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Let $\mathcal{B}(\mathbb{R})$ be the Borel-sigma algebra on $\mathbb{R}$, and let $\mathcal{L}$ be the Lebesgue sigma-algebra which is a unique extension of $\mathcal{B}(\mathbb{R})$ with respect to the length measure $m$.

I'm trying to find a counter-example for each for the following:

  1. $\mathcal{L}\neq \mathbb{P}(\mathbb{R})$
  2. $\mathcal{L}\neq \mathcal{B}(\mathbb{R})$

For $2$, I just need to find a null set (has length $0$) that is not a member of $\mathcal{B}(\mathbb{R})$. I thought about the Cantor set because it has measure zero, but it's apparently Borel measurable.

Sid Caroline
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    The Vitali set, which is the topic of the question above, is a "canonical" non-measurable set (which answers the first part of your question). To obtain a set that is Lebesgue but not Borel, the usual trick is to "hide" the Vitali set inside a null-set (say, the Cantor set) via a measurable function. – Xander Henderson Jun 14 '18 at 01:44
  • Another way of proving $\mathcal{L}\neq \mathcal{B}(\mathbb{R})$ is to prove that the cardinality of $\mathcal{L}$ is $2^{\mathfrak c}$ and the cardinality of $\mathcal{B}(\mathbb{R})$ is $\mathcal c$, where $\mathfrak c = 2^{\aleph_0}$.

    Notice that the fact that $|\mathcal{L}|=2^{\mathfrak c}$ because the cadinality of the Cantor set is $\mathfrak c$ and all its subsets belong to $\mathcal{L}$. Proving that $|\mathcal{B}(\mathbb{R})| = \mathfrak c$ is a little bit harder. Take a look at https://math.stackexchange.com/questions/70880/cardinality-of-borel-sigma-algebra.

     – Hugo Jun 14 '18 at 01:57

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