Question about Hahn's positive set lemma
$begingroup$
This lemma states:
Let $ν$ be a signed measure on the measurable space $(X,M)$ and $E$ a measurable set for which $0 < ν(E) < ∞$. Then there is a measurable subset $A$ of $E$ that is positive and of positive measure.
I'm quoting a proof taken from Royden textbook (with minor modifications).
If $E$ itself is a positive set, then the proof is complete. Otherwise, E contains sets of negative measure. Let $m_1$ be the smallest natural number for which there is a measurable set $E_1 ⊆ E$ with $ν(E_1) < -1/m_1$.
Let’s suppose natural numbers $m_1, ... ,m_{k-1}$ and measurable sets $E_1,...,E_{k-1}$ have been chosen. If $E-(E_1∪…∪E_{k-1})$ is positive the proof is complete. Otherwise let $m_k$ be the smallest natural number for which there exist an $E_k ⊆ E - ∪_{j=1}^{k-1}E_j$ such that $ν(E_k) < -1/m_k$.
If this selection process terminates, then the proof is complete. Otherwise, define $A = E - ∪_{k=1}^∞E_k$, so that $E = A ∪ [∪_{k=1}^∞E_k]$ is a disjoint decomposition of E.
Since $∪_{k=1}^∞E_k$ is a measurable subset of $E$ and $|ν(E)| < ∞$, we have $|ν(∪_{k=1}^∞E_k)| < ∞$ and, by the countable additivity of $ν$, $$-∞ < ν(∪_{k=1}^∞E_k) = Σ_{k=1}^∞ν(E_k) ≤ Σ_{k=1}^∞(-1/m_k).$$ Thus $Σ_{k=1}^∞(1/m_k) < ∞ ⇒ lim_k(1/m_k) = 0 ⇒ lim_k m_k = ∞.$
Now, if $B ⊆ A = E-(∪_{k=1}^∞E_k)$, then $B ⊆ E-(E_1∪…∪E_{k-1})$ for each $k ≥ 1$. And $m_k$ is the smallest positive integer such that there is a measurable set $E_k ⊆ E-(E_1∪…∪E_{k-1})$ satisfying $ν(E_k) < -1/m_k$.
It follow that we can't have $ν(B) < -1/(m_k-1)$, so $$ν(B) ≥ -1/(m_k-1) ⇒ ν(B) ≥ lim_k[-1/(m_k-1)] = 0.$$ Therefore A is a positive set and this proves the statement.
I have two question regarding this proof.
1) The bolded step is the only one where we use hypothesis $ν(E) < ∞$, we could get the same result by observing that $$ν(E) = ν(A) + ν(∪_{k=1}^∞E_k) > 0$$ and $$ν(∪_{k=1}^∞E_k) = Σ_{k=1}^∞ν(E_k) < Σ_{k=1}^∞(-1/m_k) < 0.$$
So $ν(∪_{k=1}^∞E_k) > -∞$ and $ν(A) > 0$. Is this correct? Because, if it is, I don't understand the reason of the additional hypothesis, it doesn't simplify the argument.
2) The choice of the sets $E_k$ seems arbitrary, does this proof require the axiom of choice? Does Hahn decomposition require axiom of choice?
measure-theory
asked Nov 25 '18 at 6:36
Alex123Alex123
524
$endgroup$
add a comment |
$begingroup$
This lemma states:
Let $ν$ be a signed measure on the measurable space $(X,M)$ and $E$ a measurable set for which $0 < ν(E) < ∞$. Then there is a measurable subset $A$ of $E$ that is positive and of positive measure.
I'm quoting a proof taken from Royden textbook (with minor modifications).
If $E$ itself is a positive set, then the proof is complete. Otherwise, E contains sets of negative measure. Let $m_1$ be the smallest natural number for which there is a measurable set $E_1 ⊆ E$ with $ν(E_1) < -1/m_1$.
