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Let $f$ be non negative and $int_Omega fdmu$ is finite then f is finite valued almost everywhere
[1] Let $f$ be non negative and $int_Omega fdmu$ is finite then f is finite valued almost everywhere.
[2]Let $E_1=(-infty,-2) bigcup (2,infty)$ $E_2=(-infty,-3)bigcup(3,infty)$ and $E_3=(-4,4)$
Define $F=chi_E -chi_E + chi_E $( ie.$E_1-E_2+E_3)$
Then prove that $int_Omega fdmu$ exists although $mu(E_1)-mu(E_2)+mu(E_3)$ is undefined
Note here that $Omega= mathbb{R} , Sigma=$ Borel class, $mu$ is the lebesgue measure on $Sigma$
Everything is as defined in the H.L ROyden
MY ATTEMPT
[1] First I write $ A=(x in Omega: f(x) =infty) $ and suppose $ mu (A)>0$
For each $n geq0$ the function $n.chi_A$ is a simple function satisfying $0leq n. chi_Aleq f$
Hence $0leq n. chi_A=int_Omega n.chi_A leq int_Omega fdmu$ for each $ngeq0$
This implies that $int_Omega fdmu =infty$
Is this correct for part one?
[2]Also for part two do I have to find the union of all and then integrate? I'm not sure exactly how to proceed.
The measure of an interval is it's length but I don't see how I can find the length of an open interval from infinity. There exits an infinite amount of open intervals and there are so many open covers.
What is the thinking behind beginning this question?
measure-theory lebesgue-integral lebesgue-measure measurable-functions
asked Nov 21 '18 at 15:00
Jason Moore
607
add a comment |
[1] Let $f$ be non negative and $int_Omega fdmu$ is finite then f is finite valued almost everywhere.
[2]Let $E_1=(-infty,-2) bigcup (2,infty)$ $E_2=(-infty,-3)bigcup(3,infty)$ and $E_3=(-4,4)$
Define $F=chi_E -chi_E + chi_E $( ie.$E_1-E_2+E_3)$
Then prove that $int_Omega fdmu$ exists although $mu(E_1)-mu(E_2)+mu(E_3)$ is undefined
Note here that $Omega= mathbb{R} , Sigma=$ Borel class, $mu$ is the lebesgue measure on $Sigma$
Everything is as defined in the H.L ROyden
MY ATTEMPT
[1] First I write $ A=(x in Omega: f(x) =infty) $ and suppose $ mu (A)>0$
For each $n geq0$ the function $n.chi_A$ is a simple function satisfying $0leq n. chi_Aleq f$
Hence $0leq n. chi_A=int_Omega n.chi_A leq int_Omega fdmu$ for each $ngeq0$
This implies that $int_Omega fdmu =infty$
Is this correct for part one?
[2]Also for part two do I have to find the union of all and then integrate? I'm not sure exactly how to proceed.
The measure of an interval is it's length but I don't see how I can find the length of an open interval from infinity. There exits an infinite amount of open intervals and there are so many open covers.
What is the thinking behind beginning this question?
measure-theory lebesgue-integral lebesgue-measure measurable-functions
asked Nov 21 '18 at 15:00
Jason Moore
607
I guess it's $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$ instead of $F=dots$? And $nmu(A)=int_Omega nchi_A$ instead of $nchi_A=dots$?
– francescop21
Nov 21 '18 at 15:14
Yes I now see your point. Thanks very much
– Jason Moore
Nov 21 '18 at 17:50
add a comment |
[1] Let $f$ be non negative and $int_Omega fdmu$ is finite then f is finite valued almost everywhere.
[2]Let $E_1=(-infty,-2) bigcup (2,infty)$ $E_2=(-infty,-3)bigcup(3,infty)$ and $E_3=(-4,4)$
Define $F=chi_E -chi_E + chi_E $( ie.$E_1-E_2+E_3)$
Then prove that $int_Omega fdmu$ exists although $mu(E_1)-mu(E_2)+mu(E_3)$ is undefined
Note here that $Omega= mathbb{R} , Sigma=$ Borel class, $mu$ is the lebesgue measure on $Sigma$
Everything is as defined in the H.L ROyden
MY ATTEMPT
[1] First I write $ A=(x in Omega: f(x) =infty) $ and suppose $ mu (A)>0$
For each $n geq0$ the function $n.chi_A$ is a simple function satisfying $0leq n. chi_Aleq f$
Hence $0leq n. chi_A=int_Omega n.chi_A leq int_Omega fdmu$ for each $ngeq0$
This implies that $int_Omega fdmu =infty$
Is this correct for part one?
