Mistake in Billingsleys book?
$begingroup$
This is about an exercise in Billingsley's book Probability and measure.
Exercise 2.15: On the field $mathscr B_0$ in $(0,1]$ define $P(A)$ to be $1$ or $0$ according as there does or does not exist some positive $epsilon_A$ (depending on $A$) such that $A$ contains the interval $(frac{1}{2},frac{1}{2} + epsilon_A]$. Show that $P$ is finitely but not countably additive.
I am able to prove that $P$ is not $sigma$-additve. If it were, then it would be continuous from above. Consider for example the sequence $A_n :=(frac{1}{2}, frac{1}{2} + frac{1}{n}]$. $A_n downarrow emptyset$, but $P(A_n) to 1 neq
0 = P(emptyset)$, since $P(A_n)=1$ for all $n$.
But I am not able to prove that $P$ is finitely additive. I believe that is, because $P$ isn't finitely additive. For example consider the sets $A = (0,1] cap mathbb Q$ and $B = (0,1] backslash mathbb Q$. They are disjoint, $P(A) = P(B) = 0$. But $Acup B = (0,1]$ and thus $P(A cup B)= 1 neq P(A) + P(B)$.
So my question is, did I make some stupid mistake? Or is there indeed a mistake in the exercise?
probability
asked Dec 27 '18 at 20:57
N.BeckN.Beck
3087
$endgroup$
add a comment |
$begingroup$
This is about an exercise in Billingsley's book Probability and measure.
Exercise 2.15: On the field $mathscr B_0$ in $(0,1]$ define $P(A)$ to be $1$ or $0$ according as there does or does not exist some positive $epsilon_A$ (depending on $A$) such that $A$ contains the interval $(frac{1}{2},frac{1}{2} + epsilon_A]$. Show that $P$ is finitely but not countably additive.
I am able to prove that $P$ is not $sigma$-additve. If it were, then it would be continuous from above. Consider for example the sequence $A_n :=(frac{1}{2}, frac{1}{2} + frac{1}{n}]$. $A_n downarrow emptyset$, but $P(A_n) to 1 neq
0 = P(emptyset)$, since $P(A_n)=1$ for all $n$.
But I am not able to prove that $P$ is finitely additive. I believe that is, because $P$ isn't finitely additive. For example consider the sets $A = (0,1] cap mathbb Q$ and $B = (0,1] backslash mathbb Q$. They are disjoint, $P(A) = P(B) = 0$. But $Acup B = (0,1]$ and thus $P(A cup B)= 1 neq P(A) + P(B)$.
So my question is, did I make some stupid mistake? Or is there indeed a mistake in the exercise?
probability
asked Dec 27 '18 at 20:57
N.BeckN.Beck
3087
$endgroup$
add a comment |
$begingroup$
This is about an exercise in Billingsley's book Probability and measure.
Exercise 2.15: On the field $mathscr B_0$ in $(0,1]$ define $P(A)$ to be $1$ or $0$ according as there does or does not exist some positive $epsilon_A$ (depending on $A$) such that $A$ contains the interval $(frac{1}{2},frac{1}{2} + epsilon_A]$. Show that $P$ is finitely but not countably additive.
I am able to prove that $P$ is not $sigma$-additve. If it were, then it would be continuous from above. Consider for example the sequence $A_n :=(frac{1}{2}, frac{1}{2} + frac{1}{n}]$. $A_n downarrow emptyset$, but $P(A_n) to 1 neq
0 = P(emptyset)$, since $P(A_n)=1$ for all $n$.
But I am not able to prove that $P$ is finitely additive. I believe that is, because $P$ isn't finitely additive. For example consider the sets $A = (0,1] cap mathbb Q$ and $B = (0,1] backslash mathbb Q$. They are disjoint, $P(A) = P(B) = 0$. But $Acup B = (0,1]$ and thus $P(A cup B)= 1 neq P(A) + P(B)$.
So my question is, did I make some stupid mistake? Or is there indeed a mistake in the exercise?
probability
asked Dec 27 '18 at 20:57
N.BeckN.Beck
3087
$endgroup$
This is about an exercise in Billingsley's book Probability and measure.
Exercise 2.15: On the field $mathscr B_0$ in $(0,1]$ define $P(A)$ to be $1$ or $0$ according as there does or does not exist some positive $epsilon_A$ (depending on $A$) such that $A$ contains the interval $(frac{1}{2},frac{1}{2} + epsilon_A]$. Show that $P$ is finitely but not countably additive.