Let’s suppose natural numbers $m_1, ... ,m_{k-1}$ and measurable sets $E_1,...,E_{k-1}$ have been chosen. If $E-(E_1∪…∪E_{k-1})$ is positive the proof is complete. Otherwise let $m_k$ be the smallest natural number for which there exist an $E_k ⊆ E - ∪_{j=1}^{k-1}E_j$ such that $ν(E_k) < -1/m_k$.
If this selection process terminates, then the proof is complete. Otherwise, define $A = E - ∪_{k=1}^∞E_k$, so that $E = A ∪ [∪_{k=1}^∞E_k]$ is a disjoint decomposition of E.
Since $∪_{k=1}^∞E_k$ is a measurable subset of $E$ and $|ν(E)| < ∞$, we have $|ν(∪_{k=1}^∞E_k)| < ∞$ and, by the countable additivity of $ν$, $$-∞ < ν(∪_{k=1}^∞E_k) = Σ_{k=1}^∞ν(E_k) ≤ Σ_{k=1}^∞(-1/m_k).$$ Thus $Σ_{k=1}^∞(1/m_k) < ∞ ⇒ lim_k(1/m_k) = 0 ⇒ lim_k m_k = ∞.$
Now, if $B ⊆ A = E-(∪_{k=1}^∞E_k)$, then $B ⊆ E-(E_1∪…∪E_{k-1})$ for each $k ≥ 1$. And $m_k$ is the smallest positive integer such that there is a measurable set $E_k ⊆ E-(E_1∪…∪E_{k-1})$ satisfying $ν(E_k) < -1/m_k$.
It follow that we can't have $ν(B) < -1/(m_k-1)$, so $$ν(B) ≥ -1/(m_k-1) ⇒ ν(B) ≥ lim_k[-1/(m_k-1)] = 0.$$ Therefore A is a positive set and this proves the statement.
I have two question regarding this proof.
1) The bolded step is the only one where we use hypothesis $ν(E) < ∞$, we could get the same result by observing that $$ν(E) = ν(A) + ν(∪_{k=1}^∞E_k) > 0$$ and $$ν(∪_{k=1}^∞E_k) = Σ_{k=1}^∞ν(E_k) < Σ_{k=1}^∞(-1/m_k) < 0.$$
So $ν(∪_{k=1}^∞E_k) > -∞$ and $ν(A) > 0$. Is this correct? Because, if it is, I don't understand the reason of the additional hypothesis, it doesn't simplify the argument.
2) The choice of the sets $E_k$ seems arbitrary, does this proof require the axiom of choice? Does Hahn decomposition require axiom of choice?
measure-theory
asked Nov 25 '18 at 6:36
Alex123Alex123
524
$endgroup$
$begingroup$
Why downvote? How could you read the post in 2 minutes?
$endgroup$
– Nosrati
Nov 25 '18 at 6:40
$begingroup$
@Nosrati I suspect that's an automatic thing connected to formatting.
$endgroup$
– Alex123
Nov 25 '18 at 6:42
$begingroup$
Not about it, I know downvoter. The moderators should remind him.
$endgroup$
– Nosrati
Nov 25 '18 at 6:43
add a comment |
$begingroup$
This lemma states:
Let $ν$ be a signed measure on the measurable space $(X,M)$ and $E$ a measurable set for which $0 < ν(E) < ∞$. Then there is a measurable subset $A$ of $E$ that is positive and of positive measure.
I'm quoting a proof taken from Royden textbook (with minor modifications).
If $E$ itself is a positive set, then the proof is complete. Otherwise, E contains sets of negative measure. Let $m_1$ be the smallest natural number for which there is a measurable set $E_1 ⊆ E$ with $ν(E_1) < -1/m_1$.
Let’s suppose natural numbers $m_1, ... ,m_{k-1}$ and measurable sets $E_1,...,E_{k-1}$ have been chosen. If $E-(E_1∪…∪E_{k-1})$ is positive the proof is complete. Otherwise let $m_k$ be the smallest natural number for which there exist an $E_k ⊆ E - ∪_{j=1}^{k-1}E_j$ such that $ν(E_k) < -1/m_k$.
