Absolute continuity of increasing functions on an interval











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I have been using Real Analysis by Royden and Fitzpatrick. I am currently stuck on problem 39 of chapter 6. The problem is stated as follows:



"Use the preceding problem to show that if f is increasing on [a, b], then f is absolutely continuous on [a, b] if and only if for each $epsilon$ , there is a $delta$ > 0 such that for a measurable subset
E of [a, b],
m*(f(E))< $epsilon$ if m(E) <$delta$"



(m* denotes outer measure and m denotes lebesgue measure)



I can see why absolute continuity implies the other property. However, I fail to see how absolute continuity follows from the fact that f is increasing and that for each $epsilon$ , there is a $delta$ > 0 such that for a measurable subset
E of [a, b],
m*(f(E))< $epsilon$ if m(E) <$delta$"



My approach is to take $delta$ responding to $epsilon$. Then for any finite collection of disjoint open interval ($a_k$,$b_k$) such that $sum_{k=1}^n (b_k-a_k) < delta$ we have that m*$(f(cup_{k=1}^n(a_k, b_k)))<epsilon$. I cannot see how this implies $sum_{k=1}^n |f(b_k)-f(a_k)|<epsilon$ as required for absolute continuity.



Is this a good approach to solve this problem? Any advice would be appreciated!



EDIT:



The previous problem (Q38):



Show that f is absolutely continuous on [a, b] if and only if for each $epsilon$ > 0, there is a $delta$ > 0 such that for every countable disjoint collection ${(a_k, b_k)}$ of open intervals in (a, b), if $sum_{k=1}^infty |b_k-a_k|<delta$ then $sum_{k=1}^infty |f(b_k)-f(a_k)|<epsilon$










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  • The fact that $f$ is increasing means that f maps intervals to intervals
    – user25959
    Nov 24 at 16:36










  • What about discontinuities f may have? Since f is increasing it would only have jump discontinuities. Although the endpoints $a_k$ and $b_k$ may be close to one another, how can we ensure that |$f(a_k)-f(b_k)$| can be made small if |$b_k-a_k$| is small enough? I can see how the proof would work in the case f were continuous.
    – ShaftSinker
    Nov 24 at 17:36












  • Sorry- I had in mind a continuous function. Perhaps it is possible to use the fact that measurable increasing functions have at most countably many discontinuities (jump discontinuities in this case).
    – user25959
    Nov 24 at 17:41






  • 1




    Can you also show the preceding problem?
    – Alex Vong
    Nov 24 at 21:19






  • 1




    maybe the question should have been, for all measurable $E$, $m^*(E) < epsilon$ if $m( f^{-1}(E)) < delta$? (ps Q38 but not Q39 is in the errata www2.math.umd.edu/~pmf/docs/Real%20Analysis.pdf )
    – Calvin Khor
    Nov 25 at 0:55

















up vote
2
down vote

favorite
1












I have been using Real Analysis by Royden and Fitzpatrick. I am currently stuck on problem 39 of chapter 6. The problem is stated as follows:



"Use the preceding problem to show that if f is increasing on [a, b], then f is absolutely continuous on [a, b] if and only if for each $epsilon$ , there is a $delta$ > 0 such that for a measurable subset
E of [a, b],
m*(f(E))< $epsilon$ if m(E) <$delta$"



(m* denotes outer measure and m denotes lebesgue measure)



I can see why absolute continuity implies the other property. However, I fail to see how absolute continuity follows from the fact that f is increasing and that for each $epsilon$ , there is a $delta$ > 0 such that for a measurable subset
E of [a, b],
m*(f(E))< $epsilon$ if m(E) <$delta$"



My approach is to take $delta$ responding to $epsilon$. Then for any finite collection of disjoint open interval ($a_k$,$b_k$) such that $sum_{k=1}^n (b_k-a_k) < delta$ we have that m*$(f(cup_{k=1}^n(a_k, b_k)))<epsilon$. I cannot see how this implies $sum_{k=1}^n |f(b_k)-f(a_k)|<epsilon$ as required for absolute continuity.



Is this a good approach to solve this problem? Any advice would be appreciated!



