Mixing
If $f$ is integrable and and $g=f$ except at fintely many points. Prove that $g$ is integrable
$begingroup$
Let $f: A rightarrow mathbb{R} $ be integrable and let $g=f$ except at fintely many points. Show that $g$ is integrable and that $int_A f = int_A g$
The question above is from Spivak's "Calculus on Manifolds" question 3-2 in the chapter on integration.
I want to prove this by using the theorem that states that a function is integrable iff $ exists p $ such that $U(f,p)-L(f,p) < epsilon$ $, forall epsilon$
Since I can assume that this condition holds for $f$ I want to find a partition $p'$ such that $U(g,p') leq U(f,p)$ and $L(g,p') > L(f,p)$.
I have tried by starting with the case where $f$ and $g$ only differ at one point. It is then clear that in the case where $g(x_0) > f(x_0)$, that we can satisfy the inequality $U(g,p') leq U(f,p)$ by refining the $p$ that satisfies $U(f,p)-L(f,p) < epsilon$.
My Question
How can I precisely pick $p'$ so that $U(g,p')=U(f,p)$ in the special case I described above? Intuitively, I can just keep refining $p$ and that partition which includes $x_0$ will tend to $0$, but I want to rigorously know at what point will $U(g,p')=U(f,p)$?
Thanks
multivariable-calculus proof-writing
asked Jan 23 at 3:26
john fowlesjohn fowles
1,208817
$endgroup$
add a comment |
$begingroup$
Let $f: A rightarrow mathbb{R} $ be integrable and let $g=f$ except at fintely many points. Show that $g$ is integrable and that $int_A f = int_A g$
The question above is from Spivak's "Calculus on Manifolds" question 3-2 in the chapter on integration.
I want to prove this by using the theorem that states that a function is integrable iff $ exists p $ such that $U(f,p)-L(f,p) < epsilon$ $, forall epsilon$
Since I can assume that this condition holds for $f$ I want to find a partition $p'$ such that $U(g,p') leq U(f,p)$ and $L(g,p') > L(f,p)$.
I have tried by starting with the case where $f$ and $g$ only differ at one point. It is then clear that in the case where $g(x_0) > f(x_0)$, that we can satisfy the inequality $U(g,p') leq U(f,p)$ by refining the $p$ that satisfies $U(f,p)-L(f,p) < epsilon$.
My Question
How can I precisely pick $p'$ so that $U(g,p')=U(f,p)$ in the special case I described above? Intuitively, I can just keep refining $p$ and that partition which includes $x_0$ will tend to $0$, but I want to rigorously know at what point will $U(g,p')=U(f,p)$?
Thanks
multivariable-calculus proof-writing
asked Jan 23 at 3:26
john fowlesjohn fowles
1,208817
$endgroup$
1
$begingroup$
The refinement may strictly decrease the upper sum, you can't ensure exact equality between them. In any case, think of the worst case scenario, where $f(x_0)$ was the maximum on its interval in $p$ and now $g(x_0)$ is a bit bigger, say by $a$. How exactly can you refine near $x_0$ so that $U(g,p')$ is close enough to $U(f,p')$? You'll need the interval containing $x_0$ to be short...how short?
$endgroup$
– Ian
Jan 23 at 3:37
$begingroup$
In the second last line of your comment, did you mean to write "close enough to $U(f,p)$ instead of "close enough to $U(f,p')$?
$endgroup$
– john fowles
Jan 23 at 5:39
$begingroup$
No, it is easier to compare on the same partition. Then $U(f,p')$ will be close enough to $U(f,p)$ because $U(f,p)$ is already quite close to the integral of $f$.
$endgroup$
– Ian
Jan 23 at 8:58
$begingroup$
Please, please be careful with your order of quantifiers.
$endgroup$
– Ted Shifrin
Jan 25 at 1:15
add a comment |
$begingroup$
Let $f: A rightarrow mathbb{R} $ be integrable and let $g=f$ except at fintely many points. Show that $g$ is integrable and that $int_A f = int_A g$
The question above is from Spivak's "Calculus on Manifolds" question 3-2 in the chapter on integration.
