Borel $sigma$-algebra on $mathbb{R}^n$
I'm reading Shiryaev's Probability Theory, and I am unable to understand the proof of a result concerning the Borel $sigma$-algebra on $mathbb{R}^n$.
He defines the Borel $sigma$-algebra $mathcal B(mathbb{R}^n)$ on $mathbb{R}^n$ as the smallest $sigma$-algebra containing the collection $mathcal{S}$ of all rectangles
$$ S = S_1 times ldots times S_n, quad S_k = (a_k, b_k]. $$
He then claims and proceeds to prove that
$$ mathcal{B}(mathbb{R}^n) = mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R}), $$
where $mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R})$ denotes the smallest $sigma$-algebra generated by the Borel rectangles
$$ B = B_1 times ldots B_n, quad B_k in mathcal{B}(mathbb{R}), $$
i.e.
$$ mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R}) = sigma(mathcal{B}(mathbb{R}) times ldots times mathcal{B}(mathbb{R})) $$
I don't understand the line of reasoning in the proof.
Before continuing, I would like to point out that I found some posts about this on MSE, namely [1] and [2], both of which provide quite nice arguments for this proof.
However, I would like to understand how the proof in the book works.
Shiryaev only proves it for $n=2$, which clearly suffices by induction.
First off, it is trivial that
$$ mathcal B(mathbb{R}^2) subset mathcal B(mathbb{R}) otimes B(mathbb{R}), $$
since every rectangle is also a Borel rectangle.
The converse is what confuses me.
He denotes
$$ tilde{mathcal{B}}_1 = mathcal{B}_1 times mathbb{R}, quad tilde{mathcal{B}}_2 = mathbb{R} times mathcal{B}_2, $$
and
$$ tilde{mathcal{S}}_1 = mathcal{S}_1 times mathbb{R}, quad tilde{mathcal{S}}_2 = mathbb{R} times mathcal{S}_2, $$
where $mathcal{S}_{1}$ and $mathcal{S}_{2}$ are the systems of half-open intervals that form the sides of the rectangles from before.
He then claims that
for all $B_1 times B_2 in mathcal{B}(mathbb{R}) times mathcal{B}(mathbb{R})$, we have
begin{align}
B_1 times B_2 = tilde B_1 cap tilde B_2 in tilde{mathcal{B}}_1 cap tilde{mathcal{B}}_2,
end{align}
where $tilde B_1 = B_1 times mathbb{R}$ and $tilde B_2 = mathbb{R} times B_2$,
which I understand. However, then he proceeds by saying
$$ tilde{mathcal{B}}_1 cap tilde{mathcal{B}}_2 color{red}{=} sigma(tilde{mathcal{S}}_1) cap tilde B_2 = sigma(tilde{mathcal{S}}_1 cap tilde B_2)
color{red}{subset} sigma(tilde{mathcal{S}}_1 cap tilde{mathcal{S}}_2) = sigma(mathcal{S}_1 times mathcal{S}_2) = mathcal{B}(mathbb{R}^2). $$
The steps marked in red are the ones that I cannot wrap my head around. Could someone clarify this for me?
probability-theory
asked Jan 4 at 14:18
MisterRiemannMisterRiemann
5,8841624
|
show 7 more comments
I'm reading Shiryaev's Probability Theory, and I am unable to understand the proof of a result concerning the Borel $sigma$-algebra on $mathbb{R}^n$.
He defines the Borel $sigma$-algebra $mathcal B(mathbb{R}^n)$ on $mathbb{R}^n$ as the smallest $sigma$-algebra containing the collection $mathcal{S}$ of all rectangles
$$ S = S_1 times ldots times S_n, quad S_k = (a_k, b_k]. $$
He then claims and proceeds to prove that
$$ mathcal{B}(mathbb{R}^n) = mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R}), $$
where $mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R})$ denotes the smallest $sigma$-algebra generated by the Borel rectangles
$$ B = B_1 times ldots B_n, quad B_k in mathcal{B}(mathbb{R}), $$
i.e.
$$ mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R}) = sigma(mathcal{B}(mathbb{R}) times ldots times mathcal{B}(mathbb{R})) $$
I don't understand the line of reasoning in the proof.
Before continuing, I would like to point out that I found some posts about this on MSE, namely [1] and [2], both of which provide quite nice arguments for this proof.
However, I would like to understand how the proof in the book works.
Shiryaev only proves it for $n=2$, which clearly suffices by induction.
First off, it is trivial that
$$ mathcal B(mathbb{R}^2) subset mathcal B(mathbb{R}) otimes B(mathbb{R}), $$
since every rectangle is also a Borel rectangle.
