In measure theory one extends the Borel sigma algebra to the Lebesgue sigma algebra by including all subsets of sets of measure zero. Why is this needed ? Why not including subsets of the sets of measure 1 or other measure ?
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2Because Every Lebesgue measurable set are of the form $B\cup N$ where $B$ is a Borel set and $N$ a nul set. – Surb Oct 07 '22 at 15:36
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Because the $\sigma$-algebra on $\mathbb R$ generated by all subsets of sets of measure $a$ (for some fixed $a>0$) is the whole set of subset of $\mathbb R$, but Lebesgue measure cannot be extended to it. – Anne Bauval Oct 07 '22 at 15:39
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Since you are able to measure a set then it is measurable and we have non-borel sets with measure zero so measurable sets are not just borel – IrbidMath Oct 07 '22 at 15:45
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Thanks. I do not understand exactly. What do you mean by 'the whole set of subset of $\mathbb{R}$' ? – user996159 Oct 07 '22 at 15:46
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I forgot an s: the set of all subsetS of $\mathbb R$. But I retract the end of that sentence: the Lebesgue measure cannot be extended to a translation invariant measure on all subsets of $\mathbb R$ but if we drop that contraint, the situation is more complicated. – Anne Bauval Oct 07 '22 at 15:57
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I again do not understand why the sigma algebra on $\mathbb{R}$ generated by all subsets of sets of measure a (for some fixed a>0) is the whole set of subsets of $\mathbb{R}$, which is the power set. Can you provide a reference or a proof for that claim ? Thanks. – user996159 Oct 07 '22 at 16:13
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I have no reference at hand, but let $A\subset\mathbb R$, then $A=\cup_{n\in\mathbb Z}(A\cap[na,(n+1)a])$. – Anne Bauval Oct 07 '22 at 16:23