Showing $f circ g$ is Riemann Integrable












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I was curious about a question that showed up on a previous qual exam. I was able to prove it using measure theory, but I was curious about a proof without measure theory. The question is:



If $text{ }f,g$ are Riemann Integrable on $[0,1]$ such that begin{equation} |g(x)-g(y)| geq alpha|x-y| end{equation} for any $x,y in [0,1]$ and for a fixed $alpha in (0,1)$. Show $f circ g$ is Riemann integrable.



I was able to prove this statement using measure theory:
We know that the set of discontinuities of $f,g$ are of measure zero, and by the reverse Lipschitz condition on $g(x)$, we know that $g(x)$ is injective, so $f circ g$ set of discontinuity has measure zero; therefore, $f circ g$ is Riemann Integrable.



I was wondering how would one prove this without measure theory.










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  • 1




    How about using Jordan measure?
    – copper.hat
    7 hours ago






  • 1




    Silly question, but how do you know if $f,g$ have discontinuity sets of measure zero and $g$ is injective implies $f circ g$ has a discontinuity set of measure zero?
    – MathematicsStudent1122
    7 hours ago










  • Let $xi$ be in the set of discontinuity of $f$, then there is at most one $x$ such that $g(x) = xi$. And $f(g(x))$ has a discontinuity if and only if $g(x)$ maps to a point of discontinuity of $f$ or $g$ has a discontinuity at $x$. So it follows that the set of discontinuity of $f circ g$ is contained in the union in the set of discontinuity of $g$ and the set where $g$ gets mapped to a discontinuity of $f$. The second set has measure zero. Union of sets of measure zero has measure zero
    – Story123
    6 hours ago












  • @copper.hat I would prefer a proof without Jordan Measure
    – Story123
    6 hours ago
















0














I was curious about a question that showed up on a previous qual exam. I was able to prove it using measure theory, but I was curious about a proof without measure theory. The question is:



If $text{ }f,g$ are Riemann Integrable on $[0,1]$ such that begin{equation} |g(x)-g(y)| geq alpha|x-y| end{equation} for any $x,y in [0,1]$ and for a fixed $alpha in (0,1)$. Show $f circ g$ is Riemann integrable.



I was able to prove this statement using measure theory:
We know that the set of discontinuities of $f,g$ are of measure zero, and by the reverse Lipschitz condition on $g(x)$, we know that $g(x)$ is injective, so $f circ g$ set of discontinuity has measure zero; therefore, $f circ g$ is Riemann Integrable.



I was wondering how would one prove this without measure theory.










share|cite|improve this question







New contributor




Story123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    How about using Jordan measure?
    – copper.hat
    7 hours ago






  • 1




    Silly question, but how do you know if $f,g$ have discontinuity sets of measure zero and $g$ is injective implies $f circ g$ has a discontinuity set of measure zero?
    – MathematicsStudent1122
    7 hours ago










  • Let $xi$ be in the set of discontinuity of $f$, then there is at most one $x$ such that $g(x) = xi$. And $f(g(x))$ has a discontinuity if and only if $g(x)$ maps to a point of discontinuity of $f$ or $g$ has a discontinuity at $x$. So it follows that the set of discontinuity of $f circ g$ is contained in the union in the set of discontinuity of $g$ and the set where $g$ gets mapped to a discontinuity of $f$. The second set has measure zero. Union of sets of measure zero has measure zero
    – Story123
    6 hours ago












  • @copper.hat I would prefer a proof without Jordan Measure
    – Story123
    6 hours ago














0












0








0







I was curious about a question that showed up on a previous qual exam. I was able to prove it using measure theory, but I was curious about a proof without measure theory. The question is:



If $text{ }f,g$ are Riemann Integrable on $[0,1]$ such that begin{equation} |g(x)-g(y)| geq alpha|x-y| end{equation} for any $x,y in [0,1]$ and for a fixed $alpha in (0,1)$. Show $f circ g$ is Riemann integrable.



I was able to prove this statement using measure theory:
We know that the set of discontinuities of $f,g$ are of measure zero, and by the reverse Lipschitz condition on $g(x)$, we know that $g(x)$ is injective, so $f circ g$ set of discontinuity has measure zero; therefore, $f circ g$ is Riemann Integrable.



