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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.

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2 Answers 2

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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.

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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 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. –  Andres Caicedo May 16 at 14:03
Excellent answer. Thanks for your time. –  Xavyar May 16 at 14:09
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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.

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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 at 0:54
I think it reads better now. –  Rene Schipperus May 16 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 at 1:17
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