Showing Lemma's Fatou for functions not necessarily not negative.
$begingroup$
Let $g$ integrable function on $E$ measurable set.
Let $(f_n)$ measurable functions and $|f_n|leq g$ for all $n$.
Show that $int_{E} liminf f_nleq liminf int_{E} f_nleq limsup int_{E} f_nleq int_{E}limsup f_n$.
I have a doubt.
with $(f_n+g)$ and Fatou, $int liminf (f_n+g)leq liminf int (f_n+g)$
Now. Is it true that $ liminf (f_n+g)= (liminf f_n)+g$? I ask this, well, if it's true then $ int liminf (f_n+g)= int [(liminf f_n)+g]=intliminf f_n+int g$ and $liminf int (f_n+g)=liminf (int f_n+int g)=liminf int f_n+int g$ and so
$int liminf f_n+int gleq liminf int f_n+int g$, and $g$ integrable, implies $int liminf f_nleq liminf int f_n$
real-analysis measure-theory lebesgue-integral
asked Dec 9 '18 at 22:20
eraldcoileraldcoil
385211
$endgroup$
add a comment |
$begingroup$
Let $g$ integrable function on $E$ measurable set.
Let $(f_n)$ measurable functions and $|f_n|leq g$ for all $n$.
Show that $int_{E} liminf f_nleq liminf int_{E} f_nleq limsup int_{E} f_nleq int_{E}limsup f_n$.
I have a doubt.
with $(f_n+g)$ and Fatou, $int liminf (f_n+g)leq liminf int (f_n+g)$
Now. Is it true that $ liminf (f_n+g)= (liminf f_n)+g$? I ask this, well, if it's true then $ int liminf (f_n+g)= int [(liminf f_n)+g]=intliminf f_n+int g$ and $liminf int (f_n+g)=liminf (int f_n+int g)=liminf int f_n+int g$ and so
$int liminf f_n+int gleq liminf int f_n+int g$, and $g$ integrable, implies $int liminf f_nleq liminf int f_n$
real-analysis measure-theory lebesgue-integral
asked Dec 9 '18 at 22:20
eraldcoileraldcoil
385211
$endgroup$
1
$begingroup$
for a proof of $liminf (f_n+g)= (liminf f_n)+g$ see proposition 2.3 in this paper. Now choose $b_n=b$ for all $ninBbb N$ and you get the stated equality
$endgroup$
– Masacroso
Dec 9 '18 at 22:28
$begingroup$
proposition 2.3. Also works with secuenque of functions?
$endgroup$
– eraldcoil
Dec 9 '18 at 22:49
1
$begingroup$
it is a point-wise sequence of functions, that is $liminf (f_n+g)= (liminf f_n)+g$ means that $liminf (f_n(x)+g(x))= (liminf f_n(x))+g(x)$ for any chosen $xin E$
$endgroup$
– Masacroso
Dec 9 '18 at 22:52
add a comment |
$begingroup$
Let $g$ integrable function on $E$ measurable set.
Let $(f_n)$ measurable functions and $|f_n|leq g$ for all $n$.
Show that $int_{E} liminf f_nleq liminf int_{E} f_nleq limsup int_{E} f_nleq int_{E}limsup f_n$.
I have a doubt.
with $(f_n+g)$ and Fatou, $int liminf (f_n+g)leq liminf int (f_n+g)$
Now. Is it true that $ liminf (f_n+g)= (liminf f_n)+g$? I ask this, well, if it's true then $ int liminf (f_n+g)= int [(liminf f_n)+g]=intliminf f_n+int g$ and $liminf int (f_n+g)=liminf (int f_n+int g)=liminf int f_n+int g$ and so
$int liminf f_n+int gleq liminf int f_n+int g$, and $g$ integrable, implies $int liminf f_nleq liminf int f_n$
real-analysis measure-theory lebesgue-integral
asked Dec 9 '18 at 22:20
eraldcoileraldcoil
385211
$endgroup$
Let $g$ integrable function on $E$ measurable set.
Let $(f_n)$ measurable functions and $|f_n|leq g$ for all $n$.
Show that $int_{E} liminf f_nleq liminf int_{E} f_nleq limsup int_{E} f_nleq int_{E}limsup f_n$.
