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?
 A: 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.
