Mixing
Prove integrability and equality
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Let $mathfrak{M}$ be a $sigma$ algebra of subsets of $X$ and let $mu:mathfrak{M}rightarrowmathbb{R}$ be a countably additive measure.$A in mathfrak{M}, mu(A) = mu(X), f in L_1(A)$. Prove that $ f in L_1(X)$ and $intlimits_{X}{fdmu} = intlimits_{A}{fdmu}$
Since $f$ is integrable on $A$ then it is measurable on this set, hence it is also measurable on $X$. Is not this enough to claim that it is also integrable on $X$?
How do I prove the equality? I do not need a complete solution. A hint is enough. Thank you
measure-theory lebesgue-integral lebesgue-measure
asked Jan 28 at 11:27
Don DraperDon Draper
87110
$endgroup$
add a comment |
$begingroup$
Let $mathfrak{M}$ be a $sigma$ algebra of subsets of $X$ and let $mu:mathfrak{M}rightarrowmathbb{R}$ be a countably additive measure.$A in mathfrak{M}, mu(A) = mu(X), f in L_1(A)$. Prove that $ f in L_1(X)$ and $intlimits_{X}{fdmu} = intlimits_{A}{fdmu}$
Since $f$ is integrable on $A$ then it is measurable on this set, hence it is also measurable on $X$. Is not this enough to claim that it is also integrable on $X$?
How do I prove the equality? I do not need a complete solution. A hint is enough. Thank you
measure-theory lebesgue-integral lebesgue-measure
asked Jan 28 at 11:27
Don DraperDon Draper
87110
$endgroup$
$begingroup$
Does $fin L_1(A)$ mean $f$ is measurable with respect to $mathfrak{M}$?
$endgroup$
– Keen-ameteur
Jan 28 at 11:33
$begingroup$
Integrability implies measurability on $A$. Am I wrong?
$endgroup$
– Don Draper
Jan 28 at 11:37
add a comment |
$begingroup$
Let $mathfrak{M}$ be a $sigma$ algebra of subsets of $X$ and let $mu:mathfrak{M}rightarrowmathbb{R}$ be a countably additive measure.$A in mathfrak{M}, mu(A) = mu(X), f in L_1(A)$. Prove that $ f in L_1(X)$ and $intlimits_{X}{fdmu} = intlimits_{A}{fdmu}$
Since $f$ is integrable on $A$ then it is measurable on this set, hence it is also measurable on $X$. Is not this enough to claim that it is also integrable on $X$?
How do I prove the equality? I do not need a complete solution. A hint is enough. Thank you
measure-theory lebesgue-integral lebesgue-measure
asked Jan 28 at 11:27
Don DraperDon Draper
87110
$endgroup$
Let $mathfrak{M}$ be a $sigma$ algebra of subsets of $X$ and let $mu:mathfrak{M}rightarrowmathbb{R}$ be a countably additive measure.$A in mathfrak{M}, mu(A) = mu(X), f in L_1(A)$. Prove that $ f in L_1(X)$ and $intlimits_{X}{fdmu} = intlimits_{A}{fdmu}$
Since $f$ is integrable on $A$ then it is measurable on this set, hence it is also measurable on $X$. Is not this enough to claim that it is also integrable on $X$?
How do I prove the equality? I do not need a complete solution. A hint is enough. Thank you
measure-theory lebesgue-integral lebesgue-measure
measure-theory lebesgue-integral lebesgue-measure
asked Jan 28 at 11:27
Don DraperDon Draper
87110
asked Jan 28 at 11:27
Don DraperDon Draper
87110
asked Jan 28 at 11:27
Don DraperDon Draper
87110
asked Jan 28 at 11:27
Don DraperDon Draper
87110
asked Jan 28 at 11:27
Don DraperDon Draper
87110
87110
$begingroup$
Does $fin L_1(A)$ mean $f$ is measurable with respect to $mathfrak{M}$?
$endgroup$
– Keen-ameteur
Jan 28 at 11:33
$begingroup$
Integrability implies measurability on $A$. Am I wrong?
$endgroup$
– Don Draper
Jan 28 at 11:37
add a comment |
$begingroup$
Does $fin L_1(A)$ mean $f$ is measurable with respect to $mathfrak{M}$?
$endgroup$
– Keen-ameteur
Jan 28 at 11:33
$begingroup$
Integrability implies measurability on $A$. Am I wrong?
$endgroup$
– Don Draper
Jan 28 at 11:37
$begingroup$
Does $fin L_1(A)$ mean $f$ is measurable with respect to $mathfrak{M}$?
