Quick question regarding the Lebesgue's Dominated Convergence Theorem
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 |
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 |
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
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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
4,115316
add a comment |
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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
4,115316
add a comment |
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
4,115316
add a comment |
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
4,115316
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
4,115316
edited Nov 25 at 13:53
answered Nov 25 at 13:46
Song
4,115316
answered Nov 25 at 13:46
Song
4,115316
answered Nov 25 at 13:46
Song
4,115316
4,115316
add a comment |
add a comment |
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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