EDIT:



The previous problem (Q38):



Show that f is absolutely continuous on [a, b] if and only if for each $epsilon$ > 0, there is a $delta$ > 0 such that for every countable disjoint collection ${(a_k, b_k)}$ of open intervals in (a, b), if $sum_{k=1}^infty |b_k-a_k|<delta$ then $sum_{k=1}^infty |f(b_k)-f(a_k)|<epsilon$










share|cite|improve this question
























  • The fact that $f$ is increasing means that f maps intervals to intervals
    – user25959
    Nov 24 at 16:36










  • What about discontinuities f may have? Since f is increasing it would only have jump discontinuities. Although the endpoints $a_k$ and $b_k$ may be close to one another, how can we ensure that |$f(a_k)-f(b_k)$| can be made small if |$b_k-a_k$| is small enough? I can see how the proof would work in the case f were continuous.
    – ShaftSinker
    Nov 24 at 17:36












  • Sorry- I had in mind a continuous function. Perhaps it is possible to use the fact that measurable increasing functions have at most countably many discontinuities (jump discontinuities in this case).
    – user25959
    Nov 24 at 17:41






  • 1




    Can you also show the preceding problem?
    – Alex Vong
    Nov 24 at 21:19






  • 1




    maybe the question should have been, for all measurable $E$, $m^*(E) < epsilon$ if $m( f^{-1}(E)) < delta$? (ps Q38 but not Q39 is in the errata www2.math.umd.edu/~pmf/docs/Real%20Analysis.pdf )
    – Calvin Khor
    Nov 25 at 0:55















up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





I have been using Real Analysis by Royden and Fitzpatrick. I am currently stuck on problem 39 of chapter 6. The problem is stated as follows:



"Use the preceding problem to show that if f is increasing on [a, b], then f is absolutely continuous on [a, b] if and only if for each $epsilon$ , there is a $delta$ > 0 such that for a measurable subset
E of [a, b],
m*(f(E))< $epsilon$ if m(E) <$delta$"



(m* denotes outer measure and m denotes lebesgue measure)



I can see why absolute continuity implies the other property. However, I fail to see how absolute continuity follows from the fact that f is increasing and that for each $epsilon$ , there is a $delta$ > 0 such that for a measurable subset
E of [a, b],
m*(f(E))< $epsilon$ if m(E) <$delta$"



My approach is to take $delta$ responding to $epsilon$. Then for any finite collection of disjoint open interval ($a_k$,$b_k$) such that $sum_{k=1}^n (b_k-a_k) < delta$ we have that m*$(f(cup_{k=1}^n(a_k, b_k)))<epsilon$. I cannot see how this implies $sum_{k=1}^n |f(b_k)-f(a_k)|<epsilon$ as required for absolute continuity.



Is this a good approach to solve this problem? Any advice would be appreciated!



EDIT:



The previous problem (Q38):



Show that f is absolutely continuous on [a, b] if and only if for each $epsilon$ > 0, there is a $delta$ > 0 such that for every countable disjoint collection ${(a_k, b_k)}$ of open intervals in (a, b), if $sum_{k=1}^infty |b_k-a_k|<delta$ then $sum_{k=1}^infty |f(b_k)-f(a_k)|<epsilon$










share|cite|improve this question















I have been using Real Analysis by Royden and Fitzpatrick. I am currently stuck on problem 39 of chapter 6. The problem is stated as follows:



"Use the preceding problem to show that if f is increasing on [a, b], then f is absolutely continuous on [a, b] if and only if for each $epsilon$ , there is a $delta$ > 0 such that for a measurable subset
E of [a, b],
m*(f(E))< $epsilon$ if m(E) <$delta$"



(m* denotes outer measure and m denotes lebesgue measure)



I can see why absolute continuity implies the other property. However, I fail to see how absolute continuity follows from the fact that f is increasing and that for each $epsilon$ , there is a $delta$ > 0 such that for a measurable subset
E of [a, b],
m*(f(E))< $epsilon$ if m(E) <$delta$"



My approach is to take $delta$ responding to $epsilon$. Then for any finite collection of disjoint open interval ($a_k$,$b_k$) such that $sum_{k=1}^n (b_k-a_k) < delta$ we have that m*$(f(cup_{k=1}^n(a_k, b_k)))<epsilon$. I cannot see how this implies $sum_{k=1}^n |f(b_k)-f(a_k)|<epsilon$ as required for absolute continuity.



Is this a good approach to solve this problem? Any advice would be appreciated!



