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Let $S=\{\{x\}\mid x \in \mathbb{R}\}$. Is $\sigma(S)$ included in the Borel $\sigma$-algebra on $\mathbb{R}$? Is $\sigma(S)$ equivalent to the Borel $\sigma$-algbrea on $\mathbb{R}$?

I think the answer to the first question is yes. Because the Borel sets include singletons, the Borel $\sigma$-algebra must contain the smallest sigma-algebra of the set of all singletons.

Here's my attempt at the second question: my answer would be no. Countable unions (or intersections) of countable sets are countable, so that any $X \in \sigma(S)$ is either countable or a compliment of a countable set. Therefore, the interval $(0,1) \not \in \sigma (S)$ but it is included in the Borel $\sigma$-algebra. My issue here is that I don't know how to formalize this statement, if it is even correct.

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2 Answers 2

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Let $$\mathscr C\equiv\{E\subseteq\mathbb R\,|\,E\text{ is countable or }E^c\text{ is countable}\}.$$ It is not difficult to check that $\mathscr C$ is a $\sigma$-algebra on $\mathbb R$. Let $\mathscr S$ denote the $\sigma$-algebra generated by singletons.

$\textbf{Proposition:}\phantom{---}$$\mathscr C=\mathscr S$.

$\textit{Proof:}\phantom{---}$ Clearly, $\{x\}\in\mathscr C$ for every $x\in\mathbb R$, so $\mathscr S\subseteq\mathscr C$, since $\mathscr C$ is a $\sigma$-algebra containing the singletons and $\mathscr S$ is, by definition, the smallest $\sigma$-algebra containing the singletons. Conversely, suppose that $E\in\mathscr C$. If $E$ is countable, then $E$ is a countable union of singletons. Therefore, $E\in\mathscr S$ (because $\mathscr S$ is $\sigma$-algebra containing the singletons). If $E^c$ is countable, then, by the same argument, $E^c\in\mathscr S$, from which it follows that $E\in\mathscr S$ (since $\sigma$-algebras are closed under complements). In conclusion, $\mathscr C\subseteq\mathscr S$. $\blacksquare$

Clearly, $(0,1)\notin\mathscr C=\mathscr S$, but $(0,1)\in\mathscr B_{\mathbb R}$. It follows that $\mathscr B_{\mathbb R}\neq\mathscr S$.

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  • $\begingroup$ Is it true that $\mathscr{S}$ is not a countably generated $\sigma$-field ? $\endgroup$ Oct 22, 2015 at 16:36
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    $\begingroup$ @FardadPouran It’s true, $\mathscr S$ is not countably generated. This result is an exercise in Billingsley’s Probability and Measure (third edition). See Problem 2.11(b) therein—hints are given at the back of the textbook. [The number is 2.10(b) according to the second edition.] $\endgroup$
    – triple_sec
    Oct 22, 2015 at 16:47
  • $\begingroup$ So neat and clean. $\endgroup$
    – ogirkar
    Mar 4, 2019 at 17:59
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    $\begingroup$ Interestingly enough, this doesn't hold for the $\sigma$-algebra generated by the class of all two point sets. For instance let $X=\{1,2\}$, then $\sigma(X)=\{X,\emptyset\}$ but $\mathscr{C}=\{\{1\},\{2\},X,\emptyset\}$ $\endgroup$ May 25, 2021 at 15:22
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    $\begingroup$ @pSrIoGcNeAsLs True, but this result seems to be an artifact specific to this example. If $X$ contains at least three distinct elements, then the $\sigma$-algebra generated by the class of all two-point sets and the $\sigma$-algebra generated by the singletons will be the same. $\endgroup$
    – triple_sec
    May 26, 2021 at 3:33
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Your proof for both parts is exactly correct as stated. There is no more formalization to be done. Just make sure to say that $( 0,1 )$ is uncountable.

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