I have a doubt.
with $(f_n+g)$ and Fatou, $int liminf (f_n+g)leq liminf int (f_n+g)$
Now. Is it true that $ liminf (f_n+g)= (liminf f_n)+g$? I ask this, well, if it's true then $ int liminf (f_n+g)= int [(liminf f_n)+g]=intliminf f_n+int g$ and $liminf int (f_n+g)=liminf (int f_n+int g)=liminf int f_n+int g$ and so
$int liminf f_n+int gleq liminf int f_n+int g$, and $g$ integrable, implies $int liminf f_nleq liminf int f_n$
real-analysis measure-theory lebesgue-integral
real-analysis measure-theory lebesgue-integral
asked Dec 9 '18 at 22:20
eraldcoileraldcoil
385211
asked Dec 9 '18 at 22:20
eraldcoileraldcoil
385211
asked Dec 9 '18 at 22:20
eraldcoileraldcoil
385211
asked Dec 9 '18 at 22:20
eraldcoileraldcoil
385211
asked Dec 9 '18 at 22:20
eraldcoileraldcoil
385211
385211
1
$begingroup$
for a proof of $liminf (f_n+g)= (liminf f_n)+g$ see proposition 2.3 in this paper. Now choose $b_n=b$ for all $ninBbb N$ and you get the stated equality
$endgroup$
– Masacroso
Dec 9 '18 at 22:28
$begingroup$
proposition 2.3. Also works with secuenque of functions?
$endgroup$
– eraldcoil
Dec 9 '18 at 22:49
1
$begingroup$
it is a point-wise sequence of functions, that is $liminf (f_n+g)= (liminf f_n)+g$ means that $liminf (f_n(x)+g(x))= (liminf f_n(x))+g(x)$ for any chosen $xin E$
$endgroup$
– Masacroso
Dec 9 '18 at 22:52
add a comment |
1
$begingroup$
for a proof of $liminf (f_n+g)= (liminf f_n)+g$ see proposition 2.3 in this paper. Now choose $b_n=b$ for all $ninBbb N$ and you get the stated equality
$endgroup$
– Masacroso
Dec 9 '18 at 22:28
$begingroup$
proposition 2.3. Also works with secuenque of functions?
$endgroup$
– eraldcoil
Dec 9 '18 at 22:49
1
$begingroup$
it is a point-wise sequence of functions, that is $liminf (f_n+g)= (liminf f_n)+g$ means that $liminf (f_n(x)+g(x))= (liminf f_n(x))+g(x)$ for any chosen $xin E$
$endgroup$
– Masacroso
Dec 9 '18 at 22:52
1
1
$begingroup$
for a proof of $liminf (f_n+g)= (liminf f_n)+g$ see proposition 2.3 in this paper. Now choose $b_n=b$ for all $ninBbb N$ and you get the stated equality
$endgroup$
– Masacroso
Dec 9 '18 at 22:28
$begingroup$
for a proof of $liminf (f_n+g)= (liminf f_n)+g$ see proposition 2.3 in this paper. Now choose $b_n=b$ for all $ninBbb N$ and you get the stated equality
$endgroup$
– Masacroso
Dec 9 '18 at 22:28
$begingroup$
proposition 2.3. Also works with secuenque of functions?
$endgroup$
– eraldcoil
Dec 9 '18 at 22:49
$begingroup$
proposition 2.3. Also works with secuenque of functions?
$endgroup$
– eraldcoil
Dec 9 '18 at 22:49
1
1
$begingroup$
it is a point-wise sequence of functions, that is $liminf (f_n+g)= (liminf f_n)+g$ means that $liminf (f_n(x)+g(x))= (liminf f_n(x))+g(x)$ for any chosen $xin E$
$endgroup$
– Masacroso
Dec 9 '18 at 22:52
$begingroup$
it is a point-wise sequence of functions, that is $liminf (f_n+g)= (liminf f_n)+g$ means that $liminf (f_n(x)+g(x))= (liminf f_n(x))+g(x)$ for any chosen $xin E$
$endgroup$
– Masacroso
Dec 9 '18 at 22:52
add a comment |
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$begingroup$
for a proof of $liminf (f_n+g)= (liminf f_n)+g$ see proposition 2.3 in this paper. Now choose $b_n=b$ for all $ninBbb N$ and you get the stated equality
$endgroup$
– Masacroso
Dec 9 '18 at 22:28
$begingroup$
proposition 2.3. Also works with secuenque of functions?
$endgroup$
– eraldcoil
Dec 9 '18 at 22:49
1
$begingroup$
it is a point-wise sequence of functions, that is $liminf (f_n+g)= (liminf f_n)+g$ means that $liminf (f_n(x)+g(x))= (liminf f_n(x))+g(x)$ for any chosen $xin E$
$endgroup$
– Masacroso
Dec 9 '18 at 22:52