The converse is what confuses me.
He denotes
$$ tilde{mathcal{B}}_1 = mathcal{B}_1 times mathbb{R}, quad tilde{mathcal{B}}_2 = mathbb{R} times mathcal{B}_2, $$
and
$$ tilde{mathcal{S}}_1 = mathcal{S}_1 times mathbb{R}, quad tilde{mathcal{S}}_2 = mathbb{R} times mathcal{S}_2, $$
where $mathcal{S}_{1}$ and $mathcal{S}_{2}$ are the systems of half-open intervals that form the sides of the rectangles from before.
He then claims that
for all $B_1 times B_2 in mathcal{B}(mathbb{R}) times mathcal{B}(mathbb{R})$, we have
begin{align}
B_1 times B_2 = tilde B_1 cap tilde B_2 in tilde{mathcal{B}}_1 cap tilde{mathcal{B}}_2,
end{align}
where $tilde B_1 = B_1 times mathbb{R}$ and $tilde B_2 = mathbb{R} times B_2$,
which I understand. However, then he proceeds by saying
$$ tilde{mathcal{B}}_1 cap tilde{mathcal{B}}_2 color{red}{=} sigma(tilde{mathcal{S}}_1) cap tilde B_2 = sigma(tilde{mathcal{S}}_1 cap tilde B_2)
color{red}{subset} sigma(tilde{mathcal{S}}_1 cap tilde{mathcal{S}}_2) = sigma(mathcal{S}_1 times mathcal{S}_2) = mathcal{B}(mathbb{R}^2). $$
The steps marked in red are the ones that I cannot wrap my head around. Could someone clarify this for me?
probability-theory
asked Jan 4 at 14:18
MisterRiemannMisterRiemann
5,8841624
1
On the first mark: do you agree with/understand $tilde{mathcal B}_1=sigma(tilde{mathcal S}_1)$?
– drhab
Jan 4 at 14:39
@drhab Yes, I can see that being true.
– MisterRiemann
Jan 4 at 14:44
1
It's a typo. The $B$ should be $mathcal B$. This proof seems overly laborious especially for the $n=2$ case.
– Matematleta
Jan 4 at 16:25
1
Thank you. The crux of this is to show that $left { Atimes mathbb R :Ain mathscr B(mathbb R)right }subseteq mathscr B(mathbb R^{2})$. You can do this by using projections, or just by checking directly, which is what Sirayev is doing I guess.
– Matematleta
Jan 4 at 17:14
1
@MisterRiemann As you say: every topological space goes along with a Borel $sigma$-algebra. I think we are on the same line here.
– drhab
2 days ago
|
show 7 more comments
I'm reading Shiryaev's Probability Theory, and I am unable to understand the proof of a result concerning the Borel $sigma$-algebra on $mathbb{R}^n$.
He defines the Borel $sigma$-algebra $mathcal B(mathbb{R}^n)$ on $mathbb{R}^n$ as the smallest $sigma$-algebra containing the collection $mathcal{S}$ of all rectangles
$$ S = S_1 times ldots times S_n, quad S_k = (a_k, b_k]. $$
He then claims and proceeds to prove that
$$ mathcal{B}(mathbb{R}^n) = mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R}), $$
where $mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R})$ denotes the smallest $sigma$-algebra generated by the Borel rectangles
$$ B = B_1 times ldots B_n, quad B_k in mathcal{B}(mathbb{R}), $$
i.e.
$$ mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R}) = sigma(mathcal{B}(mathbb{R}) times ldots times mathcal{B}(mathbb{R})) $$
I don't understand the line of reasoning in the proof.
Before continuing, I would like to point out that I found some posts about this on MSE, namely [1] and [2], both of which provide quite nice arguments for this proof.
However, I would like to understand how the proof in the book works.
Shiryaev only proves it for $n=2$, which clearly suffices by induction.
First off, it is trivial that
$$ mathcal B(mathbb{R}^2) subset mathcal B(mathbb{R}) otimes B(mathbb{R}), $$
since every rectangle is also a Borel rectangle.
The converse is what confuses me.
He denotes
$$ tilde{mathcal{B}}_1 = mathcal{B}_1 times mathbb{R}, quad tilde{mathcal{B}}_2 = mathbb{R} times mathcal{B}_2, $$
and
$$ tilde{mathcal{S}}_1 = mathcal{S}_1 times mathbb{R}, quad tilde{mathcal{S}}_2 = mathbb{R} times mathcal{S}_2, $$
where $mathcal{S}_{1}$ and $mathcal{S}_{2}$ are the systems of half-open intervals that form the sides of the rectangles from before.
