Measurable function on Ω
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$M( Omega, mathbf{R}^{+})$ is the set of measurable function on $Omega $.
let $mu ( Omega) < infty $ and $f in M( Omega, mathbf{R}^{+})$.
How can I show that?
$f$ is integrable function if only if the series $ Sigma_{n = 1}^{infty} mu ( E_n)$ is convergence, so that $E_n = { x in Omega : f(x) geq n }$.
real-analysis functional-analysis
asked Dec 2 at 13:56
joe
874
add a comment |
up vote
0
down vote
favorite
$M( Omega, mathbf{R}^{+})$ is the set of measurable function on $Omega $.
let $mu ( Omega) < infty $ and $f in M( Omega, mathbf{R}^{+})$.
How can I show that?
$f$ is integrable function if only if the series $ Sigma_{n = 1}^{infty} mu ( E_n)$ is convergence, so that $E_n = { x in Omega : f(x) geq n }$.
real-analysis functional-analysis
asked Dec 2 at 13:56
joe
874
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$M( Omega, mathbf{R}^{+})$ is the set of measurable function on $Omega $.
let $mu ( Omega) < infty $ and $f in M( Omega, mathbf{R}^{+})$.
How can I show that?
$f$ is integrable function if only if the series $ Sigma_{n = 1}^{infty} mu ( E_n)$ is convergence, so that $E_n = { x in Omega : f(x) geq n }$.
real-analysis functional-analysis
asked Dec 2 at 13:56
joe
874
$M( Omega, mathbf{R}^{+})$ is the set of measurable function on $Omega $.
let $mu ( Omega) < infty $ and $f in M( Omega, mathbf{R}^{+})$.
How can I show that?
$f$ is integrable function if only if the series $ Sigma_{n = 1}^{infty} mu ( E_n)$ is convergence, so that $E_n = { x in Omega : f(x) geq n }$.
real-analysis functional-analysis
real-analysis functional-analysis
asked Dec 2 at 13:56
joe
874
asked Dec 2 at 13:56
joe
874
asked Dec 2 at 13:56
joe
874
asked Dec 2 at 13:56
joe
874
asked Dec 2 at 13:56
joe
874
874
add a comment |
add a comment |
1 Answer
1
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1
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Let $g=sum_{n=1}^{infty}mathsf1_{E_n}$ and $h=sum_{n=0}^{infty}mathsf1_{E_n}$.
Then $h(x)=g(x)+1$ and $gleq fleq h$ so that: $$int g;dmuleqint f;dmuleqint h;dmu$$
or equivalently: $$sum_{n=1}^{infty}mu(E_n)leqint f;dmuleqsum_{n=1}^{infty}mu(E_n)+mu(Omega)$$
Since $mu(Omega)<infty$ this justifies the conclusion that $$int f;dmu<inftyiffsum_{n=1}^{infty}mu(E_n)<infty$$
answered Dec 2 at 14:15
drhab
95.3k543126
what do you mean about "$g=sum_{n=1}^{infty}mathsf1_{E_n}$ "?
– joe
Dec 2 at 14:41
Do you mean $g=sum_{n=1}^{infty}mathsfchi_{E_n}$
– joe
Dec 2 at 14:42
Yes, that is another notation for it.
– drhab
Dec 2 at 14:45
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $g=sum_{n=1}^{infty}mathsf1_{E_n}$ and $h=sum_{n=0}^{infty}mathsf1_{E_n}$.
Then $h(x)=g(x)+1$ and $gleq fleq h$ so that: $$int g;dmuleqint f;dmuleqint h;dmu$$
or equivalently: $$sum_{n=1}^{infty}mu(E_n)leqint f;dmuleqsum_{n=1}^{infty}mu(E_n)+mu(Omega)$$
Since $mu(Omega)<infty$ this justifies the conclusion that $$int f;dmu<inftyiffsum_{n=1}^{infty}mu(E_n)<infty$$
answered Dec 2 at 14:15
drhab
95.3k543126
what do you mean about "$g=sum_{n=1}^{infty}mathsf1_{E_n}$ "?
– joe
Dec 2 at 14:41
Do you mean $g=sum_{n=1}^{infty}mathsfchi_{E_n}$
– joe
Dec 2 at 14:42
Yes, that is another notation for it.
