Is there a $\sigma$ algebra $\mathfrak{B}$ such that $\mathfrak{S}\subset\mathfrak{B}$ and $\bigcup_{A\in\mathfrak{S}}A\notin\mathfrak{B}$?

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, $$\displaystyle \bigcup_{A\in\mathfrak{S}}A\subset E$$.

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

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 $$\mathfrak B$$.

• Eike Schulte, Thank you very much for your answer. Apr 7, 2021 at 23:23

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 $$\mathcal B$$.

• 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. Apr 7, 2021 at 6:13
• the Borel sigma algebra is the sigma algebra generated by the postulated topology Apr 7, 2021 at 10:04
• Imaosome, Thank you very much for your comment. Apr 7, 2021 at 23:24