He then claims that
for all $B_1 times B_2 in mathcal{B}(mathbb{R}) times mathcal{B}(mathbb{R})$, we have
begin{align}
B_1 times B_2 = tilde B_1 cap tilde B_2 in tilde{mathcal{B}}_1 cap tilde{mathcal{B}}_2,
end{align}
where $tilde B_1 = B_1 times mathbb{R}$ and $tilde B_2 = mathbb{R} times B_2$,
which I understand. However, then he proceeds by saying
$$ tilde{mathcal{B}}_1 cap tilde{mathcal{B}}_2 color{red}{=} sigma(tilde{mathcal{S}}_1) cap tilde B_2 = sigma(tilde{mathcal{S}}_1 cap tilde B_2)
color{red}{subset} sigma(tilde{mathcal{S}}_1 cap tilde{mathcal{S}}_2) = sigma(mathcal{S}_1 times mathcal{S}_2) = mathcal{B}(mathbb{R}^2). $$
The steps marked in red are the ones that I cannot wrap my head around. Could someone clarify this for me?
probability-theory
asked Jan 4 at 14:18
MisterRiemannMisterRiemann
5,8841624
I'm reading Shiryaev's Probability Theory, and I am unable to understand the proof of a result concerning the Borel $sigma$-algebra on $mathbb{R}^n$.
He defines the Borel $sigma$-algebra $mathcal B(mathbb{R}^n)$ on $mathbb{R}^n$ as the smallest $sigma$-algebra containing the collection $mathcal{S}$ of all rectangles
$$ S = S_1 times ldots times S_n, quad S_k = (a_k, b_k]. $$
He then claims and proceeds to prove that
$$ mathcal{B}(mathbb{R}^n) = mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R}), $$
where $mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R})$ denotes the smallest $sigma$-algebra generated by the Borel rectangles
$$ B = B_1 times ldots B_n, quad B_k in mathcal{B}(mathbb{R}), $$
i.e.
$$ mathcal{B}(mathbb{R}) otimes ldots otimes mathcal{B}(mathbb{R}) = sigma(mathcal{B}(mathbb{R}) times ldots times mathcal{B}(mathbb{R})) $$
I don't understand the line of reasoning in the proof.
Before continuing, I would like to point out that I found some posts about this on MSE, namely [1] and [2], both of which provide quite nice arguments for this proof.
However, I would like to understand how the proof in the book works.
Shiryaev only proves it for $n=2$, which clearly suffices by induction.
First off, it is trivial that
$$ mathcal B(mathbb{R}^2) subset mathcal B(mathbb{R}) otimes B(mathbb{R}), $$
since every rectangle is also a Borel rectangle.
The converse is what confuses me.
He denotes
$$ tilde{mathcal{B}}_1 = mathcal{B}_1 times mathbb{R}, quad tilde{mathcal{B}}_2 = mathbb{R} times mathcal{B}_2, $$
and
$$ tilde{mathcal{S}}_1 = mathcal{S}_1 times mathbb{R}, quad tilde{mathcal{S}}_2 = mathbb{R} times mathcal{S}_2, $$
where $mathcal{S}_{1}$ and $mathcal{S}_{2}$ are the systems of half-open intervals that form the sides of the rectangles from before.
He then claims that
for all $B_1 times B_2 in mathcal{B}(mathbb{R}) times mathcal{B}(mathbb{R})$, we have
begin{align}
B_1 times B_2 = tilde B_1 cap tilde B_2 in tilde{mathcal{B}}_1 cap tilde{mathcal{B}}_2,
end{align}
where $tilde B_1 = B_1 times mathbb{R}$ and $tilde B_2 = mathbb{R} times B_2$,
which I understand. However, then he proceeds by saying
$$ tilde{mathcal{B}}_1 cap tilde{mathcal{B}}_2 color{red}{=} sigma(tilde{mathcal{S}}_1) cap tilde B_2 = sigma(tilde{mathcal{S}}_1 cap tilde B_2)
color{red}{subset} sigma(tilde{mathcal{S}}_1 cap tilde{mathcal{S}}_2) = sigma(mathcal{S}_1 times mathcal{S}_2) = mathcal{B}(mathbb{R}^2). $$
The steps marked in red are the ones that I cannot wrap my head around. Could someone clarify this for me?
probability-theory
probability-theory
asked Jan 4 at 14:18
MisterRiemannMisterRiemann
5,8841624
asked Jan 4 at 14:18
MisterRiemannMisterRiemann
5,8841624
asked Jan 4 at 14:18
MisterRiemannMisterRiemann
5,8841624
asked Jan 4 at 14:18
MisterRiemannMisterRiemann
5,8841624
asked Jan 4 at 14:18
MisterRiemannMisterRiemann
5,8841624
5,8841624
1
On the first mark: do you agree with/understand $tilde{mathcal B}_1=sigma(tilde{mathcal S}_1)$?
