Do $f_n to f$ almost surely and $0<int f<infty$ imply $int f_n to int f$?
up vote
0
down vote
favorite
Let a sequence of Lebesgue integrable functions $f_n$ converge to $f$ almost surely and $0 < int f < infty$. Does this imply $int f_n to f$? If the condition $0 < int f$ is removed there is many counterexamples, For example, answers for this question provide some examples. All counterexamples I found are "delta function sequences", so $int f_n < infty$ but $f_n to 0$ almost surely. I think that intuitively the condition $0 < int f$ may prohibit the accumulation of the mass of $f_n$ in an infinitely small region.
EDIT.
I'm not sure but an additional condition $int f_n$ converges could be needed. If this additional contion is actually needed, please consider it also.
real-analysis
asked Nov 14 at 7:53
Balbadak
564
add a comment |
up vote
0
down vote
favorite
Let a sequence of Lebesgue integrable functions $f_n$ converge to $f$ almost surely and $0 < int f < infty$. Does this imply $int f_n to f$? If the condition $0 < int f$ is removed there is many counterexamples, For example, answers for this question provide some examples. All counterexamples I found are "delta function sequences", so $int f_n < infty$ but $f_n to 0$ almost surely. I think that intuitively the condition $0 < int f$ may prohibit the accumulation of the mass of $f_n$ in an infinitely small region.
EDIT.
I'm not sure but an additional condition $int f_n$ converges could be needed. If this additional contion is actually needed, please consider it also.
real-analysis
asked Nov 14 at 7:53
Balbadak
564
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let a sequence of Lebesgue integrable functions $f_n$ converge to $f$ almost surely and $0 < int f < infty$. Does this imply $int f_n to f$? If the condition $0 < int f$ is removed there is many counterexamples, For example, answers for this question provide some examples. All counterexamples I found are "delta function sequences", so $int f_n < infty$ but $f_n to 0$ almost surely. I think that intuitively the condition $0 < int f$ may prohibit the accumulation of the mass of $f_n$ in an infinitely small region.
EDIT.
I'm not sure but an additional condition $int f_n$ converges could be needed. If this additional contion is actually needed, please consider it also.
real-analysis
asked Nov 14 at 7:53
Balbadak
564
Let a sequence of Lebesgue integrable functions $f_n$ converge to $f$ almost surely and $0 < int f < infty$. Does this imply $int f_n to f$? If the condition $0 < int f$ is removed there is many counterexamples, For example, answers for this question provide some examples. All counterexamples I found are "delta function sequences", so $int f_n < infty$ but $f_n to 0$ almost surely. I think that intuitively the condition $0 < int f$ may prohibit the accumulation of the mass of $f_n$ in an infinitely small region.
EDIT.
I'm not sure but an additional condition $int f_n$ converges could be needed. If this additional contion is actually needed, please consider it also.
real-analysis
real-analysis
asked Nov 14 at 7:53
Balbadak
564
asked Nov 14 at 7:53
Balbadak
564
edited Nov 14 at 8:11
asked Nov 14 at 7:53
Balbadak
564
asked Nov 14 at 7:53
Balbadak
564
asked Nov 14 at 7:53
Balbadak
564
564
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$I_{(0,1)}+nI_{(0,frac 1 n)}to I_{(0,1)}$ almost everywhere but the integrals do not converge to the right limit. Sufficient conditions for $lim inf_n =int f$ are contained in basic theorems of measure theory: monotone convergence theorem, dominated convergence theorem etc.
answered Nov 14 at 7:55
Kavi Rama Murthy
39.9k31750
Do they not converge? I think the integrals are $2$.
– Balbadak
Nov 14 at 8:06
@rimusolem What I meant was 'integrals do not converge to the right limit'. $int f_n to int f +1$.
– Kavi Rama Murthy
Nov 14 at 8:12
I see. I missed an obvious example.
