$ limsup_{nrightarrowinfty}int_{Omega_varepsilon} f_nleqint_{Omega_varepsilon} f $ implies that of $Omega$?
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Let $f_n,fin L^1(Omega)$, all non-negative. Let $Omega_varepsilon$ be an increasing set sequence that $bigcup_{varepsilon>0}Omega_varepsilon=Omega$. If for each $varepsilon$,
$
limsup_{nrightarrowinfty}int_{Omega_varepsilon} f_nleqint_{Omega_varepsilon} f
$
Do we have
$
limsup_{nrightarrowinfty}int_{Omega} f_nleqint_{Omega} f
?$
real-analysis
asked Dec 18 '18 at 9:29
P.RP.R
184
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add a comment |
$begingroup$
Let $f_n,fin L^1(Omega)$, all non-negative. Let $Omega_varepsilon$ be an increasing set sequence that $bigcup_{varepsilon>0}Omega_varepsilon=Omega$. If for each $varepsilon$,
$
limsup_{nrightarrowinfty}int_{Omega_varepsilon} f_nleqint_{Omega_varepsilon} f
$
Do we have
$
limsup_{nrightarrowinfty}int_{Omega} f_nleqint_{Omega} f
?$
real-analysis
asked Dec 18 '18 at 9:29
P.RP.R
184
$endgroup$
$begingroup$
What kind of assumptions do you have on $Omega$? Is it a bounded open set?
$endgroup$
– BigbearZzz
Dec 18 '18 at 9:36
add a comment |
$begingroup$
Let $f_n,fin L^1(Omega)$, all non-negative. Let $Omega_varepsilon$ be an increasing set sequence that $bigcup_{varepsilon>0}Omega_varepsilon=Omega$. If for each $varepsilon$,
$
limsup_{nrightarrowinfty}int_{Omega_varepsilon} f_nleqint_{Omega_varepsilon} f
$
Do we have
$
limsup_{nrightarrowinfty}int_{Omega} f_nleqint_{Omega} f
?$
real-analysis
asked Dec 18 '18 at 9:29
P.RP.R
184
$endgroup$
Let $f_n,fin L^1(Omega)$, all non-negative. Let $Omega_varepsilon$ be an increasing set sequence that $bigcup_{varepsilon>0}Omega_varepsilon=Omega$. If for each $varepsilon$,
$
limsup_{nrightarrowinfty}int_{Omega_varepsilon} f_nleqint_{Omega_varepsilon} f
$
Do we have
$
limsup_{nrightarrowinfty}int_{Omega} f_nleqint_{Omega} f
?$
real-analysis
real-analysis
asked Dec 18 '18 at 9:29
P.RP.R
184
asked Dec 18 '18 at 9:29
P.RP.R
184
asked Dec 18 '18 at 9:29
P.RP.R
184
asked Dec 18 '18 at 9:29
P.RP.R
184
asked Dec 18 '18 at 9:29
P.RP.R
184
184
$begingroup$
What kind of assumptions do you have on $Omega$? Is it a bounded open set?
$endgroup$
– BigbearZzz
Dec 18 '18 at 9:36
add a comment |
$begingroup$
What kind of assumptions do you have on $Omega$? Is it a bounded open set?
$endgroup$
– BigbearZzz
Dec 18 '18 at 9:36
$begingroup$
What kind of assumptions do you have on $Omega$? Is it a bounded open set?
$endgroup$
– BigbearZzz
Dec 18 '18 at 9:36
$begingroup$
What kind of assumptions do you have on $Omega$? Is it a bounded open set?
$endgroup$
– BigbearZzz
Dec 18 '18 at 9:36
add a comment |
1 Answer
1
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No. Let $Omega =(0,1)$ with Lebesgue measure and $Omega_{epsilon} =(epsilon ,1)$. Let $f_n=nI_{(0,frac 1 n)}$, $f=0$. For each $epsilon$ we have $int_{Omega_{epsilon}} f_n=0$ as soon as $frac 1 n <epsilon$ but $int_{Omega} f_n=1$ for all $n$.
answered Dec 18 '18 at 9:37
Kavi Rama MurthyKavi Rama Murthy
69.7k53170
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1 Answer
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1 Answer
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$begingroup$
No. Let $Omega =(0,1)$ with Lebesgue measure and $Omega_{epsilon} =(epsilon ,1)$. Let $f_n=nI_{(0,frac 1 n)}$, $f=0$. For each $epsilon$ we have $int_{Omega_{epsilon}} f_n=0$ as soon as $frac 1 n <epsilon$ but $int_{Omega} f_n=1$ for all $n$.
answered Dec 18 '18 at 9:37
Kavi Rama MurthyKavi Rama Murthy
69.7k53170
$endgroup$
add a comment |
$begingroup$
No. Let $Omega =(0,1)$ with Lebesgue measure and $Omega_{epsilon} =(epsilon ,1)$. Let $f_n=nI_{(0,frac 1 n)}$, $f=0$. For each $epsilon$ we have $int_{Omega_{epsilon}} f_n=0$ as soon as $frac 1 n <epsilon$ but $int_{Omega} f_n=1$ for all $n$.
answered Dec 18 '18 at 9:37
Kavi Rama MurthyKavi Rama Murthy
69.7k53170
$endgroup$
add a comment |
$begingroup$
No. Let $Omega =(0,1)$ with Lebesgue measure and $Omega_{epsilon} =(epsilon ,1)$. Let $f_n=nI_{(0,frac 1 n)}$, $f=0$. For each $epsilon$ we have $int_{Omega_{epsilon}} f_n=0$ as soon as $frac 1 n <epsilon$ but $int_{Omega} f_n=1$ for all $n$.
answered Dec 18 '18 at 9:37
Kavi Rama MurthyKavi Rama Murthy
69.7k53170
$endgroup$
No. Let $Omega =(0,1)$ with Lebesgue measure and $Omega_{epsilon} =(epsilon ,1)$. Let $f_n=nI_{(0,frac 1 n)}$, $f=0$. For each $epsilon$ we have $int_{Omega_{epsilon}} f_n=0$ as soon as $frac 1 n <epsilon$ but $int_{Omega} f_n=1$ for all $n$.
answered Dec 18 '18 at 9:37
Kavi Rama MurthyKavi Rama Murthy
69.7k53170
answered Dec 18 '18 at 9:37
Kavi Rama MurthyKavi Rama Murthy
69.7k53170
answered Dec 18 '18 at 9:37
Kavi Rama MurthyKavi Rama Murthy
69.7k53170
answered Dec 18 '18 at 9:37
Kavi Rama MurthyKavi Rama Murthy
69.7k53170
69.7k53170
add a comment |
add a comment |
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$begingroup$
What kind of assumptions do you have on $Omega$? Is it a bounded open set?
$endgroup$
– BigbearZzz
Dec 18 '18 at 9:36