Riemann-integration problem
$begingroup$
Here is the exercise
Let $f:[a,b]rightarrow mathbb{R}$ be Riemann-integrable. Prove that $f^+$, $f^-$ and $|f|$ are also Riemann-integrable, when
$$f^+=begin{cases}
f(x) & f(x)geq 0 \
0 & otherwise
end{cases}$$
$$f^-=begin{cases}
-f(x) & f(x)leq 0 \
0 & otherwise
end{cases}$$
This problem seems so obvious. Why wouldn't $f^+$ be integrable? Anyway, I need to prove this using this hint:
"$f$ is Riemann-integrable if with every $epsilon>0$ there exists step functions $hleq fleq g $ such that $int h -int g <epsilon$."
I have no idea where to start...
proof-writing riemann-integration
edited Jan 25 at 15:48
mathcounterexamples.net
27k22158
asked Jan 25 at 15:42
jtejte
336
$endgroup$
add a comment |
$begingroup$
Here is the exercise
Let $f:[a,b]rightarrow mathbb{R}$ be Riemann-integrable. Prove that $f^+$, $f^-$ and $|f|$ are also Riemann-integrable, when
$$f^+=begin{cases}
f(x) & f(x)geq 0 \
0 & otherwise
end{cases}$$
$$f^-=begin{cases}
-f(x) & f(x)leq 0 \
0 & otherwise
end{cases}$$
This problem seems so obvious. Why wouldn't $f^+$ be integrable? Anyway, I need to prove this using this hint:
"$f$ is Riemann-integrable if with every $epsilon>0$ there exists step functions $hleq fleq g $ such that $int h -int g <epsilon$."
I have no idea where to start...
proof-writing riemann-integration
edited Jan 25 at 15:48
mathcounterexamples.net
27k22158
asked Jan 25 at 15:42
jtejte
336
$endgroup$
$begingroup$
I like a lot your comment This problem seems so obvious. Why wouldn't $f^+$ be integrable? Asking yourself why it should be will be better to bring you to the road of a proof! Hint build step functions $g^+, h^+$ from $h$ and $g$ such that $h^+leq f^+ leq g^+ $ and $int h^+ -int g^+ <epsilon$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 15:51
$begingroup$
Haha glad you liked my comment. Anyway, this doesn't seem to proof anything to me: Let $h$ and $g$ be step functions such that $hleq fleq g$. Now it's obvious that $h^+leq f^+leq g^+$ and as $f$ is Riemann-integrable, then $int h^+ -int g^+<epsilon$ thus making $f^+$ also Riemann-integrable.
$endgroup$
– jte
Jan 25 at 16:22
$begingroup$
@Matematleta Thanks I'll try to understand that.
$endgroup$
– jte
Jan 25 at 16:22
add a comment |
$begingroup$
Here is the exercise
Let $f:[a,b]rightarrow mathbb{R}$ be Riemann-integrable. Prove that $f^+$, $f^-$ and $|f|$ are also Riemann-integrable, when
$$f^+=begin{cases}
f(x) & f(x)geq 0 \
0 & otherwise
end{cases}$$
$$f^-=begin{cases}
-f(x) & f(x)leq 0 \
0 & otherwise
end{cases}$$
This problem seems so obvious. Why wouldn't $f^+$ be integrable? Anyway, I need to prove this using this hint:
"$f$ is Riemann-integrable if with every $epsilon>0$ there exists step functions $hleq fleq g $ such that $int h -int g <epsilon$."
I have no idea where to start...
proof-writing riemann-integration
edited Jan 25 at 15:48
mathcounterexamples.net
27k22158
asked Jan 25 at 15:42
jtejte
336
$endgroup$
Here is the exercise
Let $f:[a,b]rightarrow mathbb{R}$ be Riemann-integrable. Prove that $f^+$, $f^-$ and $|f|$ are also Riemann-integrable, when
$$f^+=begin{cases}
f(x) & f(x)geq 0 \
0 & otherwise
end{cases}$$
$$f^-=begin{cases}
-f(x) & f(x)leq 0 \
0 & otherwise
end{cases}$$
This problem seems so obvious. Why wouldn't $f^+$ be integrable? Anyway, I need to prove this using this hint:
"$f$ is Riemann-integrable if with every $epsilon>0$ there exists step functions $hleq fleq g $ such that $int h -int g <epsilon$."