– Balbadak
Nov 14 at 8:14
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
accepted
$I_{(0,1)}+nI_{(0,frac 1 n)}to I_{(0,1)}$ almost everywhere but the integrals do not converge to the right limit. Sufficient conditions for $lim inf_n =int f$ are contained in basic theorems of measure theory: monotone convergence theorem, dominated convergence theorem etc.
answered Nov 14 at 7:55
Kavi Rama Murthy
39.9k31750
Do they not converge? I think the integrals are $2$.
– Balbadak
Nov 14 at 8:06
@rimusolem What I meant was 'integrals do not converge to the right limit'. $int f_n to int f +1$.
– Kavi Rama Murthy
Nov 14 at 8:12
I see. I missed an obvious example.
– Balbadak
Nov 14 at 8:14
add a comment |
up vote
1
down vote
accepted
$I_{(0,1)}+nI_{(0,frac 1 n)}to I_{(0,1)}$ almost everywhere but the integrals do not converge to the right limit. Sufficient conditions for $lim inf_n =int f$ are contained in basic theorems of measure theory: monotone convergence theorem, dominated convergence theorem etc.
answered Nov 14 at 7:55
Kavi Rama Murthy
39.9k31750
Do they not converge? I think the integrals are $2$.
– Balbadak
Nov 14 at 8:06
@rimusolem What I meant was 'integrals do not converge to the right limit'. $int f_n to int f +1$.
– Kavi Rama Murthy
Nov 14 at 8:12
I see. I missed an obvious example.
– Balbadak
Nov 14 at 8:14
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$I_{(0,1)}+nI_{(0,frac 1 n)}to I_{(0,1)}$ almost everywhere but the integrals do not converge to the right limit. Sufficient conditions for $lim inf_n =int f$ are contained in basic theorems of measure theory: monotone convergence theorem, dominated convergence theorem etc.
answered Nov 14 at 7:55
Kavi Rama Murthy
39.9k31750
$I_{(0,1)}+nI_{(0,frac 1 n)}to I_{(0,1)}$ almost everywhere but the integrals do not converge to the right limit. Sufficient conditions for $lim inf_n =int f$ are contained in basic theorems of measure theory: monotone convergence theorem, dominated convergence theorem etc.
answered Nov 14 at 7:55
Kavi Rama Murthy
39.9k31750
edited Nov 14 at 8:14
answered Nov 14 at 7:55
Kavi Rama Murthy
39.9k31750
answered Nov 14 at 7:55
Kavi Rama Murthy
39.9k31750
answered Nov 14 at 7:55
Kavi Rama Murthy
39.9k31750
39.9k31750
Do they not converge? I think the integrals are $2$.
– Balbadak
Nov 14 at 8:06
@rimusolem What I meant was 'integrals do not converge to the right limit'. $int f_n to int f +1$.
– Kavi Rama Murthy
Nov 14 at 8:12
I see. I missed an obvious example.
– Balbadak
Nov 14 at 8:14
add a comment |
Do they not converge? I think the integrals are $2$.
– Balbadak
Nov 14 at 8:06
@rimusolem What I meant was 'integrals do not converge to the right limit'. $int f_n to int f +1$.
– Kavi Rama Murthy
Nov 14 at 8:12
I see. I missed an obvious example.
– Balbadak
Nov 14 at 8:14
Do they not converge? I think the integrals are $2$.
– Balbadak
Nov 14 at 8:06
Do they not converge? I think the integrals are $2$.
– Balbadak
Nov 14 at 8:06
@rimusolem What I meant was 'integrals do not converge to the right limit'. $int f_n to int f +1$.
– Kavi Rama Murthy
Nov 14 at 8:12
@rimusolem What I meant was 'integrals do not converge to the right limit'. $int f_n to int f +1$.
– Kavi Rama Murthy
Nov 14 at 8:12
I see. I missed an obvious example.
– Balbadak
Nov 14 at 8:14
I see. I missed an obvious example.
– Balbadak
Nov 14 at 8:14
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997944%2fdo-f-n-to-f-almost-surely-and-0-int-f-infty-imply-int-f-n-to-int-f%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
