min(f(x), 5) integrability
$begingroup$
Q: If $f(x)$ is integrable on $[a,b]$, let $g(x)=min(f(x),5)$. Is $g(x)$ integrable on $[a,b]$?
Does the number that g(x) chooses matter? (ie. can the 5 really just be any constant?)
My Start: I would think yes, just because any part where g(x) chooses f(x) is obviously integrable, and any constant is also integrable so is it just integrable everywhere?
calculus integration
asked Jan 8 at 16:34
Emma PascoeEmma Pascoe
161
$endgroup$
add a comment |
$begingroup$
Q: If $f(x)$ is integrable on $[a,b]$, let $g(x)=min(f(x),5)$. Is $g(x)$ integrable on $[a,b]$?
Does the number that g(x) chooses matter? (ie. can the 5 really just be any constant?)
My Start: I would think yes, just because any part where g(x) chooses f(x) is obviously integrable, and any constant is also integrable so is it just integrable everywhere?
calculus integration
asked Jan 8 at 16:34
Emma PascoeEmma Pascoe
161
$endgroup$
add a comment |
$begingroup$
Q: If $f(x)$ is integrable on $[a,b]$, let $g(x)=min(f(x),5)$. Is $g(x)$ integrable on $[a,b]$?
Does the number that g(x) chooses matter? (ie. can the 5 really just be any constant?)
My Start: I would think yes, just because any part where g(x) chooses f(x) is obviously integrable, and any constant is also integrable so is it just integrable everywhere?
calculus integration
asked Jan 8 at 16:34
Emma PascoeEmma Pascoe
161
$endgroup$
Q: If $f(x)$ is integrable on $[a,b]$, let $g(x)=min(f(x),5)$. Is $g(x)$ integrable on $[a,b]$?
Does the number that g(x) chooses matter? (ie. can the 5 really just be any constant?)
My Start: I would think yes, just because any part where g(x) chooses f(x) is obviously integrable, and any constant is also integrable so is it just integrable everywhere?
calculus integration
calculus integration
asked Jan 8 at 16:34
Emma PascoeEmma Pascoe
161
asked Jan 8 at 16:34
Emma PascoeEmma Pascoe
161
asked Jan 8 at 16:34
Emma PascoeEmma Pascoe
161
asked Jan 8 at 16:34
Emma PascoeEmma Pascoe
161
asked Jan 8 at 16:34
Emma PascoeEmma Pascoe
161
161
add a comment |
add a comment |
2 Answers
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$min(a, b)=dfrac{a+b-|a-b|}{2}$. If $a$ is an integrable function and $b$ is a constant, then the right hand side should be Lebesgue integrable.
answered Jan 8 at 16:41
EuxhenHEuxhenH
482210
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add a comment |
$begingroup$
I'm not very sure about my reasoning, but wouldn't $f(x):=displaystyle frac{1}{x}$ on [-1,1] make a counter-example?
answered Jan 8 at 16:46
Edward H.Edward H.
1739
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
$min(a, b)=dfrac{a+b-|a-b|}{2}$. If $a$ is an integrable function and $b$ is a constant, then the right hand side should be Lebesgue integrable.
answered Jan 8 at 16:41
EuxhenHEuxhenH
482210
$endgroup$
add a comment |
$begingroup$
$min(a, b)=dfrac{a+b-|a-b|}{2}$. If $a$ is an integrable function and $b$ is a constant, then the right hand side should be Lebesgue integrable.
answered Jan 8 at 16:41
EuxhenHEuxhenH
482210
$endgroup$
add a comment |
$begingroup$
$min(a, b)=dfrac{a+b-|a-b|}{2}$. If $a$ is an integrable function and $b$ is a constant, then the right hand side should be Lebesgue integrable.
answered Jan 8 at 16:41
EuxhenHEuxhenH
482210
$endgroup$
$min(a, b)=dfrac{a+b-|a-b|}{2}$. If $a$ is an integrable function and $b$ is a constant, then the right hand side should be Lebesgue integrable.
answered Jan 8 at 16:41
EuxhenHEuxhenH
482210
answered Jan 8 at 16:41
EuxhenHEuxhenH
482210
answered Jan 8 at 16:41
EuxhenHEuxhenH
482210
answered Jan 8 at 16:41
EuxhenHEuxhenH
482210
482210
add a comment |
add a comment |
$begingroup$
I'm not very sure about my reasoning, but wouldn't $f(x):=displaystyle frac{1}{x}$ on [-1,1] make a counter-example?
answered Jan 8 at 16:46
Edward H.Edward H.
1739
$endgroup$
add a comment |
$begingroup$
I'm not very sure about my reasoning, but wouldn't $f(x):=displaystyle frac{1}{x}$ on [-1,1] make a counter-example?
answered Jan 8 at 16:46
Edward H.Edward H.
1739
$endgroup$
add a comment |
$begingroup$
I'm not very sure about my reasoning, but wouldn't $f(x):=displaystyle frac{1}{x}$ on [-1,1] make a counter-example?
answered Jan 8 at 16:46
Edward H.Edward H.
1739
$endgroup$
I'm not very sure about my reasoning, but wouldn't $f(x):=displaystyle frac{1}{x}$ on [-1,1] make a counter-example?
answered Jan 8 at 16:46
Edward H.Edward H.
1739
answered Jan 8 at 16:46
Edward H.Edward H.
1739
answered Jan 8 at 16:46
Edward H.Edward H.
1739
answered Jan 8 at 16:46
Edward H.Edward H.
1739
1739
add a comment |
add a comment |
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