If this selection process terminates, then the proof is complete. Otherwise, define $A = E - ∪_{k=1}^∞E_k$, so that $E = A ∪ [∪_{k=1}^∞E_k]$ is a disjoint decomposition of E.
Since $∪_{k=1}^∞E_k$ is a measurable subset of $E$ and $|ν(E)| < ∞$, we have $|ν(∪_{k=1}^∞E_k)| < ∞$ and, by the countable additivity of $ν$, $$-∞ < ν(∪_{k=1}^∞E_k) = Σ_{k=1}^∞ν(E_k) ≤ Σ_{k=1}^∞(-1/m_k).$$ Thus $Σ_{k=1}^∞(1/m_k) < ∞ ⇒ lim_k(1/m_k) = 0 ⇒ lim_k m_k = ∞.$
Now, if $B ⊆ A = E-(∪_{k=1}^∞E_k)$, then $B ⊆ E-(E_1∪…∪E_{k-1})$ for each $k ≥ 1$. And $m_k$ is the smallest positive integer such that there is a measurable set $E_k ⊆ E-(E_1∪…∪E_{k-1})$ satisfying $ν(E_k) < -1/m_k$.
It follow that we can't have $ν(B) < -1/(m_k-1)$, so $$ν(B) ≥ -1/(m_k-1) ⇒ ν(B) ≥ lim_k[-1/(m_k-1)] = 0.$$ Therefore A is a positive set and this proves the statement.
I have two question regarding this proof.
1) The bolded step is the only one where we use hypothesis $ν(E) < ∞$, we could get the same result by observing that $$ν(E) = ν(A) + ν(∪_{k=1}^∞E_k) > 0$$ and $$ν(∪_{k=1}^∞E_k) = Σ_{k=1}^∞ν(E_k) < Σ_{k=1}^∞(-1/m_k) < 0.$$
So $ν(∪_{k=1}^∞E_k) > -∞$ and $ν(A) > 0$. Is this correct? Because, if it is, I don't understand the reason of the additional hypothesis, it doesn't simplify the argument.
2) The choice of the sets $E_k$ seems arbitrary, does this proof require the axiom of choice? Does Hahn decomposition require axiom of choice?
measure-theory
asked Nov 25 '18 at 6:36
Alex123Alex123
524
$endgroup$
This lemma states:
Let $ν$ be a signed measure on the measurable space $(X,M)$ and $E$ a measurable set for which $0 < ν(E) < ∞$. Then there is a measurable subset $A$ of $E$ that is positive and of positive measure.
I'm quoting a proof taken from Royden textbook (with minor modifications).
If $E$ itself is a positive set, then the proof is complete. Otherwise, E contains sets of negative measure. Let $m_1$ be the smallest natural number for which there is a measurable set $E_1 ⊆ E$ with $ν(E_1) < -1/m_1$.
Let’s suppose natural numbers $m_1, ... ,m_{k-1}$ and measurable sets $E_1,...,E_{k-1}$ have been chosen. If $E-(E_1∪…∪E_{k-1})$ is positive the proof is complete. Otherwise let $m_k$ be the smallest natural number for which there exist an $E_k ⊆ E - ∪_{j=1}^{k-1}E_j$ such that $ν(E_k) < -1/m_k$.
If this selection process terminates, then the proof is complete. Otherwise, define $A = E - ∪_{k=1}^∞E_k$, so that $E = A ∪ [∪_{k=1}^∞E_k]$ is a disjoint decomposition of E.
Since $∪_{k=1}^∞E_k$ is a measurable subset of $E$ and $|ν(E)| < ∞$, we have $|ν(∪_{k=1}^∞E_k)| < ∞$ and, by the countable additivity of $ν$, $$-∞ < ν(∪_{k=1}^∞E_k) = Σ_{k=1}^∞ν(E_k) ≤ Σ_{k=1}^∞(-1/m_k).$$ Thus $Σ_{k=1}^∞(1/m_k) < ∞ ⇒ lim_k(1/m_k) = 0 ⇒ lim_k m_k = ∞.$
Now, if $B ⊆ A = E-(∪_{k=1}^∞E_k)$, then $B ⊆ E-(E_1∪…∪E_{k-1})$ for each $k ≥ 1$. And $m_k$ is the smallest positive integer such that there is a measurable set $E_k ⊆ E-(E_1∪…∪E_{k-1})$ satisfying $ν(E_k) < -1/m_k$.
