# Borel vs Lebesgue measure - Cantor set

I am currently trying to understand the difference between Borel and Lebesgue measure.

The only difference that I found is that the Borel measure is defined only on sets which are results of countable many sum/intersection of closed sets (or equivalently open sets), while Lebesgue measure can be defined also on sets which are results of uncountable many sums/intersections.

For example, Cantor set is a result of sum of uncountable many closed sets, so it has no Borel measure, while it has Lebesgue measure (which is zero by the way).

Is this "countable assumption" the only one difference between these measures?

• More or less all of that is wrong. A Borel measure is defined on the Borel sets, but a Borel set need not be the result of countably many sum/intersection of open or closed sets. The Cantor set is a Borel set; it had better be, since it's closed. (And saying "has no Borel" measure makes no sense. There's no such thing as "Borel measure". There is such a thing as a Borel measure...) – David C. Ullrich Oct 11 '15 at 15:49
• @DavidC.Ullrich I think it was clear that they meant "the restriction of the Lebesgue measure to the Borel $\sigma$-algebra". Alternately one could say "the extension of the length premeasure to the Borel $\sigma$-algebra". – Ian Oct 11 '15 at 16:26
• @Ian If you say so. It's clear that's what the OP should have meant, but when I read "so it has no Borel measure, while it has Lebesgue measure " I really tend to think there's some confusion there... (in fact it's clear that there's a lot of confusion here on other related points, which makes my conjecture more plausible.) – David C. Ullrich Oct 11 '15 at 16:36
• @DavidC. Ullrich: Because the Borel hierarchy involves only countable ordinals, it might be argued that each Borel set can be constructed using countably many unions and complements of open sets. – John Dawkins Oct 11 '15 at 17:34
• @JohnDawkins Of course. Now, read the OP. Do you really think that's what he had in mind? I don't follow some of the comments I'm getting - the OP was confused. – David C. Ullrich Oct 11 '15 at 17:40

If you start with the Borel $\sigma$-algebra on $\mathbb R^n$ wich can be classified as the $\sigma$-algebra generated by the usual topology on $\mathbb R^n$, then by means of the Borel measure this $\sigma$-algebra can be expanded. Denote the Borel $\sigma$-algebra as $\mathcal B$ and let $\mu$ denote the Borel measure on this collection.
Let it be that $\mathcal F\subset\wp(\mathbb R^n)$ such that: $$F\in\mathcal F \text{ if and only if } B,C\in\mathcal B\text{ exist with }F\triangle B\subseteq C\text{ and }\mu C=0$$
Then it can shown that $\mathcal F$ is a $\sigma$-algebra containing $\mathcal B$ and also containing sets that are not in $\mathcal B$.
Then on $\mathcal F$ we can define measure $\lambda$ by stating that $\lambda(F):=\mu(B)$. It can be shown that like this $\lambda$ is well-defined, and actually $\lambda$ is the Lebesgue measure.
For an $F\in\mathcal F$ with $\lambda(F)=0$ it can be shown that $G\in\mathcal F$ is true for each subset of $F$. That means that $\lambda$ is a complete measure. The measure $\mu$ is not a complete measure.