Lebesgue measure

Remarks

By Carathéodory's criterion the Lebesgue measurable sets form a $\sigma$-algebra and $\lambda$ is a complete measure on it. Every Borel set is Lebesgue measurable, and $\lambda(R) = \operatorname{vol}(R)$ for every box $R$; so $\lambda([a, b]) = b - a$ on $\R$. Lebesgue measure is invariant under translations and rotations, and $\lambda(tA) = \abs{t}^n \lambda(A)$ for $t \in \R$. Up to a constant factor, it is the only translation-invariant measure on the Borel sets of $\R^n$ that is finite on bounded sets. Every countable set has measure $0$, and so does the Cantor set, which is uncountable. Lebesgue measurable sets are regular: for every measurable $E$ and $\varepsilon > 0$ there are a closed set $F$ and an open set $U$ with $F \subseteq E \subseteq U$ and $\lambda(U \setminus F) < \varepsilon$. Not every subset of $\R^n$ is Lebesgue measurable; a Vitali set is an example.