– drhab
Dec 2 at 14:45
add a comment |
up vote
1
down vote
Let $g=sum_{n=1}^{infty}mathsf1_{E_n}$ and $h=sum_{n=0}^{infty}mathsf1_{E_n}$.
Then $h(x)=g(x)+1$ and $gleq fleq h$ so that: $$int g;dmuleqint f;dmuleqint h;dmu$$
or equivalently: $$sum_{n=1}^{infty}mu(E_n)leqint f;dmuleqsum_{n=1}^{infty}mu(E_n)+mu(Omega)$$
Since $mu(Omega)<infty$ this justifies the conclusion that $$int f;dmu<inftyiffsum_{n=1}^{infty}mu(E_n)<infty$$
answered Dec 2 at 14:15
drhab
95.3k543126
what do you mean about "$g=sum_{n=1}^{infty}mathsf1_{E_n}$ "?
– joe
Dec 2 at 14:41
Do you mean $g=sum_{n=1}^{infty}mathsfchi_{E_n}$
– joe
Dec 2 at 14:42
Yes, that is another notation for it.
– drhab
Dec 2 at 14:45
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $g=sum_{n=1}^{infty}mathsf1_{E_n}$ and $h=sum_{n=0}^{infty}mathsf1_{E_n}$.
Then $h(x)=g(x)+1$ and $gleq fleq h$ so that: $$int g;dmuleqint f;dmuleqint h;dmu$$
or equivalently: $$sum_{n=1}^{infty}mu(E_n)leqint f;dmuleqsum_{n=1}^{infty}mu(E_n)+mu(Omega)$$
Since $mu(Omega)<infty$ this justifies the conclusion that $$int f;dmu<inftyiffsum_{n=1}^{infty}mu(E_n)<infty$$
answered Dec 2 at 14:15
drhab
95.3k543126
Let $g=sum_{n=1}^{infty}mathsf1_{E_n}$ and $h=sum_{n=0}^{infty}mathsf1_{E_n}$.
Then $h(x)=g(x)+1$ and $gleq fleq h$ so that: $$int g;dmuleqint f;dmuleqint h;dmu$$
or equivalently: $$sum_{n=1}^{infty}mu(E_n)leqint f;dmuleqsum_{n=1}^{infty}mu(E_n)+mu(Omega)$$
Since $mu(Omega)<infty$ this justifies the conclusion that $$int f;dmu<inftyiffsum_{n=1}^{infty}mu(E_n)<infty$$
answered Dec 2 at 14:15
drhab
95.3k543126
answered Dec 2 at 14:15
drhab
95.3k543126
answered Dec 2 at 14:15
drhab
95.3k543126
answered Dec 2 at 14:15
drhab
95.3k543126
95.3k543126
what do you mean about "$g=sum_{n=1}^{infty}mathsf1_{E_n}$ "?
– joe
Dec 2 at 14:41
Do you mean $g=sum_{n=1}^{infty}mathsfchi_{E_n}$
– joe
Dec 2 at 14:42
Yes, that is another notation for it.
– drhab
Dec 2 at 14:45
add a comment |
what do you mean about "$g=sum_{n=1}^{infty}mathsf1_{E_n}$ "?
– joe
Dec 2 at 14:41
Do you mean $g=sum_{n=1}^{infty}mathsfchi_{E_n}$
– joe
Dec 2 at 14:42
Yes, that is another notation for it.
– drhab
Dec 2 at 14:45
what do you mean about "$g=sum_{n=1}^{infty}mathsf1_{E_n}$ "?
– joe
Dec 2 at 14:41
what do you mean about "$g=sum_{n=1}^{infty}mathsf1_{E_n}$ "?
– joe
Dec 2 at 14:41
Do you mean $g=sum_{n=1}^{infty}mathsfchi_{E_n}$
– joe
Dec 2 at 14:42
Do you mean $g=sum_{n=1}^{infty}mathsfchi_{E_n}$
– joe
Dec 2 at 14:42
Yes, that is another notation for it.
– drhab
Dec 2 at 14:45
Yes, that is another notation for it.
– drhab
Dec 2 at 14:45
add a comment |
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