Monotone Convergence for Decreasing Functions
I'm trying to prove the Monotone Convergence Theorem for decreasing sequences, namely if
Let $(X,mathcal{M},mu)$ be a measure space and suppose ${f_n}$ are non-negative measurable functions decreasing pointwise to $f$. Suppose also that $f_1 in mathscr{L}(mu)$. Then $$int_X f~dmu = lim_{ntoinfty}int_X f_n~dmu.$$
Why does this statement not follow from LDCT with $f_n$ being dominated by $f_1$?
I'm also aware of the solutions with $g_n=f_1-f_n$, but the question asks to prove it using Fatou's lemma
real-analysis analysis measure-theory lebesgue-integral
asked Oct 9 '15 at 0:28
Anthony Peter
3,23711451
|
show 2 more comments
I'm trying to prove the Monotone Convergence Theorem for decreasing sequences, namely if
Let $(X,mathcal{M},mu)$ be a measure space and suppose ${f_n}$ are non-negative measurable functions decreasing pointwise to $f$. Suppose also that $f_1 in mathscr{L}(mu)$. Then $$int_X f~dmu = lim_{ntoinfty}int_X f_n~dmu.$$
Why does this statement not follow from LDCT with $f_n$ being dominated by $f_1$?
I'm also aware of the solutions with $g_n=f_1-f_n$, but the question asks to prove it using Fatou's lemma
real-analysis analysis measure-theory lebesgue-integral
asked Oct 9 '15 at 0:28
Anthony Peter
3,23711451
In Fatou's lemma, the inequality goes the other way. So the inequality $limint f_n,dmuleint f,dmu$ doesn't follow from Fatou.
– grand_chat
Oct 9 '15 at 1:26
@grand_chat why does the statement not follow from LDCT with $g=f_1$?
– Anthony Peter
Oct 9 '15 at 1:59
3
I think it does follow from LDCT.
– user99914
Oct 9 '15 at 3:01
@JohnMa Then why does every solution use the $g_n=f_1-f_n$ trick and MCT?
– Anthony Peter
Oct 9 '15 at 3:03
I guess that's because LDCT is an overkill? (It seems that LDCT is the last theorem to prove, either Fatou or MCT comes first)
– user99914
Oct 9 '15 at 3:05
|
show 2 more comments
I'm trying to prove the Monotone Convergence Theorem for decreasing sequences, namely if
Let $(X,mathcal{M},mu)$ be a measure space and suppose ${f_n}$ are non-negative measurable functions decreasing pointwise to $f$. Suppose also that $f_1 in mathscr{L}(mu)$. Then $$int_X f~dmu = lim_{ntoinfty}int_X f_n~dmu.$$
Why does this statement not follow from LDCT with $f_n$ being dominated by $f_1$?
I'm also aware of the solutions with $g_n=f_1-f_n$, but the question asks to prove it using Fatou's lemma
real-analysis analysis measure-theory lebesgue-integral
asked Oct 9 '15 at 0:28
Anthony Peter
3,23711451
I'm trying to prove the Monotone Convergence Theorem for decreasing sequences, namely if
Let $(X,mathcal{M},mu)$ be a measure space and suppose ${f_n}$ are non-negative measurable functions decreasing pointwise to $f$. Suppose also that $f_1 in mathscr{L}(mu)$. Then $$int_X f~dmu = lim_{ntoinfty}int_X f_n~dmu.$$
Why does this statement not follow from LDCT with $f_n$ being dominated by $f_1$?
I'm also aware of the solutions with $g_n=f_1-f_n$, but the question asks to prove it using Fatou's lemma
real-analysis analysis measure-theory lebesgue-integral
real-analysis analysis measure-theory lebesgue-integral
asked Oct 9 '15 at 0:28
Anthony Peter
3,23711451
asked Oct 9 '15 at 0:28
Anthony Peter
3,23711451
edited Oct 9 '15 at 2:03
asked Oct 9 '15 at 0:28
Anthony Peter
3,23711451
asked Oct 9 '15 at 0:28
Anthony Peter
3,23711451
asked Oct 9 '15 at 0:28
Anthony Peter
3,23711451
3,23711451
In Fatou's lemma, the inequality goes the other way. So the inequality $limint f_n,dmuleint f,dmu$ doesn't follow from Fatou.
– grand_chat
Oct 9 '15 at 1:26
@grand_chat why does the statement not follow from LDCT with $g=f_1$?
– Anthony Peter
Oct 9 '15 at 1:59
3
I think it does follow from LDCT.
– user99914
Oct 9 '15 at 3:01
@JohnMa Then why does every solution use the $g_n=f_1-f_n$ trick and MCT?
– Anthony Peter
Oct 9 '15 at 3:03
I guess that's because LDCT is an overkill? (It seems that LDCT is the last theorem to prove, either Fatou or MCT comes first)
– user99914
Oct 9 '15 at 3:05
|
show 2 more comments
In Fatou's lemma, the inequality goes the other way. So the inequality $limint f_n,dmuleint f,dmu$ doesn't follow from Fatou.