[2]Also for part two do I have to find the union of all and then integrate? I'm not sure exactly how to proceed.
The measure of an interval is it's length but I don't see how I can find the length of an open interval from infinity. There exits an infinite amount of open intervals and there are so many open covers.
What is the thinking behind beginning this question?
measure-theory lebesgue-integral lebesgue-measure measurable-functions
asked Nov 21 '18 at 15:00
Jason Moore
607
[1] Let $f$ be non negative and $int_Omega fdmu$ is finite then f is finite valued almost everywhere.
[2]Let $E_1=(-infty,-2) bigcup (2,infty)$ $E_2=(-infty,-3)bigcup(3,infty)$ and $E_3=(-4,4)$
Define $F=chi_E -chi_E + chi_E $( ie.$E_1-E_2+E_3)$
Then prove that $int_Omega fdmu$ exists although $mu(E_1)-mu(E_2)+mu(E_3)$ is undefined
Note here that $Omega= mathbb{R} , Sigma=$ Borel class, $mu$ is the lebesgue measure on $Sigma$
Everything is as defined in the H.L ROyden
MY ATTEMPT
[1] First I write $ A=(x in Omega: f(x) =infty) $ and suppose $ mu (A)>0$
For each $n geq0$ the function $n.chi_A$ is a simple function satisfying $0leq n. chi_Aleq f$
Hence $0leq n. chi_A=int_Omega n.chi_A leq int_Omega fdmu$ for each $ngeq0$
This implies that $int_Omega fdmu =infty$
Is this correct for part one?
[2]Also for part two do I have to find the union of all and then integrate? I'm not sure exactly how to proceed.
The measure of an interval is it's length but I don't see how I can find the length of an open interval from infinity. There exits an infinite amount of open intervals and there are so many open covers.
What is the thinking behind beginning this question?
measure-theory lebesgue-integral lebesgue-measure measurable-functions
measure-theory lebesgue-integral lebesgue-measure measurable-functions
asked Nov 21 '18 at 15:00
Jason Moore
607
asked Nov 21 '18 at 15:00
Jason Moore
607
asked Nov 21 '18 at 15:00
Jason Moore
607
asked Nov 21 '18 at 15:00
Jason Moore
607
asked Nov 21 '18 at 15:00
Jason Moore
607
607
I guess it's $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$ instead of $F=dots$? And $nmu(A)=int_Omega nchi_A$ instead of $nchi_A=dots$?
– francescop21
Nov 21 '18 at 15:14
Yes I now see your point. Thanks very much
– Jason Moore
Nov 21 '18 at 17:50
add a comment |
I guess it's $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$ instead of $F=dots$? And $nmu(A)=int_Omega nchi_A$ instead of $nchi_A=dots$?
– francescop21
Nov 21 '18 at 15:14
Yes I now see your point. Thanks very much
– Jason Moore
Nov 21 '18 at 17:50
I guess it's $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$ instead of $F=dots$? And $nmu(A)=int_Omega nchi_A$ instead of $nchi_A=dots$?
– francescop21
Nov 21 '18 at 15:14
I guess it's $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$ instead of $F=dots$? And $nmu(A)=int_Omega nchi_A$ instead of $nchi_A=dots$?
– francescop21
Nov 21 '18 at 15:14
Yes I now see your point. Thanks very much
– Jason Moore
Nov 21 '18 at 17:50
Yes I now see your point. Thanks very much
– Jason Moore
Nov 21 '18 at 17:50
add a comment |
1 Answer
1
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votes
[1] correct (except what is remarked by @francescop21 in a comment).
[2]
I preassume that you forgot the indices and meant: $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$.