I have no idea where to start...
proof-writing riemann-integration
proof-writing riemann-integration
edited Jan 25 at 15:48
mathcounterexamples.net
27k22158
asked Jan 25 at 15:42
jtejte
336
edited Jan 25 at 15:48
mathcounterexamples.net
27k22158
asked Jan 25 at 15:42
jtejte
336
edited Jan 25 at 15:48
mathcounterexamples.net
27k22158
edited Jan 25 at 15:48
mathcounterexamples.net
27k22158
edited Jan 25 at 15:48
mathcounterexamples.net
27k22158
27k22158
asked Jan 25 at 15:42
jtejte
336
asked Jan 25 at 15:42
jtejte
336
asked Jan 25 at 15:42
jtejte
336
336
$begingroup$
I like a lot your comment This problem seems so obvious. Why wouldn't $f^+$ be integrable? Asking yourself why it should be will be better to bring you to the road of a proof! Hint build step functions $g^+, h^+$ from $h$ and $g$ such that $h^+leq f^+ leq g^+ $ and $int h^+ -int g^+ <epsilon$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 15:51
$begingroup$
Haha glad you liked my comment. Anyway, this doesn't seem to proof anything to me: Let $h$ and $g$ be step functions such that $hleq fleq g$. Now it's obvious that $h^+leq f^+leq g^+$ and as $f$ is Riemann-integrable, then $int h^+ -int g^+<epsilon$ thus making $f^+$ also Riemann-integrable.
$endgroup$
– jte
Jan 25 at 16:22
$begingroup$
@Matematleta Thanks I'll try to understand that.
$endgroup$
– jte
Jan 25 at 16:22
add a comment |
$begingroup$
I like a lot your comment This problem seems so obvious. Why wouldn't $f^+$ be integrable? Asking yourself why it should be will be better to bring you to the road of a proof! Hint build step functions $g^+, h^+$ from $h$ and $g$ such that $h^+leq f^+ leq g^+ $ and $int h^+ -int g^+ <epsilon$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 15:51
$begingroup$
Haha glad you liked my comment. Anyway, this doesn't seem to proof anything to me: Let $h$ and $g$ be step functions such that $hleq fleq g$. Now it's obvious that $h^+leq f^+leq g^+$ and as $f$ is Riemann-integrable, then $int h^+ -int g^+<epsilon$ thus making $f^+$ also Riemann-integrable.
$endgroup$
– jte
Jan 25 at 16:22
$begingroup$
@Matematleta Thanks I'll try to understand that.
$endgroup$
– jte
Jan 25 at 16:22
$begingroup$
I like a lot your comment This problem seems so obvious. Why wouldn't $f^+$ be integrable? Asking yourself why it should be will be better to bring you to the road of a proof! Hint build step functions $g^+, h^+$ from $h$ and $g$ such that $h^+leq f^+ leq g^+ $ and $int h^+ -int g^+ <epsilon$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 15:51
$begingroup$
I like a lot your comment This problem seems so obvious. Why wouldn't $f^+$ be integrable? Asking yourself why it should be will be better to bring you to the road of a proof! Hint build step functions $g^+, h^+$ from $h$ and $g$ such that $h^+leq f^+ leq g^+ $ and $int h^+ -int g^+ <epsilon$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 15:51
$begingroup$
Haha glad you liked my comment. Anyway, this doesn't seem to proof anything to me: Let $h$ and $g$ be step functions such that $hleq fleq g$. Now it's obvious that $h^+leq f^+leq g^+$ and as $f$ is Riemann-integrable, then $int h^+ -int g^+<epsilon$ thus making $f^+$ also Riemann-integrable.
$endgroup$
– jte
Jan 25 at 16:22
$begingroup$
Haha glad you liked my comment. Anyway, this doesn't seem to proof anything to me: Let $h$ and $g$ be step functions such that $hleq fleq g$. Now it's obvious that $h^+leq f^+leq g^+$ and as $f$ is Riemann-integrable, then $int h^+ -int g^+<epsilon$ thus making $f^+$ also Riemann-integrable.
$endgroup$
– jte
Jan 25 at 16:22
$begingroup$
@Matematleta Thanks I'll try to understand that.
$endgroup$
– jte
Jan 25 at 16:22
$begingroup$
@Matematleta Thanks I'll try to understand that.
$endgroup$
– jte
Jan 25 at 16:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We prove the Riemann integrability of $f^+$. A similar proof can be done for $f^-$.
As $f$ is Riemann-integrable, for all $epsilon>0$ there exists step functions $h leq fleq g $ such that $int h -int g <epsilon$.
Now define $h^+ = max(h,0)$ and $g^+ = max(g,0)$. You can verify that:
$h^+, g^+$ are step functions.- You have $h^+ le f^+ le g^+$ for all $x in [a,b]$.
And also $g^+ - h^+ le g-h$ for all $x in [a,b]$. This implies $0 le int (g^+-h^+) le int (g-h) le epsilon$
And concludes the proof.
answered Jan 25 at 16:21
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
I guess this is as simple as it gets as I immediately understood it. Thanks.
