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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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    @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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    That too. But pickles...2012-11-04