Mixing
If $mu(E)=0$ then $int_E f dmu=0$
$begingroup$
Prove that if $f,ginmathscr{M}^+$ then if $mu(E)=0$ then $int_E f dmu=0$ even when $f(x)=infty$ for all $xin E$.
This is a little bit contradictory. If $mu$ is the Lebesgue measure in the space $(mathbb{R},mathscr{B}_{mathbb{R}})$ then I can define a measurable function for example $x=5$ whose integral is infinite and the Lebesgue measure of its domain is $0$.
Question:
Could someone provide me a hint for this proof?
Is the problem I raised a contradiction? If not. Why not?
Thanks in advance!
real-analysis measure-theory lebesgue-integral
asked Jan 1 at 15:26
Pedro GomesPedro Gomes
1,7452721
$endgroup$
add a comment |
$begingroup$
Prove that if $f,ginmathscr{M}^+$ then if $mu(E)=0$ then $int_E f dmu=0$ even when $f(x)=infty$ for all $xin E$.
This is a little bit contradictory. If $mu$ is the Lebesgue measure in the space $(mathbb{R},mathscr{B}_{mathbb{R}})$ then I can define a measurable function for example $x=5$ whose integral is infinite and the Lebesgue measure of its domain is $0$.
Question:
Could someone provide me a hint for this proof?
Is the problem I raised a contradiction? If not. Why not?
Thanks in advance!
real-analysis measure-theory lebesgue-integral
asked Jan 1 at 15:26
Pedro GomesPedro Gomes
1,7452721
$endgroup$
add a comment |
$begingroup$
Prove that if $f,ginmathscr{M}^+$ then if $mu(E)=0$ then $int_E f dmu=0$ even when $f(x)=infty$ for all $xin E$.
This is a little bit contradictory. If $mu$ is the Lebesgue measure in the space $(mathbb{R},mathscr{B}_{mathbb{R}})$ then I can define a measurable function for example $x=5$ whose integral is infinite and the Lebesgue measure of its domain is $0$.
Question:
Could someone provide me a hint for this proof?
Is the problem I raised a contradiction? If not. Why not?
Thanks in advance!
real-analysis measure-theory lebesgue-integral
asked Jan 1 at 15:26
Pedro GomesPedro Gomes
1,7452721
$endgroup$
Prove that if $f,ginmathscr{M}^+$ then if $mu(E)=0$ then $int_E f dmu=0$ even when $f(x)=infty$ for all $xin E$.
This is a little bit contradictory. If $mu$ is the Lebesgue measure in the space $(mathbb{R},mathscr{B}_{mathbb{R}})$ then I can define a measurable function for example $x=5$ whose integral is infinite and the Lebesgue measure of its domain is $0$.
Question:
Could someone provide me a hint for this proof?
Is the problem I raised a contradiction? If not. Why not?
Thanks in advance!
real-analysis measure-theory lebesgue-integral
real-analysis measure-theory lebesgue-integral
asked Jan 1 at 15:26
Pedro GomesPedro Gomes
1,7452721
asked Jan 1 at 15:26
Pedro GomesPedro Gomes
1,7452721
asked Jan 1 at 15:26
Pedro GomesPedro Gomes
1,7452721
asked Jan 1 at 15:26
Pedro GomesPedro Gomes
1,7452721
asked Jan 1 at 15:26
Pedro GomesPedro Gomes
1,7452721
1,7452721
add a comment |
add a comment |
1 Answer
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By definition, $int_E f , dmu = int chi_E ,f , dmu = sup { int psi , dmu mid psi text{ simple function with } 0 leq psi leq chi_E , f }.$
What is the value of $int psi , dmu$ if $psi$ is a simple function with $0 leq psi leq chi_E , f$?
What makes you think that $int_E 5 , dmu = infty$ if $mu(E) = 0$?
answered Jan 1 at 15:35
md2perpemd2perpe
7,73611028
$endgroup$
$begingroup$
$x=5$ is a line segment going up to infinity. $infty$ when x=5 and $emptyset$ when $xneq 5$
$endgroup$
– Pedro Gomes
Jan 1 at 15:43
$begingroup$
@PedroGomes, We have $int_E f mathrm{d}m = 0$ even when $f = infty$ and $m(E) = 0$. Try to apply the definition appearing in this answer.
$endgroup$
– Sangchul Lee
Jan 1 at 15:48
add a comment |
Your Answer
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By definition, $int_E f , dmu = int chi_E ,f , dmu = sup { int psi , dmu mid psi text{ simple function with } 0 leq psi leq chi_E , f }.$
What is the value of $int psi , dmu$ if $psi$ is a simple function with $0 leq psi leq chi_E , f$?
