Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $B$ is the Boolean algebra of all Lebesgue measurable sets in $I=[0,1]$ modulo Null sets.

I am asking

(1) What will be the cardinality of $B$. Does it have to be $|B|=\mathfrak c$.

(2) Is there any $b\in B$ which is not Borel set.

share|cite|improve this question
up vote 5 down vote accepted

The cardinality is $\mathfrak c$: First of all, it is at least $\mathfrak c$, as the sets $(0,x)$ are all inequivalent for different values of $x\in[0,1]$. Second, Lebesgue measure is regular, so any measurable set contains a $\sigma$-compact subset of the same measure, and is contained in a $G_\delta$ superset of the same measure. This shows that each equivalence class has a Borel representative. But there are only $\mathfrak c$ Borel sets.

share|cite|improve this answer
thanks for the answer. As I understand that there is no Lebesgue measurable set of positive measure that is not Borel. Am I correct? – Xavyar May 16 '14 at 14:00
Not quite. A simple counterexample is to take a non-Borel subset of $[0,1]$ of measure $0$ (which exists, since any subset of the Cantor set has measure $0$), and consider its union with the interval $[1,2]$. What is true is that any measurable set is the union of a Borel set and a (not necessarily Borel) measure zero set. – Andrés E. Caicedo May 16 '14 at 14:03
Excellent answer. Thanks for your time. – Xavyar May 16 '14 at 14:09

Yes given $X$ Lebesgue measurable take the intersection of a countable sequence of open sets $O_n \supseteq X$ such that $m(O_n) \leq m(X) + \frac{1}{n}$. Then $m(\bigcap O_n-X)=0$ and $X\equiv \bigcap O_n$ in $B$. This means that any equivalence class in $B$ contains a $G_{\delta}$ set.

share|cite|improve this answer
You want to consider the intersection of the open sets $O_n$, not the union, right? And in any case I don't think you want to assert that it is equal to $X$. – Trevor Wilson May 16 '14 at 0:54
I think it reads better now. – Rene Schipperus May 16 '14 at 1:06
One more thing: I think you want to say $m(\bigcap O_n - X) = 0$ and not the other way around. Of course what you wrote is also correct, but trivial. – Trevor Wilson May 16 '14 at 1:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.