If so then note that $chi_{E_1}-chi_{E_2}=chi_{[-3,-2)}+chi_{(2,3]}$ so that $f=chi_{[-3,-2)}+chi_{(2,3]}+chi_{(-4,4)}$ which is a nonnegative measurable function.
That is enough for the existence of $int fdmu$.
If moreover $mu$ is the Lebesguemeasure then $f$ is also integrable with:$$int fdmu=lambda([-3,-2))+lambda((-2,-3])+lambda((-4,4))=1+1+8=10$$
answered Nov 21 '18 at 15:15
drhab
98.3k544129
add a comment |
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1 Answer
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[1] correct (except what is remarked by @francescop21 in a comment).
[2]
I preassume that you forgot the indices and meant: $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$.
If so then note that $chi_{E_1}-chi_{E_2}=chi_{[-3,-2)}+chi_{(2,3]}$ so that $f=chi_{[-3,-2)}+chi_{(2,3]}+chi_{(-4,4)}$ which is a nonnegative measurable function.
That is enough for the existence of $int fdmu$.
If moreover $mu$ is the Lebesguemeasure then $f$ is also integrable with:$$int fdmu=lambda([-3,-2))+lambda((-2,-3])+lambda((-4,4))=1+1+8=10$$
answered Nov 21 '18 at 15:15
drhab
98.3k544129
add a comment |
[1] correct (except what is remarked by @francescop21 in a comment).
[2]
I preassume that you forgot the indices and meant: $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$.
If so then note that $chi_{E_1}-chi_{E_2}=chi_{[-3,-2)}+chi_{(2,3]}$ so that $f=chi_{[-3,-2)}+chi_{(2,3]}+chi_{(-4,4)}$ which is a nonnegative measurable function.
That is enough for the existence of $int fdmu$.
If moreover $mu$ is the Lebesguemeasure then $f$ is also integrable with:$$int fdmu=lambda([-3,-2))+lambda((-2,-3])+lambda((-4,4))=1+1+8=10$$
answered Nov 21 '18 at 15:15
drhab
98.3k544129
add a comment |
[1] correct (except what is remarked by @francescop21 in a comment).
[2]
I preassume that you forgot the indices and meant: $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$.
If so then note that $chi_{E_1}-chi_{E_2}=chi_{[-3,-2)}+chi_{(2,3]}$ so that $f=chi_{[-3,-2)}+chi_{(2,3]}+chi_{(-4,4)}$ which is a nonnegative measurable function.
That is enough for the existence of $int fdmu$.
If moreover $mu$ is the Lebesguemeasure then $f$ is also integrable with:$$int fdmu=lambda([-3,-2))+lambda((-2,-3])+lambda((-4,4))=1+1+8=10$$
answered Nov 21 '18 at 15:15
drhab
98.3k544129
[1] correct (except what is remarked by @francescop21 in a comment).
[2]
I preassume that you forgot the indices and meant: $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$.
If so then note that $chi_{E_1}-chi_{E_2}=chi_{[-3,-2)}+chi_{(2,3]}$ so that $f=chi_{[-3,-2)}+chi_{(2,3]}+chi_{(-4,4)}$ which is a nonnegative measurable function.
That is enough for the existence of $int fdmu$.
If moreover $mu$ is the Lebesguemeasure then $f$ is also integrable with:$$int fdmu=lambda([-3,-2))+lambda((-2,-3])+lambda((-4,4))=1+1+8=10$$
answered Nov 21 '18 at 15:15
drhab
98.3k544129
answered Nov 21 '18 at 15:15
drhab
98.3k544129
answered Nov 21 '18 at 15:15
drhab
98.3k544129
answered Nov 21 '18 at 15:15
drhab
98.3k544129
98.3k544129
add a comment |
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I guess it's $f=chi_{E_1}-chi_{E_2}+chi_{E_3}$ instead of $F=dots$? And $nmu(A)=int_Omega nchi_A$ instead of $nchi_A=dots$?
– francescop21
Nov 21 '18 at 15:14
Yes I now see your point. Thanks very much
– Jason Moore
Nov 21 '18 at 17:50