– drhab
Jan 4 at 14:39
@drhab Yes, I can see that being true.
– MisterRiemann
Jan 4 at 14:44
1
It's a typo. The $B$ should be $mathcal B$. This proof seems overly laborious especially for the $n=2$ case.
– Matematleta
Jan 4 at 16:25
1
Thank you. The crux of this is to show that $left { Atimes mathbb R :Ain mathscr B(mathbb R)right }subseteq mathscr B(mathbb R^{2})$. You can do this by using projections, or just by checking directly, which is what Sirayev is doing I guess.
– Matematleta
Jan 4 at 17:14
1
@MisterRiemann As you say: every topological space goes along with a Borel $sigma$-algebra. I think we are on the same line here.
– drhab
2 days ago
|
show 7 more comments
1
On the first mark: do you agree with/understand $tilde{mathcal B}_1=sigma(tilde{mathcal S}_1)$?
– drhab
Jan 4 at 14:39
@drhab Yes, I can see that being true.
– MisterRiemann
Jan 4 at 14:44
1
It's a typo. The $B$ should be $mathcal B$. This proof seems overly laborious especially for the $n=2$ case.
– Matematleta
Jan 4 at 16:25
1
Thank you. The crux of this is to show that $left { Atimes mathbb R :Ain mathscr B(mathbb R)right }subseteq mathscr B(mathbb R^{2})$. You can do this by using projections, or just by checking directly, which is what Sirayev is doing I guess.
– Matematleta
Jan 4 at 17:14
1
@MisterRiemann As you say: every topological space goes along with a Borel $sigma$-algebra. I think we are on the same line here.
– drhab
2 days ago
1
1
On the first mark: do you agree with/understand $tilde{mathcal B}_1=sigma(tilde{mathcal S}_1)$?
– drhab
Jan 4 at 14:39
On the first mark: do you agree with/understand $tilde{mathcal B}_1=sigma(tilde{mathcal S}_1)$?
– drhab
Jan 4 at 14:39
@drhab Yes, I can see that being true.
– MisterRiemann
Jan 4 at 14:44
@drhab Yes, I can see that being true.
– MisterRiemann
Jan 4 at 14:44
1
1
It's a typo. The $B$ should be $mathcal B$. This proof seems overly laborious especially for the $n=2$ case.
– Matematleta
Jan 4 at 16:25
It's a typo. The $B$ should be $mathcal B$. This proof seems overly laborious especially for the $n=2$ case.
– Matematleta
Jan 4 at 16:25
1
1
Thank you. The crux of this is to show that $left { Atimes mathbb R :Ain mathscr B(mathbb R)right }subseteq mathscr B(mathbb R^{2})$. You can do this by using projections, or just by checking directly, which is what Sirayev is doing I guess.
– Matematleta
Jan 4 at 17:14
Thank you. The crux of this is to show that $left { Atimes mathbb R :Ain mathscr B(mathbb R)right }subseteq mathscr B(mathbb R^{2})$. You can do this by using projections, or just by checking directly, which is what Sirayev is doing I guess.
– Matematleta
Jan 4 at 17:14
1
1
@MisterRiemann As you say: every topological space goes along with a Borel $sigma$-algebra. I think we are on the same line here.
– drhab
2 days ago
@MisterRiemann As you say: every topological space goes along with a Borel $sigma$-algebra. I think we are on the same line here.
– drhab
2 days ago
|
show 7 more comments
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1
On the first mark: do you agree with/understand $tilde{mathcal B}_1=sigma(tilde{mathcal S}_1)$?
– drhab
Jan 4 at 14:39
@drhab Yes, I can see that being true.
– MisterRiemann
Jan 4 at 14:44
1
It's a typo. The $B$ should be $mathcal B$. This proof seems overly laborious especially for the $n=2$ case.
– Matematleta
Jan 4 at 16:25
1
Thank you. The crux of this is to show that $left { Atimes mathbb R :Ain mathscr B(mathbb R)right }subseteq mathscr B(mathbb R^{2})$. You can do this by using projections, or just by checking directly, which is what Sirayev is doing I guess.
– Matematleta
Jan 4 at 17:14
1
@MisterRiemann As you say: every topological space goes along with a Borel $sigma$-algebra. I think we are on the same line here.
– drhab
2 days ago