I want to prove this by using the theorem that states that a function is integrable iff $ exists p $ such that $U(f,p)-L(f,p) < epsilon$ $, forall epsilon$
Since I can assume that this condition holds for $f$ I want to find a partition $p'$ such that $U(g,p') leq U(f,p)$ and $L(g,p') > L(f,p)$.
I have tried by starting with the case where $f$ and $g$ only differ at one point. It is then clear that in the case where $g(x_0) > f(x_0)$, that we can satisfy the inequality $U(g,p') leq U(f,p)$ by refining the $p$ that satisfies $U(f,p)-L(f,p) < epsilon$.
My Question
How can I precisely pick $p'$ so that $U(g,p')=U(f,p)$ in the special case I described above? Intuitively, I can just keep refining $p$ and that partition which includes $x_0$ will tend to $0$, but I want to rigorously know at what point will $U(g,p')=U(f,p)$?
Thanks
multivariable-calculus proof-writing
asked Jan 23 at 3:26
john fowlesjohn fowles
1,208817
$endgroup$
Let $f: A rightarrow mathbb{R} $ be integrable and let $g=f$ except at fintely many points. Show that $g$ is integrable and that $int_A f = int_A g$
The question above is from Spivak's "Calculus on Manifolds" question 3-2 in the chapter on integration.
I want to prove this by using the theorem that states that a function is integrable iff $ exists p $ such that $U(f,p)-L(f,p) < epsilon$ $, forall epsilon$
Since I can assume that this condition holds for $f$ I want to find a partition $p'$ such that $U(g,p') leq U(f,p)$ and $L(g,p') > L(f,p)$.
I have tried by starting with the case where $f$ and $g$ only differ at one point. It is then clear that in the case where $g(x_0) > f(x_0)$, that we can satisfy the inequality $U(g,p') leq U(f,p)$ by refining the $p$ that satisfies $U(f,p)-L(f,p) < epsilon$.
My Question
How can I precisely pick $p'$ so that $U(g,p')=U(f,p)$ in the special case I described above? Intuitively, I can just keep refining $p$ and that partition which includes $x_0$ will tend to $0$, but I want to rigorously know at what point will $U(g,p')=U(f,p)$?
Thanks
multivariable-calculus proof-writing
multivariable-calculus proof-writing
asked Jan 23 at 3:26
john fowlesjohn fowles
1,208817
asked Jan 23 at 3:26
john fowlesjohn fowles
1,208817
asked Jan 23 at 3:26
john fowlesjohn fowles
1,208817
asked Jan 23 at 3:26
john fowlesjohn fowles
1,208817
asked Jan 23 at 3:26
john fowlesjohn fowles
1,208817
1,208817
1
$begingroup$
The refinement may strictly decrease the upper sum, you can't ensure exact equality between them. In any case, think of the worst case scenario, where $f(x_0)$ was the maximum on its interval in $p$ and now $g(x_0)$ is a bit bigger, say by $a$. How exactly can you refine near $x_0$ so that $U(g,p')$ is close enough to $U(f,p')$? You'll need the interval containing $x_0$ to be short...how short?
$endgroup$
– Ian
Jan 23 at 3:37
$begingroup$
In the second last line of your comment, did you mean to write "close enough to $U(f,p)$ instead of "close enough to $U(f,p')$?
$endgroup$
– john fowles
Jan 23 at 5:39
$begingroup$
No, it is easier to compare on the same partition. Then $U(f,p')$ will be close enough to $U(f,p)$ because $U(f,p)$ is already quite close to the integral of $f$.
$endgroup$
– Ian
Jan 23 at 8:58
$begingroup$
Please, please be careful with your order of quantifiers.
$endgroup$
– Ted Shifrin
Jan 25 at 1:15
add a comment |
1
$begingroup$
The refinement may strictly decrease the upper sum, you can't ensure exact equality between them. In any case, think of the worst case scenario, where $f(x_0)$ was the maximum on its interval in $p$ and now $g(x_0)$ is a bit bigger, say by $a$. How exactly can you refine near $x_0$ so that $U(g,p')$ is close enough to $U(f,p')$? You'll need the interval containing $x_0$ to be short...how short?
