Constructing sigma algebras in countably many steps

Question

I'm learning about $\sigma$-algebras and was interested when my textbook briefly mentioned the impossibility of constructing the Borel $\sigma$-algebra of $\mathbb{R}$, $\mathcal{B}(\mathbb{R})$, from the open intervals in countably many steps. More precisely, let $\mathcal{L}_0$ be the collection of all open intervals $(a,b)\subseteq\mathbb{R}$. Given $\mathcal{L}_i$, we define $$\mathcal{L}_{i+1}=\bigg\{\bigcup_{k\in\mathbb{N}}A_k, \Big(\bigcup_{k\in\mathbb{N}}A_k\Big)^{c}\ :\ A_k\in \mathcal{L}_{i} \bigg\}\ \ \ \text{ and } \ \ \ \hat{\mathcal{L}}=\bigcup_{i\in\mathbb{N}}\mathcal{L}_i$$

Then my textbook says that $\hat{\mathcal{L}}\subsetneqq\mathcal{B}(\mathbb{R})$.

My question is this: can one (easily) exhibit a Borel set not in $\hat{\mathcal{L}}$? Is it necessary to use heavy machinery like AC or CH in order to construct such a set? I'm quite stuck.

Many thanks!

"Easily"? I doubt it; seems llke whatever $E$ is, showing that $E$ has not appeared at some finite stage is not going to be "easy". You might find a proof that such an example exists by looking for information on "descriptive set theory" or more specifically "Baire classes". For example the external references at https://en.wikipedia.org/wiki/Baire_function — David C. Ullrich, Jan 16 '19 at 14:31
CH is not involved in this, nor in the higher countable levels of the Borel hierarchy. — DanielWainfleet, Jan 16 '19 at 21:23
For reference: I think the textbook that contains this statement is "Measures, Integrals and Martingales" by René Schilling, chapter 3. — Robert Hönig, Jun 11 '21 at 14:55

score 8 · Accepted Answer · answered Jan 16 '19 at 15:45

One can find proofs in many set theory texts for the stronger result that, for each countable ordinal level of the Borel hierarchy, there exist Borel sets not belonging to that level. Look in book indexes for "universal set". You can also google Borel + "universal set".

However, I've found very few published proofs that limit themselves to the finite levels situation that most measure theory texts state and which you have stated in your question. For this reason I’ve made note of such proofs when I’ve encountered them, and in case it could be of use to you or others, below are the only three such references that I currently know about.

[1] Patrick Paul Billingsley, Probability and Measure, 3rd edition, Wiley Series in Probability and Mathematical Statistics, John Wiley and Sons, 1995 [reprinted as “Anniversary Edition” in 2012], xiv + 593 pages.

On pp. 30-32 there is a detailed construction of a Borel set that does not belong to any of the finite Borel classes.

[2] Mikls Laczkovich, Conjecture and Proof, Classroom Resource Materials, Mathematical Association of America, 2001, x + 118 pages.

See pp. 98-101 and p. 105 and Exercise 17.7 on p. 106.

[3] Eric M. Vestrup, The Theory of Measures and Integration, Wiley Series in Probability and Statistics, Wiley-Interscience, 2003, xviii + 594 pages.

See Section 1.7 (pp. 28-34). Vestrup’s construction is an expanded version of the construction given in the 3rd edition of Billingsley’s book [1]. For the stronger version involving transfinite Borel classes, see Exercise 2.23 on p. 36 (and Notes on the Problems for 2.23, on p. 555).

I believe that the stronger version, that no countable level of the Borel hierarchy is all the real Borel sets, is due to Lebesgue or to Borel. — DanielWainfleet, Jan 16 '19 at 21:25
@DanielWainfleet: The result is due to Lebesgue, who proved this in a lengthy 1905 paper that is well known for having one of the most famous errors ever made in a mathematics paper. The error is famous not just because it was by someone like Lebesgue or the fact that it's rather elementary, but rather because it was probably the single most significant driving force behind the development of descriptive set theory in the 1920s and 1930s. Lebesgue's error is discussed in detail in this 29 July 2006 sci.math post archived at Math Forum. — Dave L. Renfro, Jan 17 '19 at 08:24
Relevant to my previous comment is Is the image of a $G_\delta$ set under a continuous mapping of $\mathbb R^n$ a Borel set AND Continuous images of open sets are Borel. Also relevant is Borel hierarchy doesn't "collapse" before $\omega_1$. — Dave L. Renfro, Jan 17 '19 at 08:35
I have heard of Lebesgue's famous error and that its detection led to the discovery of the class of projective sets, but I had not seen his "proof" before, nor its rebuttal. — DanielWainfleet, Jan 17 '19 at 13:55
Math Forum took down their archive of hundreds of thousands of math posts a year or two ago, so the 29 July 2006 sci.math post link in my earlier comment no longer works. The corresponding google sci.math posts are still available --- biography of Suslin and details of Lebesgue's error (both posted 29 July 2006). — Dave L. Renfro, Apr 06 '21 at 06:01

score -1 · Answer 2 · answered Jun 16 '19 at 21:05

I published the proof given in (Billingsley, Probability and Measure, 3rd edition, pages 30-32) and people sent me to this thread. I have questions to this proof. I do not understand it. Somebody, please help if still interested in the topic. Problem from the book Probability and Measure (by Patrick Billingsley), on constructing σ-Fields.

Constructing sigma algebras in countably many steps

2 Answers2

Linked