I was wondering how would one prove this without measure theory.










share|cite|improve this question







New contributor




Story123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I was curious about a question that showed up on a previous qual exam. I was able to prove it using measure theory, but I was curious about a proof without measure theory. The question is:



If $text{ }f,g$ are Riemann Integrable on $[0,1]$ such that begin{equation} |g(x)-g(y)| geq alpha|x-y| end{equation} for any $x,y in [0,1]$ and for a fixed $alpha in (0,1)$. Show $f circ g$ is Riemann integrable.



I was able to prove this statement using measure theory:
We know that the set of discontinuities of $f,g$ are of measure zero, and by the reverse Lipschitz condition on $g(x)$, we know that $g(x)$ is injective, so $f circ g$ set of discontinuity has measure zero; therefore, $f circ g$ is Riemann Integrable.



I was wondering how would one prove this without measure theory.







real-analysis integration riemann-integration






share|cite|improve this question







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Story123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Story123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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asked 7 hours ago









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Story123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Story123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    How about using Jordan measure?
    – copper.hat
    7 hours ago






  • 1




    Silly question, but how do you know if $f,g$ have discontinuity sets of measure zero and $g$ is injective implies $f circ g$ has a discontinuity set of measure zero?
    – MathematicsStudent1122
    7 hours ago










  • Let $xi$ be in the set of discontinuity of $f$, then there is at most one $x$ such that $g(x) = xi$. And $f(g(x))$ has a discontinuity if and only if $g(x)$ maps to a point of discontinuity of $f$ or $g$ has a discontinuity at $x$. So it follows that the set of discontinuity of $f circ g$ is contained in the union in the set of discontinuity of $g$ and the set where $g$ gets mapped to a discontinuity of $f$. The second set has measure zero. Union of sets of measure zero has measure zero
    – Story123
    6 hours ago












  • @copper.hat I would prefer a proof without Jordan Measure
    – Story123
    6 hours ago














  • 1




    How about using Jordan measure?
    – copper.hat
    7 hours ago






  • 1




    Silly question, but how do you know if $f,g$ have discontinuity sets of measure zero and $g$ is injective implies $f circ g$ has a discontinuity set of measure zero?
    – MathematicsStudent1122
    7 hours ago










  • Let $xi$ be in the set of discontinuity of $f$, then there is at most one $x$ such that $g(x) = xi$. And $f(g(x))$ has a discontinuity if and only if $g(x)$ maps to a point of discontinuity of $f$ or $g$ has a discontinuity at $x$. So it follows that the set of discontinuity of $f circ g$ is contained in the union in the set of discontinuity of $g$ and the set where $g$ gets mapped to a discontinuity of $f$. The second set has measure zero. Union of sets of measure zero has measure zero
    – Story123
    6 hours ago












  • @copper.hat I would prefer a proof without Jordan Measure
    – Story123
    6 hours ago








1




1




How about using Jordan measure?
– copper.hat
7 hours ago




How about using Jordan measure?
– copper.hat
7 hours ago




1




1




Silly question, but how do you know if $f,g$ have discontinuity sets of measure zero and $g$ is injective implies $f circ g$ has a discontinuity set of measure zero?
– MathematicsStudent1122
7 hours ago




Silly question, but how do you know if $f,g$ have discontinuity sets of measure zero and $g$ is injective implies $f circ g$ has a discontinuity set of measure zero?
– MathematicsStudent1122
7 hours ago












Let $xi$ be in the set of discontinuity of $f$, then there is at most one $x$ such that $g(x) = xi$. And $f(g(x))$ has a discontinuity if and only if $g(x)$ maps to a point of discontinuity of $f$ or $g$ has a discontinuity at $x$. So it follows that the set of discontinuity of $f circ g$ is contained in the union in the set of discontinuity of $g$ and the set where $g$ gets mapped to a discontinuity of $f$. The second set has measure zero. Union of sets of measure zero has measure zero
– Story123
6 hours ago






Let $xi$ be in the set of discontinuity of $f$, then there is at most one $x$ such that $g(x) = xi$. And $f(g(x))$ has a discontinuity if and only if $g(x)$ maps to a point of discontinuity of $f$ or $g$ has a discontinuity at $x$. So it follows that the set of discontinuity of $f circ g$ is contained in the union in the set of discontinuity of $g$ and the set where $g$ gets mapped to a discontinuity of $f$. The second set has measure zero. Union of sets of measure zero has measure zero
– Story123
6 hours ago














@copper.hat I would prefer a proof without Jordan Measure
– Story123
6 hours ago




@copper.hat I would prefer a proof without Jordan Measure
– Story123
6 hours ago















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