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I am reading a famous book by Kolmogorov and Fomin (4th Edition, translated from Russian to Japanese).

I asked the following question:
Is $\displaystyle \bigcap_{\mathfrak{B}\in\Sigma}\mathfrak{B}$ a $\sigma$ algebra? Is there always really the maximum element $E$?

In this book, the definition of $\sigma$ algebra is the following:

Definition:
Let $\mathfrak{B}$ be a non-empty set of sets.
$\mathfrak{B}$ is called a $\sigma$ algebra when $\mathfrak{B}$ satisfies the following conditions:

  • If $A\in\mathfrak{B}, B\in\mathfrak{B}$, then $A\triangle B\in\mathfrak{B}, A\cap B\in\mathfrak{B}$.
  • If $A_1,A_2,\dots,A_n,\dots$ are elements of $\mathfrak{B}$, then $\displaystyle \bigcup_{n=1}^{\infty} A_n\in\mathfrak{B}$.
  • There is an element $E\in\mathfrak{B}$ such that $A\cap E=A$ for any element $A\in\mathfrak{B}$.

I doubt the definition of $\sigma$ algebra in this book is different from the standard definition of $\sigma$ algebra e.g. on Wikipedia.

Let $\mathfrak{B_1}$ and $\mathfrak{B_2}$ be two $\sigma$-algebras.
Maybe the maximum element $E_1$ for $\mathfrak{B_1}$ is not equal to the the maximum element $E_2$ for $\mathfrak{B_2}$.

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    $\begingroup$ Normally, one defines a $\sigma$-algebra on some set $X$, i.e. all elements of the $\sigma$-algebra are subsets of $X$ and $X$ is an element of the $\sigma$-algebra. Your definition avoids choosing $X$ first; however, the $E$ in the last condition plays that role (note that $A \cap E = A$ is equivalent to $A \subseteq E$, so all elements of $\mathfrak B$ are subsets of $E$). $\endgroup$ Apr 7, 2021 at 7:18
  • $\begingroup$ @EikeSchulte Thank you very much for your comment. $\endgroup$
    – tchappy ha
    Apr 7, 2021 at 8:10

2 Answers 2

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A set $\mathfrak B$ satisfies the definition in your book if and only if $\mathfrak B$ is a $\sigma$-algebra (in the wikipedia sense) on $\cup\mathfrak B$: for the noteworthy algebraic detail, notice that $A\setminus B=A\triangle (A\cap B)$. This is also equivalent to the property of $\mathfrak B$ being a $\sigma$-algebra in the wikipedia sense over some set. So yes, the set on which $\mathfrak B$ is a wikipedia-$\sigma$-algebra is not given beforehand, but it's uniquely determined by $\mathfrak B$. It's a slight change of perspective, which is reminiscent of what some authors do in elementary set theory with functions: they define a (partial) function as a set $f$ containing only pairs and such that yada-yada-yada, and then they recover the domain of $f$ as the class $\{a\,:\,\exists b, (a,b)\in f\}$, which turns out to be a set because reasons.

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  • $\begingroup$ Thank you very much for your answer, Gae. S. $\endgroup$
    – tchappy ha
    Apr 7, 2021 at 8:10
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The quoted definiton is like giving each $\sigma$-algebra its own "universe" of discourse. It states that $A \subseteq E$ for all $A$ in it. Probably the reason why it doesn't mention complements either (normally these are taken wrt the given universe..). So the quoted definition even for a fixed universe, is not equivalent to the standard one, it seems as complements need not be included.

And I don't think that $\bigcap_{\mathcal{B} \in \Sigma} \mathcal{B}$ actually needs to be a $\sigma$-algebra in their sense. Do they claim it to be?

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  • $\begingroup$ Thank you very much Henno Brandsma. The authors don't claim $\cap_{\mathfrak{B}\in\Sigma} \mathfrak{B}$ is a $\sigma$-algebra in their book. $\endgroup$
    – tchappy ha
    Apr 7, 2021 at 8:09
  • $\begingroup$ Henno Brandsma, I want an example for which $\cap_{\mathfrak{B}\in\Sigma}\mathfrak{B}$ is not a $\sigma$ algebra. $\endgroup$
    – tchappy ha
    Apr 7, 2021 at 8:13
  • $\begingroup$ I want an example for which $\cap_{\mathfrak{B}\in\Sigma} \mathfrak{B}$ doesn't have a maximum element. $\endgroup$
    – tchappy ha
    Apr 7, 2021 at 9:06
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    $\begingroup$ We can show that the quoted definition with set $E$ is equivalent to standard definition of $\sigma$-algebras on the set $\Omega$, by choosing $E=\Omega$. The 'countable union'-part is equal in both, as well as the 'containing the universe'-part by choosing $E=\Omega$. It remains to show that under the other two properties in the definition, the stability under complements is eqvalent to the stability under intersection and symmetric difference. This can be done as for two sets $A,B$ holds $B^{\complement A}=A\setminus B=A\triangle(A\cap B)$ and $A\triangle B=(A\setminus B)\cup(B\setminus A)$. $\endgroup$
    – mag
    Apr 7, 2021 at 11:38

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