What sets are Lebesgue measurable?

I cannot detect the fallacy in the set of the following statements in my inconsistent notes:

1. A sigma algebra is a set of the sets in the generating set closed under the set operations countable union, countable intersection, set difference, relative complement.

2. A Borel set is a subset of $\mathbb{R}$ constructed from open and closed intervals in $\mathbb{R}$ by taking the operations countable union and intersection.

3. The set of Lebesgue measurable sets on $\mathbb{R}$ is a sigma algebra generated from the open and closed intervals in $\mathbb{R}$.

4. By the statements 2 and 3 every Lebesgue measurable set on $\mathbb{R}$ is a Borel set.

5. There is a Lebesgue measurable set that is not Borel.

6. By 4 and 5, falsity.

Which statements in my notes are not true and why?

• 3, namely that the Lebesgue $\sigma$-algebra 'generated' by the open intervals (and rays), might be misunderstood. It's not the smallest $\sigma$-algebra containing all intervals; that would be the Borel $\sigma$-algebra. However, it is the $\sigma$-algebra generated from them by the process of Carathéodory's extension theorem. – Jonathan Y. Nov 8 '13 at 10:35

1 Answer

The problem is statement (3).

The Lebesgue measure algebra is indeed a $\sigma$-algebra, but it is generated by completing the Borel $\sigma$-algebra with respect to the null sets ideal.

One can prove that there are only $2^{\aleph_0}$ Borel sets, but since the Cantor set is Borel, and of measure zero, every subset of the Cantor set is measurable. But then again the Cantor set has cardinality $2^{\aleph_0}$, so it has $2^{2^{\aleph_0}}$ subsets, all of which are Lebesgue measurable; and so most of them are not even Borel sets.