Hi All: I finally found a proof of the Kolmogorov extension theorem that I've been able to follow for the most part ( it's at the link below ). But recently I got stuck. At the beginning of section 1.4, they start out with "Let $F$ be the set of all measurable cylinder sets in $R^{\infty}$. Then, a few sentences down, they state that $F$ is not a sigma algebra but $F_{n}$ is ? I didn't know to write the $F$ that they use but my question is why $F$ is not a sigma algebra but $F_{n}$ is. They seem like the same thing to me ? I realize that it's a lot to ask and the help is much appreciated. Thanks a lot.
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Why $\mathcal{F}_n$ is a sigma algebra.
- $\mathcal{F}_n$ contains $\mathbb{R}^n \times \mathbb{R} \times \mathbb{R} \times \cdots = \mathbb{R}^\infty$.
- If $(E \times \mathbb{R} \times \mathbb{R} \times \cdots) \in \mathbb{F}_n$, then its complement is $E^c \times \mathbb{R} \times \mathbb{R} \times \cdots$ which is also in $\mathbb{F}_n$.
- Similarly check that $\mathcal{F}_n$ is closed under countable unions by writing out members of $\mathcal {F}_n$ in the form $E \times \mathbb{R} \times \mathbb{R} \times \cdots$.
Why $\mathcal{F}$ is not a sigma algebra.
The text says
$\mathcal{F}$ is not a sigma algebra since it is not closed under countably infinite unions.
Consider $$\bigcup_{k=1}^\infty \left(\underbrace{\mathbb{R} \times \cdots \times \mathbb{R}}_{k-1} \times \{0\} \times \mathbb{R} \times \mathbb{R} \times \cdots\right).$$ This is the union of sets in $\mathcal{F}$, since it can be written as $$\bigcup_{k=1}^\infty \left(E_k \times \mathbb{R} \times \mathbb{R} \times \cdots \right), \text{ where $E_k := \underbrace{\mathbb{R} \times \cdots \times \mathbb{R}}_{k-1} \times \{0\} \subseteq \mathbb{R}^k$}.$$ But the union cannot be written as $E \times \mathbb{R} \times \mathbb{R} \times \cdots$ for some $E \in \mathbb{R}^n$.
angryavian
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Thank you very much. I will print out and go through it and then give a check. – mark leeds May 04 '22 at 22:46
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@angryvian: a couple of questions since the notation ( not yours ) confuses me. 1) in the why $F_{n}$ is a sigma algebra section, doesn't one need to take the complement of the R's also ? Why do you only take the complement of $E$ ? In the section Why $F$ is not a sigma algebra section, it stated that $F$ contains sets of the form $E \times R \times R \cdots$ with $E$ a Borel set in some $R^{n}$. Then you have $(k-1)$ R's inside the parentheses along with a null set. Why is there not an $E$ in the countable union expression ? Also, is $n$ equal to $(k-1)$ in your countable union expression. – mark leeds May 04 '22 at 23:13
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@markleeds For your first question, try drawing a picture in two dimensions of $A \times \mathbb{R}$ for some set $A$, and figure out what the complement is. For your second question, I updated my answer to show you why the sets in the union belong to $\mathbb{F}$. – angryavian May 05 '22 at 00:13
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okay. I'll try that. I think one of my ( many ) problems is visualizing these things. thanks. I'll also go through your updated answer. – mark leeds May 05 '22 at 01:04
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I see now why the complement is only taken for A and not R. If $X_{1}$ is [0,1] then $X_{1} \times R$ is a rectangle with base [0,1] and infinite height so the complement is everything else so it's $X_{1}^{c} \times R$. That makes sense. Thanks. – mark leeds May 05 '22 at 01:28
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angryavian: this is my last bother I promise. Could you give me either an analytical or intutitve explanation of the difference between $F$ and $F_{n}$. Maybe that will help because I still don't see it. Thanks. – mark leeds May 05 '22 at 01:41
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@markleeds Are you asking about the difference in terms of the definition? $\mathcal{F}_1 \subset \mathcal{F}_2 \subset \cdots$ is an increasing sequencee of sigma algebras. $\mathcal{F}$ is their union. If you are asking about the difference in terms of $\mathcal{F}_n$ being a sigma algebra but $\mathcal{F}$ not being one: in general the union of sigma algebras is not necessarily a sigma algebra. You can find other examples here. – angryavian May 05 '22 at 02:13
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Thanks angryavian: I'll check out the link. it sounds helpful. But what I was really asking was if you compare $F$ to say $F_{n}$ where $n= 38$ for example, what's the difference between them in an intuitive or visual sense ? – mark leeds May 05 '22 at 02:31
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I think that I'm close to getting it but not quite. Let me try to explain my confusion. $F$ is the collection of all cylinders in infinite dimensions so $R \times R \times R \times \cdots R$. What confuses me is $F_{n}$. If it's $R^{n} \times R \times R \times R \times \cdots R$, then what is the difference between $F_{n}$ and $F_{n+1}$. $F_{n+1}$ is, by definition $R^{n+1} \times R \times R \times \cdots R$ so, since the product measure is infinite, doesn't it ends up with the same dimension as $F_{n}$ so they are then identical rather than increasing ? – mark leeds May 05 '22 at 03:38
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@markleeds $F$ is a collection of sets; it is not $R \times R \times \cdots$. Similarly, $F_n$ is a collection of sets of the form $E \times R \times \cdots$ where $E \subseteq R^n$, but it is not $R^n \times R \times \cdots$. An example of a set in $F_2$ but not in $F_1$ is $[0,1] \times {3} \times R \times R \times \cdots$. – angryavian May 05 '22 at 14:26
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Oh, okay. The $F_{n}$ are sets of products with an infinite number of R's but the first n of its sets can be any set and not necessarily $R^n$. I really appreciate your help and learned a lot and I think I understand it now. – mark leeds May 05 '22 at 15:37
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Also, the link you pointed to in unions of sigma algebras was incredibly enlightening also. I forgot to thank you for that. thanks. – mark leeds May 05 '22 at 15:41