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

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 $A\in\mathfrak{B}$.

My question is the following:

Let $\mathfrak{S}$ be a non-empty set of sets.
Is there a $\sigma$ algebra $\mathfrak{B}$ such that $\mathfrak{S}\subset\mathfrak{B}$ and $\displaystyle \bigcup_{A\in\mathfrak{S}}A\notin\mathfrak{B}$?

Are there a non-empty set of sets $\mathfrak{S}$ and a $\sigma$ algebra $\mathfrak{B}$ such that $\mathfrak{S}\subset\mathfrak{B}$ and $\displaystyle \bigcup_{A\in\mathfrak{S}}A\notin\mathfrak{B}$?

By the way, if $\mathfrak{B}$ is a $\sigma$ algebra, there is an element $E\in\mathfrak{B}$ such that $A\subset E$ for any element $A\in\mathfrak{B}$.
So, $\displaystyle \bigcup_{A\in\mathfrak{S}}A\subset E$.

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    $\begingroup$ There are examples where the union belongs to $\mathcal B$ and examples wher it does not belong. What exactly is your question? $\endgroup$ Apr 7, 2021 at 5:15
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    $\begingroup$ Obviously the answer is no in general, for instance if $\mathfrak{S}$ has only one element. $\endgroup$ Apr 7, 2021 at 5:20
  • $\begingroup$ @KaviRamaMurthy Thank you very much for your comment. I edited my question, but I am afraid lest my question is still nonsense. $\endgroup$
    – tchappy ha
    Apr 7, 2021 at 5:58
  • $\begingroup$ @EricWofsey Thank you very much for your comment. $\endgroup$
    – tchappy ha
    Apr 7, 2021 at 5:59

2 Answers 2

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The precise definition of the Borel $\sigma$-algebra in Kavi Rama Murthy’s answer is not important. You only need a $\sigma$-algebra $\mathfrak B$ with maximal element $E$ such that

  • not all subsets of $E$ are elements of $\mathfrak B$ but
  • every one-element subset of $E$ is an element of $\mathfrak B$.

Pick any subset $A$ of $E$ that is not in $\mathfrak B$. The collection $\mathfrak S = \{ \{ x \} | x \in A\}$ has union $A$ and $\mathfrak S \subseteq \mathfrak B$, but $A$ is not an element of $\mathfrak B$.

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  • $\begingroup$ Eike Schulte, Thank you very much for your answer. $\endgroup$
    – tchappy ha
    Apr 7, 2021 at 23:23
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Let $\mathcal B$ be the Borel sigma algebra of $\mathbb R$ and $E$ be a set which is not a Borel set. Consider the collection of all singleton sets $\{x\}$ with $ x\in E$. The union of these does not belong to $\mathcal B$.

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  • $\begingroup$ I am very new to measure theory, so I don't know "Borel sigma algebra" now. But I believe your answer is the best answer for this question. $\endgroup$
    – tchappy ha
    Apr 7, 2021 at 6:13
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    $\begingroup$ the Borel sigma algebra is the sigma algebra generated by the postulated topology $\endgroup$
    – lmaosome
    Apr 7, 2021 at 10:04
  • $\begingroup$ Imaosome, Thank you very much for your comment. $\endgroup$
    – tchappy ha
    Apr 7, 2021 at 23:24

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