$endgroup$
– Keen-ameteur
Jan 28 at 11:33
$begingroup$
Does $fin L_1(A)$ mean $f$ is measurable with respect to $mathfrak{M}$?
$endgroup$
– Keen-ameteur
Jan 28 at 11:33
$begingroup$
Integrability implies measurability on $A$. Am I wrong?
$endgroup$
– Don Draper
Jan 28 at 11:37
$begingroup$
Integrability implies measurability on $A$. Am I wrong?
$endgroup$
– Don Draper
Jan 28 at 11:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have to assume that $f$ is measurable on $X$. $int_X f, dmu=int_A f ,dmu+int_{Xsetminus A} f ,dmu$ and the second term is $0$. [Any measurable function is integrable over a measurable set of measure $0$ and value of the integral is $0$].
answered Jan 28 at 11:54
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
$endgroup$
$begingroup$
Thank you. I should have noted this myself, but I am still too bad at solving such problems...
$endgroup$
– Don Draper
Jan 28 at 12:03
$begingroup$
do we need to know that $mu(A) = mu(X)$ to infer that $f$ is integrable on $X$?
$endgroup$
– Don Draper
Jan 28 at 14:11
$begingroup$
@DonDraper Yes. Otherwise there is no reason for $f$ to be intergable on $X$.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:09
$begingroup$
another question if you don't mind. Does the equality you in your post imply the integrability of $f$ on $X$? Or do we need to prove the integrability first and only then we have the right to make the conclusion? | If so, what is the reasoning behind integrability on $X$?
$endgroup$
– Don Draper
Jan 29 at 3:02
1
$begingroup$
@DonDraper Befor writing the equation you have to use the fact that any function is integrable over a set of measure $0$. If $f$ is integrable over $A$ and its complement then it is intergable over $X$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:24
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You have to assume that $f$ is measurable on $X$. $int_X f, dmu=int_A f ,dmu+int_{Xsetminus A} f ,dmu$ and the second term is $0$. [Any measurable function is integrable over a measurable set of measure $0$ and value of the integral is $0$].
answered Jan 28 at 11:54
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
$endgroup$
$begingroup$
Thank you. I should have noted this myself, but I am still too bad at solving such problems...
$endgroup$
– Don Draper
Jan 28 at 12:03
$begingroup$
do we need to know that $mu(A) = mu(X)$ to infer that $f$ is integrable on $X$?
$endgroup$
– Don Draper
Jan 28 at 14:11
$begingroup$
@DonDraper Yes. Otherwise there is no reason for $f$ to be intergable on $X$.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:09
$begingroup$
another question if you don't mind. Does the equality you in your post imply the integrability of $f$ on $X$? Or do we need to prove the integrability first and only then we have the right to make the conclusion? | If so, what is the reasoning behind integrability on $X$?
$endgroup$
– Don Draper
Jan 29 at 3:02
1
$begingroup$
@DonDraper Befor writing the equation you have to use the fact that any function is integrable over a set of measure $0$. If $f$ is integrable over $A$ and its complement then it is intergable over $X$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:24
add a comment |
$begingroup$
You have to assume that $f$ is measurable on $X$. $int_X f, dmu=int_A f ,dmu+int_{Xsetminus A} f ,dmu$ and the second term is $0$. [Any measurable function is integrable over a measurable set of measure $0$ and value of the integral is $0$].
answered Jan 28 at 11:54
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
$endgroup$
$begingroup$
Thank you. I should have noted this myself, but I am still too bad at solving such problems...
$endgroup$
– Don Draper
Jan 28 at 12:03
$begingroup$
do we need to know that $mu(A) = mu(X)$ to infer that $f$ is integrable on $X$?
$endgroup$
– Don Draper
Jan 28 at 14:11
$begingroup$
@DonDraper Yes. Otherwise there is no reason for $f$ to be intergable on $X$.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:09
$begingroup$
another question if you don't mind. Does the equality you in your post imply the integrability of $f$ on $X$? Or do we need to prove the integrability first and only then we have the right to make the conclusion? | If so, what is the reasoning behind integrability on $X$?
$endgroup$
– Don Draper
Jan 29 at 3:02
1
$begingroup$
@DonDraper Befor writing the equation you have to use the fact that any function is integrable over a set of measure $0$. If $f$ is integrable over $A$ and its complement then it is intergable over $X$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:24
add a comment |
$begingroup$
You have to assume that $f$ is measurable on $X$. $int_X f, dmu=int_A f ,dmu+int_{Xsetminus A} f ,dmu$ and the second term is $0$. [Any measurable function is integrable over a measurable set of measure $0$ and value of the integral is $0$].
answered Jan 28 at 11:54
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
$endgroup$
You have to assume that $f$ is measurable on $X$. $int_X f, dmu=int_A f ,dmu+int_{Xsetminus A} f ,dmu$ and the second term is $0$. [Any measurable function is integrable over a measurable set of measure $0$ and value of the integral is $0$].
answered Jan 28 at 11:54
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
answered Jan 28 at 11:54
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
answered Jan 28 at 11:54
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
answered Jan 28 at 11:54
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
70.7k53170
$begingroup$
Thank you. I should have noted this myself, but I am still too bad at solving such problems...
