Let on $\mathbb{R}^n$ be $\mathcal{L}$ - Lebesgue measurable sets, and $\mathcal{B}$ - Borel measurable sets. Why not $\mathcal{L} = \mathcal{B}$?

Question

It appears to me, at least because we are in $\mathbb{R}^n$, that they should be the same, and not - how we've been told in our lecture - that $\mathcal{B}(\mathbb{R}^n) \subset \mathcal{L}(\mathbb{R}^n)$

I mean, what should you do differently here? Sets in $\mathbb{R}^n$ which are Borel measurable are from the sigma algebra generated by the open sets in $\mathbb{R}^n$ which are of the form $(a_1, b_1) \times ... \times (a_n, b_n)$

So what difference does it make whether I choose a set from $\mathcal{L}$ or $\mathcal{B}$?

Well simply because there are sets is $\mathcal L$ that are not in $\mathcal B$, see this for example: https://metaphor.ethz.ch/x/2021/fs/401-2284-00L/sc/notes_exclass04.pdf — miguelsxvi, Oct 23 '21 at 10:00
There are some posts on this site about sets which are Lebesgue measurable but not Borel measurable: Lebesgue measurable set that is not a Borel measurable set. Also cardinality argument could be used: Cardinality of Borel sigma algebra. — Martin Sleziak, Oct 23 '21 at 10:01
That's clear, but I am talking specifically about sets on $\mathbb{R}^n$. Do you maybe know an example for that? Or is that example on the Cantor set the example I'm looking for? — anon, Oct 23 '21 at 10:16
The Lebesgue sigma algebra is obtained completing the Borrel one. Take any non Borrel set which is contained in a Lebesgue (measure) negligible set. — blamethelag, Oct 23 '21 at 10:24
"Take any non Borrel set which is contained in a Lebesgue (measure) negligible set." Yes, but like what for example? That's my question basically, I want to know which set (in $\mathbb{R}^n$) is Lebesgue measurable, but not Borel measurable — anon, Oct 23 '21 at 10:27
miguelsxvi generously gave you an exemple for $n = 1$. It means he offered you $A \subset \mathbb R$ not Borel such that there exists $N \in \mathcal B (\mathbb R) $ with $A \subset N$ and $\lambda(N) = 0$. If you want an example in dimension $n$ considere $A^n$ and use the definition of the product measure. — blamethelag, Oct 23 '21 at 10:35
Of strong related interest is Is there a $\sigma$-algebra on $\mathbb{R}$ strictly between the Borel and Lebesgue algebras? and this 5 June 2001 sci.math post. Regarding “explicit natural examples”, the following from the lower fourth of p. 756 of Continuity and Baire functions by Lorch (1971) indicates the situation at even almost the very beginning of the Borel hierarchy: “$\ldots$ the reader is reminded of the fact (continued) — Dave L. Renfro, Oct 23 '21 at 10:52
that sets which are of type $F_{\sigma \delta \sigma}$ or $G_{\delta \sigma \delta}$ and not of lower type—with respect to any of the classic topologies—are very thinly scattered through the literature. In fact, looking for them is almost like hunting for unicorns.” — Dave L. Renfro, Oct 23 '21 at 10:52
@blamethelag My tab in my browser crashes all the time if I try to open the link miguelsxvi sent me, and I also cannot download that PDF, too — anon, Oct 23 '21 at 10:54

Let on $\mathbb{R}^n$ be $\mathcal{L}$ - Lebesgue measurable sets, and $\mathcal{B}$ - Borel measurable sets. Why not $\mathcal{L} = \mathcal{B}$?

0 Answers0