# Is there a $\sigma$-algebra on $\mathbb{R}$ strictly between the Borel and Lebesgue algebras?

So, after proving that $\mathfrak{B}(\mathbb{R})\subset \mathfrak{L}(\mathbb{R})$, I asked myself, and now asking you, is there a set $\mathfrak{S}(\mathbb{R})$, which satisfies:

$$\mathfrak{B}(\mathbb{R} )\subset \mathfrak{S}(\mathbb{R})\subset \mathfrak{L}(\mathbb{R})$$

($\mathfrak{B}(\mathbb{R})$ is the Borel's set on $\mathbb{R}$ ; $\mathfrak{L}(\mathbb{R})$ is family of sets which are Lebesgue's measurable - which is a $\sigma$ algebra.)

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I am confused, do you require $\frak G(\Bbb R)$ to have some special property? –  Asaf Karagila May 7 '12 at 21:12
@AsafKaragila: "special property?" What do you mean in that? –  Salech Alhasov May 7 '12 at 21:16
Should it be a $\sigma$-algebra? Just the fact that most sets are not Borel ensures that there are things between them. Do you want "special" things to be there too? –  Asaf Karagila May 7 '12 at 21:17
I have added [descriptive-set-theory] because it fits perfectly, maybe even better than [set-theory], but I was reluctant to remove the latter. If someone thinks it should be removed - please take matters into your own hands! –  Asaf Karagila May 7 '12 at 21:40

In ZFC the cardinality of $\frak B(\Bbb R)$ is $2^{\aleph_0}$ while the cardinality of $\frak L(\Bbb R)$ is $2^{2^{\aleph_0}}$. By that virtue alone there are plenty of other sets in between.

If you wish to take $\frak G(\Bbb R)$ to be some sort of a $\sigma$-algebra, and not just any set, let $\cal A\subseteq\frak L(\Bbb R)$ such that $|{\cal A}|<2^{2^{\aleph_0}}$, and consider $\frak G(\Bbb R)$ to be the $\sigma$-algebra generated by $(\frak B(\Bbb R)\cup\cal A)$. If we took $\cal A$ such that $\cal A\nsubseteq\frak B(\Bbb R)$ this would be a $\sigma$-algebra which strictly contains the Borel sets and is strictly contained in the Lebesgue measurable sets.

Edit:

Some time after Byron's comment below I realized that indeed this may not be accurate. I suddenly realized that I cannot be certain that $\frak L(\Bbb R)$ is not generated by a set of less than $2^{2^{\aleph_0}}$ many elements. It's not that bad, though. One can still bound it with some certainty:

We require that $|{\cal A}|^{\aleph_0}<2^{2^{\aleph_0}}$. If $|{\cal A}|\leq\frak c$ then this is indeed true, however one can come up with models where this need not be true for any subset of $\cal P(\Bbb R)$.

For what it's worth, I asked a question on MathOverflow some time ago, but did not receive any answer regarding this question yet. I still believe this is true, though.

There is a class of sets called analytic sets which are defined to be the continuous image of Borel sets (in fact $G_\delta$ sets are enough, but it turns out to be the same thing). It is a theorem that analytic sets properly contain the Borel sets and they are Lebesgue measurable.

Therefore the complement of analytic (co-analytic) sets are also Lebesgue measurable sets (they also contain all the Borel sets, and an amazing result is that a set is Borel if and only if it is both analytic and co-analytic.)

If you consider the $\sigma$-algebra generated by the union of all analytic and co-analytic sets, you will find yourself still inside the Lebesgue measurable universe, but with a strictly larger family of sets.

Further reading on how $\sigma$-algebras are born:

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You have to be a tad careful here. A continuous image of a closed set of reals is the countable union of compact sets, so it is Borel. I mentioned this here: mathoverflow.net/questions/23478/… It is a natural confusion, as we tend to exchange the Euclidean set of reals with all its other set theoretic manifestations. But taking continuous images of complements of continuous images of closed sets gives you all analytic sets. –  Andres Caicedo May 7 '12 at 21:46
@Andres: So taking continuous image of all Borel gets me covered, or do I go beyond the analytic sets that way? –  Asaf Karagila May 7 '12 at 21:47
Borel is fine: All continuous images of closed sets are $F_\sigma$, their complements are $G_\delta$, and $G_\delta$ already suffices. And you do not overspill, you get precisely the boldface $\Sigma^1_1$ sets whether you take continuous images of $G_\delta$ sets, or of all Borel sets, or of all boldface $\Sigma^1_1$ sets. –  Andres Caicedo May 7 '12 at 21:52
@Andres: Thank you very much for this correct and useful information for future reference! –  Asaf Karagila May 7 '12 at 21:53
Related threads: math.stackexchange.com/q/20421/5363 and math.stackexchange.com/q/18702/5363, math.stackexchange.com/questions/137277 and links therein, for example –  t.b. May 8 '12 at 3:27

The class of Borel sets has the same size as the set of reals. The class of Lebesgue measurable sets has the same size as the power set of the reals.

To see this: Any subset of the Cantor set is Lebesgue measurable, with measure 0. Since the Cantor set has the same size as the reals, this shows that there are at least $|{\mathcal P}({\mathbb R})|$ many Lebesgue measurable sets, but of course there can be no more.