– grand_chat
Oct 9 '15 at 1:26
@grand_chat why does the statement not follow from LDCT with $g=f_1$?
– Anthony Peter
Oct 9 '15 at 1:59
3
I think it does follow from LDCT.
– user99914
Oct 9 '15 at 3:01
@JohnMa Then why does every solution use the $g_n=f_1-f_n$ trick and MCT?
– Anthony Peter
Oct 9 '15 at 3:03
I guess that's because LDCT is an overkill? (It seems that LDCT is the last theorem to prove, either Fatou or MCT comes first)
– user99914
Oct 9 '15 at 3:05
In Fatou's lemma, the inequality goes the other way. So the inequality $limint f_n,dmuleint f,dmu$ doesn't follow from Fatou.
– grand_chat
Oct 9 '15 at 1:26
In Fatou's lemma, the inequality goes the other way. So the inequality $limint f_n,dmuleint f,dmu$ doesn't follow from Fatou.
– grand_chat
Oct 9 '15 at 1:26
@grand_chat why does the statement not follow from LDCT with $g=f_1$?
– Anthony Peter
Oct 9 '15 at 1:59
@grand_chat why does the statement not follow from LDCT with $g=f_1$?
– Anthony Peter
Oct 9 '15 at 1:59
3
3
I think it does follow from LDCT.
– user99914
Oct 9 '15 at 3:01
I think it does follow from LDCT.
– user99914
Oct 9 '15 at 3:01
@JohnMa Then why does every solution use the $g_n=f_1-f_n$ trick and MCT?
– Anthony Peter
Oct 9 '15 at 3:03
@JohnMa Then why does every solution use the $g_n=f_1-f_n$ trick and MCT?
– Anthony Peter
Oct 9 '15 at 3:03
I guess that's because LDCT is an overkill? (It seems that LDCT is the last theorem to prove, either Fatou or MCT comes first)
– user99914
Oct 9 '15 at 3:05
I guess that's because LDCT is an overkill? (It seems that LDCT is the last theorem to prove, either Fatou or MCT comes first)
– user99914
Oct 9 '15 at 3:05
|
show 2 more comments
1 Answer
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As the OP says in the comments, the problem is posed at the end of the first chapter of Rudin's Real and Complex Analysis, so it is not clear whether he intends the student to use the Monotone Convergence Theorem, Fatou's Lemma or the Dominated Convergence Theorem.
You are right that the proof is a trivial application of DCT. However, it is certainly an instructive exercise to prove it using just MCT or Fatou's Lemma.
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As the OP says in the comments, the problem is posed at the end of the first chapter of Rudin's Real and Complex Analysis, so it is not clear whether he intends the student to use the Monotone Convergence Theorem, Fatou's Lemma or the Dominated Convergence Theorem.
You are right that the proof is a trivial application of DCT. However, it is certainly an instructive exercise to prove it using just MCT or Fatou's Lemma.
add a comment |
As the OP says in the comments, the problem is posed at the end of the first chapter of Rudin's Real and Complex Analysis, so it is not clear whether he intends the student to use the Monotone Convergence Theorem, Fatou's Lemma or the Dominated Convergence Theorem.
You are right that the proof is a trivial application of DCT. However, it is certainly an instructive exercise to prove it using just MCT or Fatou's Lemma.
add a comment |
As the OP says in the comments, the problem is posed at the end of the first chapter of Rudin's Real and Complex Analysis, so it is not clear whether he intends the student to use the Monotone Convergence Theorem, Fatou's Lemma or the Dominated Convergence Theorem.
You are right that the proof is a trivial application of DCT. However, it is certainly an instructive exercise to prove it using just MCT or Fatou's Lemma.
As the OP says in the comments, the problem is posed at the end of the first chapter of Rudin's Real and Complex Analysis, so it is not clear whether he intends the student to use the Monotone Convergence Theorem, Fatou's Lemma or the Dominated Convergence Theorem.
You are right that the proof is a trivial application of DCT. However, it is certainly an instructive exercise to prove it using just MCT or Fatou's Lemma.
edited Nov 26 '18 at 14:07
community wiki
2 revs
Brahadeesh
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In Fatou's lemma, the inequality goes the other way. So the inequality $limint f_n,dmuleint f,dmu$ doesn't follow from Fatou.
– grand_chat
Oct 9 '15 at 1:26
@grand_chat why does the statement not follow from LDCT with $g=f_1$?
– Anthony Peter
Oct 9 '15 at 1:59
3
I think it does follow from LDCT.
– user99914
Oct 9 '15 at 3:01
@JohnMa Then why does every solution use the $g_n=f_1-f_n$ trick and MCT?
– Anthony Peter
Oct 9 '15 at 3:03
I guess that's because LDCT is an overkill? (It seems that LDCT is the last theorem to prove, either Fatou or MCT comes first)
– user99914
Oct 9 '15 at 3:05