$endgroup$
– jte
Jan 25 at 16:26
add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
votes
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oldest
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oldest
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$begingroup$
We prove the Riemann integrability of $f^+$. A similar proof can be done for $f^-$.
As $f$ is Riemann-integrable, for all $epsilon>0$ there exists step functions $h leq fleq g $ such that $int h -int g <epsilon$.
Now define $h^+ = max(h,0)$ and $g^+ = max(g,0)$. You can verify that:
$h^+, g^+$ are step functions.- You have $h^+ le f^+ le g^+$ for all $x in [a,b]$.
And also $g^+ - h^+ le g-h$ for all $x in [a,b]$. This implies $0 le int (g^+-h^+) le int (g-h) le epsilon$
And concludes the proof.
answered Jan 25 at 16:21
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
I guess this is as simple as it gets as I immediately understood it. Thanks.
$endgroup$
– jte
Jan 25 at 16:26
add a comment |
$begingroup$
We prove the Riemann integrability of $f^+$. A similar proof can be done for $f^-$.
As $f$ is Riemann-integrable, for all $epsilon>0$ there exists step functions $h leq fleq g $ such that $int h -int g <epsilon$.
Now define $h^+ = max(h,0)$ and $g^+ = max(g,0)$. You can verify that:
$h^+, g^+$ are step functions.- You have $h^+ le f^+ le g^+$ for all $x in [a,b]$.
And also $g^+ - h^+ le g-h$ for all $x in [a,b]$. This implies $0 le int (g^+-h^+) le int (g-h) le epsilon$
And concludes the proof.
answered Jan 25 at 16:21
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
$begingroup$
I guess this is as simple as it gets as I immediately understood it. Thanks.
$endgroup$
– jte
Jan 25 at 16:26
add a comment |
$begingroup$
We prove the Riemann integrability of $f^+$. A similar proof can be done for $f^-$.
As $f$ is Riemann-integrable, for all $epsilon>0$ there exists step functions $h leq fleq g $ such that $int h -int g <epsilon$.
Now define $h^+ = max(h,0)$ and $g^+ = max(g,0)$. You can verify that:
$h^+, g^+$ are step functions.- You have $h^+ le f^+ le g^+$ for all $x in [a,b]$.
And also $g^+ - h^+ le g-h$ for all $x in [a,b]$. This implies $0 le int (g^+-h^+) le int (g-h) le epsilon$
And concludes the proof.
answered Jan 25 at 16:21
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
We prove the Riemann integrability of $f^+$. A similar proof can be done for $f^-$.
As $f$ is Riemann-integrable, for all $epsilon>0$ there exists step functions $h leq fleq g $ such that $int h -int g <epsilon$.
Now define $h^+ = max(h,0)$ and $g^+ = max(g,0)$. You can verify that:
$h^+, g^+$ are step functions.- You have $h^+ le f^+ le g^+$ for all $x in [a,b]$.
And also $g^+ - h^+ le g-h$ for all $x in [a,b]$. This implies $0 le int (g^+-h^+) le int (g-h) le epsilon$
And concludes the proof.
answered Jan 25 at 16:21
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 25 at 16:21
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 25 at 16:21
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 25 at 16:21
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
$begingroup$
I guess this is as simple as it gets as I immediately understood it. Thanks.
$endgroup$
– jte
Jan 25 at 16:26
add a comment |
$begingroup$
I guess this is as simple as it gets as I immediately understood it. Thanks.
$endgroup$
– jte
Jan 25 at 16:26
$begingroup$
I guess this is as simple as it gets as I immediately understood it. Thanks.
$endgroup$
– jte
Jan 25 at 16:26
$begingroup$
I guess this is as simple as it gets as I immediately understood it. Thanks.
$endgroup$
– jte
Jan 25 at 16:26
add a comment |
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$begingroup$
I like a lot your comment This problem seems so obvious. Why wouldn't $f^+$ be integrable? Asking yourself why it should be will be better to bring you to the road of a proof! Hint build step functions $g^+, h^+$ from $h$ and $g$ such that $h^+leq f^+ leq g^+ $ and $int h^+ -int g^+ <epsilon$.
$endgroup$
– mathcounterexamples.net
Jan 25 at 15:51
$begingroup$
Haha glad you liked my comment. Anyway, this doesn't seem to proof anything to me: Let $h$ and $g$ be step functions such that $hleq fleq g$. Now it's obvious that $h^+leq f^+leq g^+$ and as $f$ is Riemann-integrable, then $int h^+ -int g^+<epsilon$ thus making $f^+$ also Riemann-integrable.
$endgroup$
– jte
Jan 25 at 16:22
$begingroup$
@Matematleta Thanks I'll try to understand that.
$endgroup$
– jte
Jan 25 at 16:22