I am able to prove that $P$ is not $sigma$-additve. If it were, then it would be continuous from above. Consider for example the sequence $A_n :=(frac{1}{2}, frac{1}{2} + frac{1}{n}]$. $A_n downarrow emptyset$, but $P(A_n) to 1 neq
0 = P(emptyset)$, since $P(A_n)=1$ for all $n$.
But I am not able to prove that $P$ is finitely additive. I believe that is, because $P$ isn't finitely additive. For example consider the sets $A = (0,1] cap mathbb Q$ and $B = (0,1] backslash mathbb Q$. They are disjoint, $P(A) = P(B) = 0$. But $Acup B = (0,1]$ and thus $P(A cup B)= 1 neq P(A) + P(B)$.
So my question is, did I make some stupid mistake? Or is there indeed a mistake in the exercise?
probability
probability
asked Dec 27 '18 at 20:57
N.BeckN.Beck
3087
asked Dec 27 '18 at 20:57
N.BeckN.Beck
3087
asked Dec 27 '18 at 20:57
N.BeckN.Beck
3087
asked Dec 27 '18 at 20:57
N.BeckN.Beck
3087
asked Dec 27 '18 at 20:57
N.BeckN.Beck
3087
3087
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your mistake: Recall that $mathscr{B}_0$ is defined as the set of "finite disjoint unions of intervals in $(0,1]$." The issue with your counterexample is simply that the $A,B$ you use do not belong to $mathscr{B}_0$.
Now, take any two disjoint $A,Binmathscr{B}_0$: by assumption, there exist $n,mgeq 1$ and disjoint non-empty intervals $I_1,dots,I_n,J_1,dots,J_msubseteq (0,1]$ such that
$$
A = bigcup_{i=1}^n I_i , qquad B = bigcup_{i=1}^m J_i
$$
We have the following cases cases:
Clearly, if $P(A)=1$ and $P(B)=0$, then $P(Acup B)=1$. Similarly if $P(A)=0$ and $P(B)=1$.
One cannot have $P(A)=P(B)= 1$, since then by definition $1/2in Acap B$; but $A,B$ are taken disjoint.
Suppose $P(A)=P(B)=0$, and by contradiction assume $P(Acup B)=1$. This means $1/2in Acup B$, so wlog assume $1/2in A$, say $1/2in I_1$ (again wlog). Since $P(Acup B)=1$, there exists $varepsilon >0$ such that $(1/2, 1/2+varepsilon] subseteq Acup B$: consider $(1/2, 1/2+varepsilon] cap I_1$. It's a non-empty intersection of intervals, so it's a non-empty interval. It is immediate to see it's of the form $(1/2, 1/2+varepsilon']$, and it's contained in $A$: so $P(A)=1$, contradiction.
Therefore, in all possible cases we have $P(A)+P(B)=P(Acup B)$. This shows $P$ is finitely additive.
answered Dec 27 '18 at 21:18
Clement C.Clement C.
50.4k33890
$endgroup$
$begingroup$
Oh, ok . Thanks a lot. :) I thought $ mathscr B_0$ is the completion of the Borel $sigma$-algebra, as it is usual.
$endgroup$
– N.Beck
Dec 27 '18 at 22:35
$begingroup$
@N.Beck Yes, I am not sure it's entirely standard notation (It is, though, in that book). Glad to help!
$endgroup$
– Clement C.
Dec 27 '18 at 22:51
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Your mistake: Recall that $mathscr{B}_0$ is defined as the set of "finite disjoint unions of intervals in $(0,1]$." The issue with your counterexample is simply that the $A,B$ you use do not belong to $mathscr{B}_0$.
Now, take any two disjoint $A,Binmathscr{B}_0$: by assumption, there exist $n,mgeq 1$ and disjoint non-empty intervals $I_1,dots,I_n,J_1,dots,J_msubseteq (0,1]$ such that
$$
A = bigcup_{i=1}^n I_i , qquad B = bigcup_{i=1}^m J_i
$$
We have the following cases cases:
Clearly, if $P(A)=1$ and $P(B)=0$, then $P(Acup B)=1$. Similarly if $P(A)=0$ and $P(B)=1$.
One cannot have $P(A)=P(B)= 1$, since then by definition $1/2in Acap B$; but $A,B$ are taken disjoint.