What makes you think that $int_E 5 , dmu = infty$ if $mu(E) = 0$?
answered Jan 1 at 15:35
md2perpemd2perpe
7,73611028
$endgroup$
$begingroup$
$x=5$ is a line segment going up to infinity. $infty$ when x=5 and $emptyset$ when $xneq 5$
$endgroup$
– Pedro Gomes
Jan 1 at 15:43
$begingroup$
@PedroGomes, We have $int_E f mathrm{d}m = 0$ even when $f = infty$ and $m(E) = 0$. Try to apply the definition appearing in this answer.
$endgroup$
– Sangchul Lee
Jan 1 at 15:48
add a comment |
$begingroup$
By definition, $int_E f , dmu = int chi_E ,f , dmu = sup { int psi , dmu mid psi text{ simple function with } 0 leq psi leq chi_E , f }.$
What is the value of $int psi , dmu$ if $psi$ is a simple function with $0 leq psi leq chi_E , f$?
What makes you think that $int_E 5 , dmu = infty$ if $mu(E) = 0$?
answered Jan 1 at 15:35
md2perpemd2perpe
7,73611028
$endgroup$
$begingroup$
$x=5$ is a line segment going up to infinity. $infty$ when x=5 and $emptyset$ when $xneq 5$
$endgroup$
– Pedro Gomes
Jan 1 at 15:43
$begingroup$
@PedroGomes, We have $int_E f mathrm{d}m = 0$ even when $f = infty$ and $m(E) = 0$. Try to apply the definition appearing in this answer.
$endgroup$
– Sangchul Lee
Jan 1 at 15:48
add a comment |
$begingroup$
By definition, $int_E f , dmu = int chi_E ,f , dmu = sup { int psi , dmu mid psi text{ simple function with } 0 leq psi leq chi_E , f }.$
What is the value of $int psi , dmu$ if $psi$ is a simple function with $0 leq psi leq chi_E , f$?
What makes you think that $int_E 5 , dmu = infty$ if $mu(E) = 0$?
answered Jan 1 at 15:35
md2perpemd2perpe
7,73611028
$endgroup$
By definition, $int_E f , dmu = int chi_E ,f , dmu = sup { int psi , dmu mid psi text{ simple function with } 0 leq psi leq chi_E , f }.$
What is the value of $int psi , dmu$ if $psi$ is a simple function with $0 leq psi leq chi_E , f$?
What makes you think that $int_E 5 , dmu = infty$ if $mu(E) = 0$?
answered Jan 1 at 15:35
md2perpemd2perpe
7,73611028
answered Jan 1 at 15:35
md2perpemd2perpe
7,73611028
answered Jan 1 at 15:35
md2perpemd2perpe
7,73611028
answered Jan 1 at 15:35
md2perpemd2perpe
7,73611028
7,73611028
$begingroup$
$x=5$ is a line segment going up to infinity. $infty$ when x=5 and $emptyset$ when $xneq 5$
$endgroup$
– Pedro Gomes
Jan 1 at 15:43
$begingroup$
@PedroGomes, We have $int_E f mathrm{d}m = 0$ even when $f = infty$ and $m(E) = 0$. Try to apply the definition appearing in this answer.
$endgroup$
– Sangchul Lee
Jan 1 at 15:48
add a comment |
$begingroup$
$x=5$ is a line segment going up to infinity. $infty$ when x=5 and $emptyset$ when $xneq 5$
$endgroup$
– Pedro Gomes
Jan 1 at 15:43
$begingroup$
@PedroGomes, We have $int_E f mathrm{d}m = 0$ even when $f = infty$ and $m(E) = 0$. Try to apply the definition appearing in this answer.
$endgroup$
– Sangchul Lee
Jan 1 at 15:48
$begingroup$
$x=5$ is a line segment going up to infinity. $infty$ when x=5 and $emptyset$ when $xneq 5$
$endgroup$
– Pedro Gomes
Jan 1 at 15:43
$begingroup$
$x=5$ is a line segment going up to infinity. $infty$ when x=5 and $emptyset$ when $xneq 5$
$endgroup$
– Pedro Gomes
Jan 1 at 15:43
$begingroup$
@PedroGomes, We have $int_E f mathrm{d}m = 0$ even when $f = infty$ and $m(E) = 0$. Try to apply the definition appearing in this answer.
$endgroup$
– Sangchul Lee
Jan 1 at 15:48
$begingroup$
@PedroGomes, We have $int_E f mathrm{d}m = 0$ even when $f = infty$ and $m(E) = 0$. Try to apply the definition appearing in this answer.
$endgroup$
– Sangchul Lee
Jan 1 at 15:48
add a comment |
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