31









          RamiroRamiro

          7,36421535




          7,36421535







          • 1




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












          • 1




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







          1




          1




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




          $begingroup$
          Yes, I also saw this after you gave the detailed explanation in the other question
          $endgroup$
          – High GPA
          Mar 28 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 $sumlimits_k=1^N |f(b_k)-f(a_k)| < epsilon /2$ for each $N$. Let $N to infty$ to complete the proof.






          share|cite|improve this answer









          $endgroup$












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
















          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 $sumlimits_k=1^N |f(b_k)-f(a_k)| < epsilon /2$ for each $N$. Let $N to infty$ to complete the proof.






          share|cite|improve this answer









          $endgroup$












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














          2












          2








          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 $sumlimits_k=1^N |f(b_k)-f(a_k)| < epsilon /2$ for each $N$. Let $N to infty$ to complete the proof.






          share|cite|improve this answer









          $endgroup$



          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 $sumlimits_k=1^N |f(b_k)-f(a_k)| < epsilon /2$ for each $N$. Let $N to infty$ to complete the proof.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 25 at 8:24









          Kavi Rama MurthyKavi Rama Murthy

          74.6k53270




          74.6k53270











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

















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
















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





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


















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