Suppose $P(A)=P(B)=0$, and by contradiction assume $P(Acup B)=1$. This means $1/2in Acup B$, so wlog assume $1/2in A$, say $1/2in I_1$ (again wlog). Since $P(Acup B)=1$, there exists $varepsilon >0$ such that $(1/2, 1/2+varepsilon] subseteq Acup B$: consider $(1/2, 1/2+varepsilon] cap I_1$. It's a non-empty intersection of intervals, so it's a non-empty interval. It is immediate to see it's of the form $(1/2, 1/2+varepsilon']$, and it's contained in $A$: so $P(A)=1$, contradiction.
Therefore, in all possible cases we have $P(A)+P(B)=P(Acup B)$. This shows $P$ is finitely additive.
answered Dec 27 '18 at 21:18
Clement C.Clement C.
50.4k33890
$endgroup$
$begingroup$
Oh, ok . Thanks a lot. :) I thought $ mathscr B_0$ is the completion of the Borel $sigma$-algebra, as it is usual.
$endgroup$
– N.Beck
Dec 27 '18 at 22:35
$begingroup$
@N.Beck Yes, I am not sure it's entirely standard notation (It is, though, in that book). Glad to help!
$endgroup$
– Clement C.
Dec 27 '18 at 22:51
add a comment |
$begingroup$
Your mistake: Recall that $mathscr{B}_0$ is defined as the set of "finite disjoint unions of intervals in $(0,1]$." The issue with your counterexample is simply that the $A,B$ you use do not belong to $mathscr{B}_0$.
Now, take any two disjoint $A,Binmathscr{B}_0$: by assumption, there exist $n,mgeq 1$ and disjoint non-empty intervals $I_1,dots,I_n,J_1,dots,J_msubseteq (0,1]$ such that
$$
A = bigcup_{i=1}^n I_i , qquad B = bigcup_{i=1}^m J_i
$$
We have the following cases cases:
Clearly, if $P(A)=1$ and $P(B)=0$, then $P(Acup B)=1$. Similarly if $P(A)=0$ and $P(B)=1$.
One cannot have $P(A)=P(B)= 1$, since then by definition $1/2in Acap B$; but $A,B$ are taken disjoint.
Suppose $P(A)=P(B)=0$, and by contradiction assume $P(Acup B)=1$. This means $1/2in Acup B$, so wlog assume $1/2in A$, say $1/2in I_1$ (again wlog). Since $P(Acup B)=1$, there exists $varepsilon >0$ such that $(1/2, 1/2+varepsilon] subseteq Acup B$: consider $(1/2, 1/2+varepsilon] cap I_1$. It's a non-empty intersection of intervals, so it's a non-empty interval. It is immediate to see it's of the form $(1/2, 1/2+varepsilon']$, and it's contained in $A$: so $P(A)=1$, contradiction.
Therefore, in all possible cases we have $P(A)+P(B)=P(Acup B)$. This shows $P$ is finitely additive.
answered Dec 27 '18 at 21:18
Clement C.Clement C.
50.4k33890
$endgroup$
$begingroup$
Oh, ok . Thanks a lot. :) I thought $ mathscr B_0$ is the completion of the Borel $sigma$-algebra, as it is usual.
$endgroup$
– N.Beck
Dec 27 '18 at 22:35
$begingroup$
@N.Beck Yes, I am not sure it's entirely standard notation (It is, though, in that book). Glad to help!
$endgroup$
– Clement C.
Dec 27 '18 at 22:51
add a comment |
$begingroup$
Your mistake: Recall that $mathscr{B}_0$ is defined as the set of "finite disjoint unions of intervals in $(0,1]$." The issue with your counterexample is simply that the $A,B$ you use do not belong to $mathscr{B}_0$.
Now, take any two disjoint $A,Binmathscr{B}_0$: by assumption, there exist $n,mgeq 1$ and disjoint non-empty intervals $I_1,dots,I_n,J_1,dots,J_msubseteq (0,1]$ such that
$$
A = bigcup_{i=1}^n I_i , qquad B = bigcup_{i=1}^m J_i
$$
We have the following cases cases:
Clearly, if $P(A)=1$ and $P(B)=0$, then $P(Acup B)=1$. Similarly if $P(A)=0$ and $P(B)=1$.
One cannot have $P(A)=P(B)= 1$, since then by definition $1/2in Acap B$; but $A,B$ are taken disjoint.
