Showing $f circ g$ is Riemann Integrable
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
asked 7 hours ago
Story123
1577
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add a comment |
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
asked 7 hours ago
Story123
1577
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
add a comment |
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
asked 7 hours ago
Story123
1577
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
real-analysis integration riemann-integration
asked 7 hours ago
Story123
1577
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.
asked 7 hours ago
Story123
1577
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.
asked 7 hours ago
Story123
1577
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.
asked 7 hours ago
Story123
1577
asked 7 hours ago
Story123
1577
1577
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.
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.
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
add a comment |
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
add a comment |
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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