$endgroup$
– Ian
Jan 23 at 3:37
$begingroup$
In the second last line of your comment, did you mean to write "close enough to $U(f,p)$ instead of "close enough to $U(f,p')$?
$endgroup$
– john fowles
Jan 23 at 5:39
$begingroup$
No, it is easier to compare on the same partition. Then $U(f,p')$ will be close enough to $U(f,p)$ because $U(f,p)$ is already quite close to the integral of $f$.
$endgroup$
– Ian
Jan 23 at 8:58
$begingroup$
Please, please be careful with your order of quantifiers.
$endgroup$
– Ted Shifrin
Jan 25 at 1:15
1
1
$begingroup$
The refinement may strictly decrease the upper sum, you can't ensure exact equality between them. In any case, think of the worst case scenario, where $f(x_0)$ was the maximum on its interval in $p$ and now $g(x_0)$ is a bit bigger, say by $a$. How exactly can you refine near $x_0$ so that $U(g,p')$ is close enough to $U(f,p')$? You'll need the interval containing $x_0$ to be short...how short?
$endgroup$
– Ian
Jan 23 at 3:37
$begingroup$
The refinement may strictly decrease the upper sum, you can't ensure exact equality between them. In any case, think of the worst case scenario, where $f(x_0)$ was the maximum on its interval in $p$ and now $g(x_0)$ is a bit bigger, say by $a$. How exactly can you refine near $x_0$ so that $U(g,p')$ is close enough to $U(f,p')$? You'll need the interval containing $x_0$ to be short...how short?
$endgroup$
– Ian
Jan 23 at 3:37
$begingroup$
In the second last line of your comment, did you mean to write "close enough to $U(f,p)$ instead of "close enough to $U(f,p')$?
$endgroup$
– john fowles
Jan 23 at 5:39
$begingroup$
In the second last line of your comment, did you mean to write "close enough to $U(f,p)$ instead of "close enough to $U(f,p')$?
$endgroup$
– john fowles
Jan 23 at 5:39
$begingroup$
No, it is easier to compare on the same partition. Then $U(f,p')$ will be close enough to $U(f,p)$ because $U(f,p)$ is already quite close to the integral of $f$.
$endgroup$
– Ian
Jan 23 at 8:58
$begingroup$
No, it is easier to compare on the same partition. Then $U(f,p')$ will be close enough to $U(f,p)$ because $U(f,p)$ is already quite close to the integral of $f$.
$endgroup$
– Ian
Jan 23 at 8:58
$begingroup$
Please, please be careful with your order of quantifiers.
$endgroup$
– Ted Shifrin
Jan 25 at 1:15
$begingroup$
Please, please be careful with your order of quantifiers.
$endgroup$
– Ted Shifrin
Jan 25 at 1:15
add a comment |
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$begingroup$
The refinement may strictly decrease the upper sum, you can't ensure exact equality between them. In any case, think of the worst case scenario, where $f(x_0)$ was the maximum on its interval in $p$ and now $g(x_0)$ is a bit bigger, say by $a$. How exactly can you refine near $x_0$ so that $U(g,p')$ is close enough to $U(f,p')$? You'll need the interval containing $x_0$ to be short...how short?
$endgroup$
– Ian
Jan 23 at 3:37
$begingroup$
In the second last line of your comment, did you mean to write "close enough to $U(f,p)$ instead of "close enough to $U(f,p')$?
$endgroup$
– john fowles
Jan 23 at 5:39
$begingroup$
No, it is easier to compare on the same partition. Then $U(f,p')$ will be close enough to $U(f,p)$ because $U(f,p)$ is already quite close to the integral of $f$.
$endgroup$
– Ian
Jan 23 at 8:58
$begingroup$
Please, please be careful with your order of quantifiers.
$endgroup$
– Ted Shifrin
Jan 25 at 1:15