$endgroup$
– Don Draper
Jan 28 at 12:03
$begingroup$
do we need to know that $mu(A) = mu(X)$ to infer that $f$ is integrable on $X$?
$endgroup$
– Don Draper
Jan 28 at 14:11
$begingroup$
@DonDraper Yes. Otherwise there is no reason for $f$ to be intergable on $X$.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:09
$begingroup$
another question if you don't mind. Does the equality you in your post imply the integrability of $f$ on $X$? Or do we need to prove the integrability first and only then we have the right to make the conclusion? | If so, what is the reasoning behind integrability on $X$?
$endgroup$
– Don Draper
Jan 29 at 3:02
1
$begingroup$
@DonDraper Befor writing the equation you have to use the fact that any function is integrable over a set of measure $0$. If $f$ is integrable over $A$ and its complement then it is intergable over $X$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:24
add a comment |
$begingroup$
Thank you. I should have noted this myself, but I am still too bad at solving such problems...
$endgroup$
– Don Draper
Jan 28 at 12:03
$begingroup$
do we need to know that $mu(A) = mu(X)$ to infer that $f$ is integrable on $X$?
$endgroup$
– Don Draper
Jan 28 at 14:11
$begingroup$
@DonDraper Yes. Otherwise there is no reason for $f$ to be intergable on $X$.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:09
$begingroup$
another question if you don't mind. Does the equality you in your post imply the integrability of $f$ on $X$? Or do we need to prove the integrability first and only then we have the right to make the conclusion? | If so, what is the reasoning behind integrability on $X$?
$endgroup$
– Don Draper
Jan 29 at 3:02
1
$begingroup$
@DonDraper Befor writing the equation you have to use the fact that any function is integrable over a set of measure $0$. If $f$ is integrable over $A$ and its complement then it is intergable over $X$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:24
$begingroup$
Thank you. I should have noted this myself, but I am still too bad at solving such problems...
$endgroup$
– Don Draper
Jan 28 at 12:03
$begingroup$
Thank you. I should have noted this myself, but I am still too bad at solving such problems...
$endgroup$
– Don Draper
Jan 28 at 12:03
$begingroup$
do we need to know that $mu(A) = mu(X)$ to infer that $f$ is integrable on $X$?
$endgroup$
– Don Draper
Jan 28 at 14:11
$begingroup$
do we need to know that $mu(A) = mu(X)$ to infer that $f$ is integrable on $X$?
$endgroup$
– Don Draper
Jan 28 at 14:11
$begingroup$
@DonDraper Yes. Otherwise there is no reason for $f$ to be intergable on $X$.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:09
$begingroup$
@DonDraper Yes. Otherwise there is no reason for $f$ to be intergable on $X$.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:09
$begingroup$
another question if you don't mind. Does the equality you in your post imply the integrability of $f$ on $X$? Or do we need to prove the integrability first and only then we have the right to make the conclusion? | If so, what is the reasoning behind integrability on $X$?
$endgroup$
– Don Draper
Jan 29 at 3:02
$begingroup$
another question if you don't mind. Does the equality you in your post imply the integrability of $f$ on $X$? Or do we need to prove the integrability first and only then we have the right to make the conclusion? | If so, what is the reasoning behind integrability on $X$?
$endgroup$
– Don Draper
Jan 29 at 3:02
1
1
$begingroup$
@DonDraper Befor writing the equation you have to use the fact that any function is integrable over a set of measure $0$. If $f$ is integrable over $A$ and its complement then it is intergable over $X$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:24
$begingroup$
@DonDraper Befor writing the equation you have to use the fact that any function is integrable over a set of measure $0$. If $f$ is integrable over $A$ and its complement then it is intergable over $X$.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:24
add a comment |
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$begingroup$
Does $fin L_1(A)$ mean $f$ is measurable with respect to $mathfrak{M}$?
$endgroup$
– Keen-ameteur
Jan 28 at 11:33
$begingroup$
Integrability implies measurability on $A$. Am I wrong?
$endgroup$
– Don Draper
Jan 28 at 11:37