Quick question regarding the Lebesgue's Dominated Convergence Theorem
up vote
1
down vote
favorite
I'm reading Rudin's Real and Complex Analysis book. In the statement where we require $$|f_n(x)| le g(x)$$ for all $n$. Could this be relaxed to all, but finitely many $n$? I was looking at his proof and couldn't find any reason why not, but I thought I go ahead and ask to make sure.
measure-theory lebesgue-integral
asked Nov 25 at 13:39
daniel
463
add a comment |
up vote
1
down vote
favorite
I'm reading Rudin's Real and Complex Analysis book. In the statement where we require $$|f_n(x)| le g(x)$$ for all $n$. Could this be relaxed to all, but finitely many $n$? I was looking at his proof and couldn't find any reason why not, but I thought I go ahead and ask to make sure.
measure-theory lebesgue-integral
asked Nov 25 at 13:39
daniel
463
Yes, absolutely, since the statement of the theorem is concerned with limits, which "only care" about the $f_{n}$ after a certain $n_{0}$.
– Matija Sreckovic
Nov 25 at 13:42
1
I think so too.. just paranoid sometimes :)
– daniel
Nov 25 at 13:44
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm reading Rudin's Real and Complex Analysis book. In the statement where we require $$|f_n(x)| le g(x)$$ for all $n$. Could this be relaxed to all, but finitely many $n$? I was looking at his proof and couldn't find any reason why not, but I thought I go ahead and ask to make sure.
measure-theory lebesgue-integral
asked Nov 25 at 13:39
daniel
463
I'm reading Rudin's Real and Complex Analysis book. In the statement where we require $$|f_n(x)| le g(x)$$ for all $n$. Could this be relaxed to all, but finitely many $n$? I was looking at his proof and couldn't find any reason why not, but I thought I go ahead and ask to make sure.
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Nov 25 at 13:39
daniel
463
asked Nov 25 at 13:39
daniel
463
asked Nov 25 at 13:39
daniel
463
asked Nov 25 at 13:39
daniel
463
asked Nov 25 at 13:39
daniel
463
463
Yes, absolutely, since the statement of the theorem is concerned with limits, which "only care" about the $f_{n}$ after a certain $n_{0}$.
– Matija Sreckovic
Nov 25 at 13:42
1
I think so too.. just paranoid sometimes :)
– daniel
Nov 25 at 13:44
add a comment |
Yes, absolutely, since the statement of the theorem is concerned with limits, which "only care" about the $f_{n}$ after a certain $n_{0}$.
– Matija Sreckovic
Nov 25 at 13:42
1
I think so too.. just paranoid sometimes :)
– daniel
Nov 25 at 13:44
Yes, absolutely, since the statement of the theorem is concerned with limits, which "only care" about the $f_{n}$ after a certain $n_{0}$.
– Matija Sreckovic
Nov 25 at 13:42
Yes, absolutely, since the statement of the theorem is concerned with limits, which "only care" about the $f_{n}$ after a certain $n_{0}$.
– Matija Sreckovic
Nov 25 at 13:42
1
1
I think so too.. just paranoid sometimes :)
– daniel
Nov 25 at 13:44
I think so too.. just paranoid sometimes :)
– daniel
Nov 25 at 13:44
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
Of course you can. If $|f_n(x)|leq g(x)$ for $n> N$, you may throw away first $N$ terms since it does not affect the limit. If $f_n in L^1$ for all $n$, you may take $$
g' = g + |f_1| +cdots + |f_{N}|$$ as a majorant explicitly.
Note: Actually, condition of single majorant $g$ can be replaced with existence of $L^1$-sequence ${g_n, g }$ such that $|f_n| leq g_n$, $|lim_n f_n|leq g$, and $int g_n dmuto int gdmu.$
answered Nov 25 at 13:46
Song
1,39512
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
Of course you can. If $|f_n(x)|leq g(x)$ for $n> N$, you may throw away first $N$ terms since it does not affect the limit. If $f_n in L^1$ for all $n$, you may take $$
g' = g + |f_1| +cdots + |f_{N}|$$ as a majorant explicitly.
Note: Actually, condition of single majorant $g$ can be replaced with existence of $L^1$-sequence ${g_n, g }$ such that $|f_n| leq g_n$, $|lim_n f_n|leq g$, and $int g_n dmuto int gdmu.$
answered Nov 25 at 13:46
Song
1,39512
add a comment |
up vote
1
down vote
Of course you can. If $|f_n(x)|leq g(x)$ for $n> N$, you may throw away first $N$ terms since it does not affect the limit. If $f_n in L^1$ for all $n$, you may take $$
g' = g + |f_1| +cdots + |f_{N}|$$ as a majorant explicitly.
Note: Actually, condition of single majorant $g$ can be replaced with existence of $L^1$-sequence ${g_n, g }$ such that $|f_n| leq g_n$, $|lim_n f_n|leq g$, and $int g_n dmuto int gdmu.$
answered Nov 25 at 13:46
Song
1,39512
add a comment |
up vote
1
down vote
up vote
1
down vote
Of course you can. If $|f_n(x)|leq g(x)$ for $n> N$, you may throw away first $N$ terms since it does not affect the limit. If $f_n in L^1$ for all $n$, you may take $$
g' = g + |f_1| +cdots + |f_{N}|$$ as a majorant explicitly.
Note: Actually, condition of single majorant $g$ can be replaced with existence of $L^1$-sequence ${g_n, g }$ such that $|f_n| leq g_n$, $|lim_n f_n|leq g$, and $int g_n dmuto int gdmu.$
answered Nov 25 at 13:46
Song
1,39512
Of course you can. If $|f_n(x)|leq g(x)$ for $n> N$, you may throw away first $N$ terms since it does not affect the limit. If $f_n in L^1$ for all $n$, you may take $$
g' = g + |f_1| +cdots + |f_{N}|$$ as a majorant explicitly.
Note: Actually, condition of single majorant $g$ can be replaced with existence of $L^1$-sequence ${g_n, g }$ such that $|f_n| leq g_n$, $|lim_n f_n|leq g$, and $int g_n dmuto int gdmu.$
answered Nov 25 at 13:46
Song
1,39512
edited Nov 25 at 13:53
answered Nov 25 at 13:46
Song
1,39512
answered Nov 25 at 13:46
Song
1,39512
answered Nov 25 at 13:46
Song
1,39512
1,39512
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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%2f3012847%2fquick-question-regarding-the-lebesgues-dominated-convergence-theorem%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
Yes, absolutely, since the statement of the theorem is concerned with limits, which "only care" about the $f_{n}$ after a certain $n_{0}$.
– Matija Sreckovic
Nov 25 at 13:42
1
I think so too.. just paranoid sometimes :)
– daniel
Nov 25 at 13:44