It follow that we can't have $ν(B) < -1/(m_k-1)$, so $$ν(B) ≥ -1/(m_k-1) ⇒ ν(B) ≥ lim_k[-1/(m_k-1)] = 0.$$ Therefore A is a positive set and this proves the statement.
I have two question regarding this proof.
1) The bolded step is the only one where we use hypothesis $ν(E) < ∞$, we could get the same result by observing that $$ν(E) = ν(A) + ν(∪_{k=1}^∞E_k) > 0$$ and $$ν(∪_{k=1}^∞E_k) = Σ_{k=1}^∞ν(E_k) < Σ_{k=1}^∞(-1/m_k) < 0.$$
So $ν(∪_{k=1}^∞E_k) > -∞$ and $ν(A) > 0$. Is this correct? Because, if it is, I don't understand the reason of the additional hypothesis, it doesn't simplify the argument.
2) The choice of the sets $E_k$ seems arbitrary, does this proof require the axiom of choice? Does Hahn decomposition require axiom of choice?
measure-theory
measure-theory
asked Nov 25 '18 at 6:36
Alex123Alex123
524
asked Nov 25 '18 at 6:36
Alex123Alex123
524
edited Nov 25 '18 at 10:01
Alex123
asked Nov 25 '18 at 6:36
Alex123Alex123
524
asked Nov 25 '18 at 6:36
Alex123Alex123
524
asked Nov 25 '18 at 6:36
Alex123Alex123
524
524
$begingroup$
Why downvote? How could you read the post in 2 minutes?
$endgroup$
– Nosrati
Nov 25 '18 at 6:40
$begingroup$
@Nosrati I suspect that's an automatic thing connected to formatting.
$endgroup$
– Alex123
Nov 25 '18 at 6:42
$begingroup$
Not about it, I know downvoter. The moderators should remind him.
$endgroup$
– Nosrati
Nov 25 '18 at 6:43
add a comment |
$begingroup$
Why downvote? How could you read the post in 2 minutes?
$endgroup$
– Nosrati
Nov 25 '18 at 6:40
$begingroup$
@Nosrati I suspect that's an automatic thing connected to formatting.
$endgroup$
– Alex123
Nov 25 '18 at 6:42
$begingroup$
Not about it, I know downvoter. The moderators should remind him.
$endgroup$
– Nosrati
Nov 25 '18 at 6:43
$begingroup$
Why downvote? How could you read the post in 2 minutes?
$endgroup$
– Nosrati
Nov 25 '18 at 6:40
$begingroup$
Why downvote? How could you read the post in 2 minutes?
$endgroup$
– Nosrati
Nov 25 '18 at 6:40
$begingroup$
@Nosrati I suspect that's an automatic thing connected to formatting.
$endgroup$
– Alex123
Nov 25 '18 at 6:42
$begingroup$
@Nosrati I suspect that's an automatic thing connected to formatting.
$endgroup$
– Alex123
Nov 25 '18 at 6:42
$begingroup$
Not about it, I know downvoter. The moderators should remind him.
$endgroup$
– Nosrati
Nov 25 '18 at 6:43
$begingroup$
Not about it, I know downvoter. The moderators should remind him.
$endgroup$
– Nosrati
Nov 25 '18 at 6:43
add a comment |
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$begingroup$
Why downvote? How could you read the post in 2 minutes?
$endgroup$
– Nosrati
Nov 25 '18 at 6:40
$begingroup$
@Nosrati I suspect that's an automatic thing connected to formatting.
$endgroup$
– Alex123
Nov 25 '18 at 6:42
$begingroup$
Not about it, I know downvoter. The moderators should remind him.
$endgroup$
– Nosrati
Nov 25 '18 at 6:43