Suppose $P(A)=P(B)=0$, and by contradiction assume $P(Acup B)=1$. This means $1/2in Acup B$, so wlog assume $1/2in A$, say $1/2in I_1$ (again wlog). Since $P(Acup B)=1$, there exists $varepsilon >0$ such that $(1/2, 1/2+varepsilon] subseteq Acup B$: consider $(1/2, 1/2+varepsilon] cap I_1$. It's a non-empty intersection of intervals, so it's a non-empty interval. It is immediate to see it's of the form $(1/2, 1/2+varepsilon']$, and it's contained in $A$: so $P(A)=1$, contradiction.
Therefore, in all possible cases we have $P(A)+P(B)=P(Acup B)$. This shows $P$ is finitely additive.
answered Dec 27 '18 at 21:18
Clement C.Clement C.
50.4k33890
$endgroup$
Your mistake: Recall that $mathscr{B}_0$ is defined as the set of "finite disjoint unions of intervals in $(0,1]$." The issue with your counterexample is simply that the $A,B$ you use do not belong to $mathscr{B}_0$.
Now, take any two disjoint $A,Binmathscr{B}_0$: by assumption, there exist $n,mgeq 1$ and disjoint non-empty intervals $I_1,dots,I_n,J_1,dots,J_msubseteq (0,1]$ such that
$$
A = bigcup_{i=1}^n I_i , qquad B = bigcup_{i=1}^m J_i
$$
We have the following cases cases:
Clearly, if $P(A)=1$ and $P(B)=0$, then $P(Acup B)=1$. Similarly if $P(A)=0$ and $P(B)=1$.
One cannot have $P(A)=P(B)= 1$, since then by definition $1/2in Acap B$; but $A,B$ are taken disjoint.
Suppose $P(A)=P(B)=0$, and by contradiction assume $P(Acup B)=1$. This means $1/2in Acup B$, so wlog assume $1/2in A$, say $1/2in I_1$ (again wlog). Since $P(Acup B)=1$, there exists $varepsilon >0$ such that $(1/2, 1/2+varepsilon] subseteq Acup B$: consider $(1/2, 1/2+varepsilon] cap I_1$. It's a non-empty intersection of intervals, so it's a non-empty interval. It is immediate to see it's of the form $(1/2, 1/2+varepsilon']$, and it's contained in $A$: so $P(A)=1$, contradiction.
Therefore, in all possible cases we have $P(A)+P(B)=P(Acup B)$. This shows $P$ is finitely additive.
answered Dec 27 '18 at 21:18
Clement C.Clement C.
50.4k33890
answered Dec 27 '18 at 21:18
Clement C.Clement C.
50.4k33890
answered Dec 27 '18 at 21:18
Clement C.Clement C.
50.4k33890
answered Dec 27 '18 at 21:18
Clement C.Clement C.
50.4k33890
50.4k33890
$begingroup$
Oh, ok . Thanks a lot. :) I thought $ mathscr B_0$ is the completion of the Borel $sigma$-algebra, as it is usual.
$endgroup$
– N.Beck
Dec 27 '18 at 22:35
$begingroup$
@N.Beck Yes, I am not sure it's entirely standard notation (It is, though, in that book). Glad to help!
$endgroup$
– Clement C.
Dec 27 '18 at 22:51
add a comment |
$begingroup$
Oh, ok . Thanks a lot. :) I thought $ mathscr B_0$ is the completion of the Borel $sigma$-algebra, as it is usual.
$endgroup$
– N.Beck
Dec 27 '18 at 22:35
$begingroup$
@N.Beck Yes, I am not sure it's entirely standard notation (It is, though, in that book). Glad to help!
$endgroup$
– Clement C.
Dec 27 '18 at 22:51
$begingroup$
Oh, ok . Thanks a lot. :) I thought $ mathscr B_0$ is the completion of the Borel $sigma$-algebra, as it is usual.
$endgroup$
– N.Beck
Dec 27 '18 at 22:35
$begingroup$
Oh, ok . Thanks a lot. :) I thought $ mathscr B_0$ is the completion of the Borel $sigma$-algebra, as it is usual.
$endgroup$
– N.Beck
Dec 27 '18 at 22:35
$begingroup$
@N.Beck Yes, I am not sure it's entirely standard notation (It is, though, in that book). Glad to help!
$endgroup$
– Clement C.
Dec 27 '18 at 22:51
$begingroup$
@N.Beck Yes, I am not sure it's entirely standard notation (It is, though, in that book). Glad to help!
$endgroup$
– Clement C.
Dec 27 '18 at 22:51
add a comment |
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