EDIT:



The previous problem (Q38):



Show that f is absolutely continuous on [a, b] if and only if for each $epsilon$ > 0, there is a $delta$ > 0 such that for every countable disjoint collection ${(a_k, b_k)}$ of open intervals in (a, b), if $sum_{k=1}^infty |b_k-a_k|<delta$ then $sum_{k=1}^infty |f(b_k)-f(a_k)|<epsilon$







analysis measure-theory absolute-continuity






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share|cite|improve this question













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edited Nov 25 at 0:21

























asked Nov 24 at 16:09









ShaftSinker

112




112












  • The fact that $f$ is increasing means that f maps intervals to intervals
    – user25959
    Nov 24 at 16:36










  • What about discontinuities f may have? Since f is increasing it would only have jump discontinuities. Although the endpoints $a_k$ and $b_k$ may be close to one another, how can we ensure that |$f(a_k)-f(b_k)$| can be made small if |$b_k-a_k$| is small enough? I can see how the proof would work in the case f were continuous.
    – ShaftSinker
    Nov 24 at 17:36












  • Sorry- I had in mind a continuous function. Perhaps it is possible to use the fact that measurable increasing functions have at most countably many discontinuities (jump discontinuities in this case).
    – user25959
    Nov 24 at 17:41






  • 1




    Can you also show the preceding problem?
    – Alex Vong
    Nov 24 at 21:19






  • 1




    maybe the question should have been, for all measurable $E$, $m^*(E) < epsilon$ if $m( f^{-1}(E)) < delta$? (ps Q38 but not Q39 is in the errata www2.math.umd.edu/~pmf/docs/Real%20Analysis.pdf )
    – Calvin Khor
    Nov 25 at 0:55




















  • The fact that $f$ is increasing means that f maps intervals to intervals
    – user25959
    Nov 24 at 16:36










  • What about discontinuities f may have? Since f is increasing it would only have jump discontinuities. Although the endpoints $a_k$ and $b_k$ may be close to one another, how can we ensure that |$f(a_k)-f(b_k)$| can be made small if |$b_k-a_k$| is small enough? I can see how the proof would work in the case f were continuous.
    – ShaftSinker
    Nov 24 at 17:36












  • Sorry- I had in mind a continuous function. Perhaps it is possible to use the fact that measurable increasing functions have at most countably many discontinuities (jump discontinuities in this case).
    – user25959
    Nov 24 at 17:41






  • 1




    Can you also show the preceding problem?
    – Alex Vong
    Nov 24 at 21:19






  • 1




    maybe the question should have been, for all measurable $E$, $m^*(E) < epsilon$ if $m( f^{-1}(E)) < delta$? (ps Q38 but not Q39 is in the errata www2.math.umd.edu/~pmf/docs/Real%20Analysis.pdf )
    – Calvin Khor
    Nov 25 at 0:55


















The fact that $f$ is increasing means that f maps intervals to intervals
– user25959
Nov 24 at 16:36




The fact that $f$ is increasing means that f maps intervals to intervals
– user25959
Nov 24 at 16:36












What about discontinuities f may have? Since f is increasing it would only have jump discontinuities. Although the endpoints $a_k$ and $b_k$ may be close to one another, how can we ensure that |$f(a_k)-f(b_k)$| can be made small if |$b_k-a_k$| is small enough? I can see how the proof would work in the case f were continuous.
– ShaftSinker
Nov 24 at 17:36






What about discontinuities f may have? Since f is increasing it would only have jump discontinuities. Although the endpoints $a_k$ and $b_k$ may be close to one another, how can we ensure that |$f(a_k)-f(b_k)$| can be made small if |$b_k-a_k$| is small enough? I can see how the proof would work in the case f were continuous.
– ShaftSinker
Nov 24 at 17:36














Sorry- I had in mind a continuous function. Perhaps it is possible to use the fact that measurable increasing functions have at most countably many discontinuities (jump discontinuities in this case).
– user25959
Nov 24 at 17:41




Sorry- I had in mind a continuous function. Perhaps it is possible to use the fact that measurable increasing functions have at most countably many discontinuities (jump discontinuities in this case).
– user25959
Nov 24 at 17:41




1




1




Can you also show the preceding problem?
– Alex Vong
Nov 24 at 21:19




Can you also show the preceding problem?
– Alex Vong
Nov 24 at 21:19




1




1




maybe the question should have been, for all measurable $E$, $m^*(E) < epsilon$ if $m( f^{-1}(E)) < delta$? (ps Q38 but not Q39 is in the errata www2.math.umd.edu/~pmf/docs/Real%20Analysis.pdf )
– Calvin Khor
Nov 25 at 0:55






maybe the question should have been, for all measurable $E$, $m^*(E) < epsilon$ if $m( f^{-1}(E)) < delta$? (ps Q38 but not Q39 is in the errata www2.math.umd.edu/~pmf/docs/Real%20Analysis.pdf )
– Calvin Khor
Nov 25 at 0:55

















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