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The borel algebra on the topological space R is defined as the σ-algebra generated by the open sets (or, equivalently, by the closed sets). Logically, I thought that since this includes all the open sets (a,b) where a and b are real numbers, then, this would be equivalent to the power set. For example, the set (0.001, 0.0231) would be included as well as (-12, 19029) correct? I can't think of any set that would not be included. However, I have read that the Borel σ-algebra is not, in general, the whole power set.

Can anyone give a gentle explanation as to why this is the case?

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    There are lots of subsets of $\mathbb{R}$. What makes you think that every subset of $\mathbb{R}$ can be written as a countable union of countable intersections of open and closed subsets?2012-10-20
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    @ZhenLin, well if the borel algebral included all open subsets then wouldn't that mean that every subset of R should be included? like (3,5) would be there, so would (4,10000) or (0.001, 0.0213) correct?2012-10-20
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    For a set which isn't even lebesgue measureable, check the [vitali set](https://en.wikipedia.org/wiki/Vitali_set). Its construction requires the axiom of choice.2012-10-20
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    And to see why it's difficult to give a completely explicit example of a set which isn't in the Borel $\sigma$-algebra, see this [MO answer](http://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice/32746#32746).2012-10-20
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    This does not answer your main question, but there is a whole big hierarchy of classes of sets of real numbers, each strictly bigger than the previous. First, there is the set of open sets, in this context usually written $\mathcal{G}$. Then there are the countable intersections of such, called $\mathcal{G}_{\delta}$ (which includes all closed sets). Then there is $\mathcal{G}_{\delta\sigma}$, defined as the countable intersection of $\mathcal{G}_{\delta}$ sets. Then there is $\mathcal{G}_{\delta\sigma\delta}$, $\mathcal{G}_{\delta\sigma\delta\sigma}$, …2012-10-20
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    Why should having all of the open sets give you all subsets of $\Bbb R$? Most subsets of $\Bbb R$ are neither open nor closed. Indeed, $\Bbb R$ has only $2^\omega$ open and closed subsets, but it has $2^{2^\omega}>2^\omega$ subsets.2012-10-20
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    For more on the hierarchy I mentioned (and beyond), see the MO answer referenced by @ZhenLin.2012-10-20
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    The set of Borel sets has the same cardinality as $\mathbb R$, but the cardinality of the powerset of $\mathbb R$ is strictly larger.2012-10-20
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    @K.Stm. so does the powerset of the reals include these non-measurable sets (like the Vitali set)?2012-10-20
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    @BrianM.Scott. Thanks, so so are you saying that clopen sets like (3, 32.33] would not be included in the borel algebra? Because I know that all open or closed sets are included (not sure about clopen sets though)2012-10-20
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    @BYS2 1. The power set of a given set $S$ contains every subset of $S$. Since the Vitali Set is a set of reals, it is a subset of the reals, so an element in the powerset of the reals. 2. A set of the form (0,1] is not _clopen_ (= closed and open), but neither closed nor open so especially and certainly not clopen. You may call them half-open, half-closed. Such sets are in the borel sigma algebra, since e.g. $(0,1] = (0,\infty) \cap [-\infty,1] = (0,\infty) \cap (1,\infty)^c$.2012-10-20
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    @BYS2: The set $(3,32.33]$ is not clopen; *clopen* means *closed and open*, and the only clopen subsets of $\Bbb R$ are $\varnothing$ and $\Bbb R$. All half-open intervals **are** Borel sets, so $(3,32.33]$ is indeed a Borel set.2012-10-20
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    @K.Stm. thank you very much for your help! I have limited background in this stuff and I appreciate your clarifications!2012-10-20
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    @BrianM.Scott, thanks for your reply! But what exactly did you mean by 'Most subsets of R are neither open nor closed' then? call me simplistic but most of the basic sets I learned about were either open, closed or half-open :S... like (3,4), [5,6] and the likes.. are you talking about non-measurable sets then?2012-10-20
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    @BYS2: I was simply counting. $\Bbb R$ has $2^{\mathfrak c}$ subsets, and only $\mathfrak c$ of them are Borel, so $\Bbb R$ has $2^{\mathfrak c}$ non-Borel subsets. Every non-measurable set is also non-Borel, but there are measurable non-Borel sets. There are [explicit constructions of non-Borel sets](http://en.wikipedia.org/wiki/Borel_algebra#Non-Borel_sets), though they’re rather complicated.2012-10-20
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    @BrianM.Scott, ok thanks a bunch!2012-10-20
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    @Brian: Even though these are explicit constructions, these sets are not provably Borel when the axiom of choice fails. In particular it is consistent without the axiom of choice that every set of real numbers is Borel.2012-10-22

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You can show that there are $\mathfrak{c} = 2^{\aleph_0}$ Borel subsets of the real line, and so by Cantor's Theorem ($|X| < | \mathcal{P} (X)|$) it follows that there are non-Borel subsets of $\mathbb{R}$.

To see that there are $\mathfrak{c}$-many Borel subsets of $\mathbb{R}$, we can proceed as follows:

  1. define $\Sigma_1^0$ to be the family of all open subsets of $\mathbb{R}$;
  2. for $0 < \alpha < \omega_1$ define $\Pi_\alpha^0$ to be the family of all complements of sets in $\Sigma_\alpha^0$ (so that $\Pi_1^0$ consists of all closed subsets of $\mathbb{R}$);
  3. for $1 < \alpha < \omega_1$ define $\Sigma_\alpha^0$ to be the family of all countable unions of sets in $\bigcup_{\xi < \alpha} \Pi_\xi^0$.

Then you can show that $B = \bigcup_{\alpha < \omega_1} \Sigma_\alpha^0 = \bigcup_{\alpha < \omega_1} \Pi_\alpha^0$ is the family of all Borel subsets of $\mathbb{R}$. Furthermore, transfinite induction will show that $| \Sigma_\alpha^0 | = \mathfrak{c}$ for all $\alpha < \omega_1$, which implies that $\mathfrak{c} \leq | B | \leq \aleph_1 \cdot \mathfrak{c} = \mathfrak{c}$.

Specific examples of non-Borel sets are in general difficult to describe. Perhaps the easiest to describe is a Vitali set, obtained by taking a representative from each equivalence class of the relation $x \sim y \Leftrightarrow x -y \in \mathbb{Q}$. Such a set is not Lebesgue measurable, and hence not Borel. Another example, due to Lusin, is given in Wikipedia.

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