On the other hand, the Borel sets can be seen as the product of a transfinite construction of length $\aleph_1$, the first uncountable cardinal. Note that $\aleph_1\le|{\mathbb R}|$. In this construction, we start with the open sets (there are $|{\mathbb R}|$ many such sets), and at each stage add complements, and countable unions of sets from the previous stage. At limits, we take the union of the stages so far considered. None of these stages increases the cardinality of the class of sets collected so far, since $|{\mathbb R}|=|{\mathbb R}|^{\aleph_0}$.

Suppose now that $\Sigma$ is a $\sigma$-algebra of sets of reals, and $|\Sigma|^{\aleph_0}=|\Sigma|$. Let $A$ be any set of reals not in $\Sigma$, and let $\Lambda$ be the $\sigma$-algebra generated by $\{A\}\cup\Sigma$. Then $|\Lambda|=|\Sigma|$.

This shows that there are many $\sigma$-algebras in between the Borel sets and the Lebesgue measurable sets; in fact, you can create a very long chain of $\sigma$-algebras in the Lebesgue sets, $$\Sigma_0\subsetneq \Sigma_1\subsetneq\dots\subsetneq\Sigma_\alpha\subsetneq\dots,$$ with $\Sigma_0$ the algebra of Borel sets: For $\alpha<|{\mathbb R}|^+$, the algebras all have the same size as the reals. You may go longer or stop when you reach $|{\mathbb R}|^+$, depending on the cardinality of the power set of the reals.

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I will race you: who can write about projective sets faster? :-) –  Asaf Karagila May 7 '12 at 21:30
Well, we could also talk about definable sets of reals, etc. But I fear this may all be too technical to the original poster. –  Andres Caicedo May 7 '12 at 21:36
I guess you right. I tried to give a simple example within the confines of ZFC! –  Asaf Karagila May 7 '12 at 21:36

A less sophisticated, but potentially useful answer: You know that the intersection of $\sigma$-algebras is again a $\sigma$-algebra. If you start with $\mathfrak{B}(\mathbb{R})$ you can put some probability measure on it and take the completion under this probability measure. Doing this for every probability measure, you get a large family of $\sigma$-algebras extending $\mathfrak{B}(\mathbb{R})$. Their intersection, which you might call $\mathfrak{U}(\mathbb{R})$ is the $\sigma$-algebra of universally measurable sets. The analytic sets mentioned above are all universally measurable. This is an extremely useful fact.

In some applications such as statistics, you want t work with a whole family of measures on a $\sigma$-algebra. Now by construction, you can extend every probability measure on $\mathfrak{B}(\mathbb{R})$ in a unique way to all of $\mathfrak{U}(\mathbb{R})$. If $A\subseteq\mathfrak{B}(\mathbb{R}^2)$, the projection on the first coordinate will in general not be in $\mathfrak{B}(\mathbb{R})$, but it will be analytic and hence in $\mathfrak{U}(\mathbb{R})$. It is possible to rewrite optimization problems in terms of such projections. So let $f:A\to\mathbb{R}$ be a measurable function. Define a function $m:\mathbb{R}\to\mathbb{R}$ by $m(r)=\inf\{t:(r,t)\in A\}$. The infimum is understood to be in the extended real numbers so that $m(r)=\infty$ if there i no $t$ such that $(r,t)\in A$. In general, $m$ will not be $\mathfrak{B}(\mathbb{R})$-measurable, but $\mathfrak{U}(\mathbb{R})$-measurable. To see the latter, we note that we can rewrite $\{r:m(r)<\alpha\}$ as the projection of $A\cap\mathbb{R}\times(-\infty,\alpha)$ onto the first coordinate, a set that is analytic and hence in $\mathfrak{U}(\mathbb{R})$. This idea has very important applications. In stochastic dynamic programming, it can be that the value function is not $\mathfrak{B}(\mathbb{R})$-measurable, but $\mathfrak{U}(\mathbb{R})$-measurable for essentially the reason I just gave.

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Beginning in the mid 1920s several (mostly) Russian mathematicians studied quite a number of $\aleph_{1}$-length hierarchies of sigma-algebras that fit between the analytic sets and the lowest projective set levels. I researched this literature fairly extensively about 5 years ago, but I've forgotten the details and don't have my findings available with me right now. It also seems I never wrote a sci.math essay about this topic, at least I can't find one. However, the following suggestions should allow you (and others interested) to begin looking into this topic, and when I get the chance (it may be a while), I'll add a lot more detail.

A useful survey paper is

Vladimir G Kanovei, Kolmogorov's ideas in the theory of operations on sets, Russian Mathematical Surveys 43 #6 (1988), 93-128.

Unfortunately, this paper does not seem to be freely available on the internet. However, 3 useful papers published in 1983 in Fundamenta Mathematica, by John P. Burgess and titled Classical hierarchies from a modern standpoint (Part I, Part II, Part III), are all freely available on the internet.

Finally, also worth searching for are the C sets of E. Selivanovskii (search also using the spelling "Selivanovskij"), sets that were first introduced and studied by Selivanovskii in a paper (written in Russian) published in pp. 379-413 of Matematiceskij Sbornik 35 (1928).

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I will look for those! Dave, thank you very much. –  Salech Alhasov May 8 '12 at 17:58