Problem with the definition of Lebesgue Measure and borel sets! The definition of Lebesgue measure (in my textbook):

The set-function $\lambda^{n}$ on  ($\mathbb{R}^{n}, \mathcal{B}(\mathbb{R}^{n})$) that assigns every half-open $[[a,b)) = [a_{1},b_{1}) \times \dots \times [a_{n},b_{n})\in\mathcal{J}$ the value:
$ \lambda^{n}([[a,b))):=\prod_{j=1}^{n}(b_{j}-a_{j}) $ is called n-dimensional Lebesgue measure.

On the next page the book mentions that the Lebesgue measure is a measure on the Borel sets $\mathcal{B}(\mathbb{R})$
My question/problem:
According to the definition of a measure in the book. The measure must be a map between $\mathcal{B}(\mathbb{R})$ and $[0,\infty]$. But how can the defined $\lambda^{n}$ maps all the elements in $\mathcal{B}(\mathbb{R})$? When it only maps half-open rectangles? Can $\lambda^{n}$ map other kinds of set in $\mathcal{B}(\mathbb{R})$?
 A: The set of half open Intervals are a so called half ring. $\lambda^n$ is defined as a pre-measure on that half ring. By a method of Carathéodory such a premeasure on a half ring can be extended to the $\sigma$-Algebra created by that half ring (in our case that is the Borel-Algebra).
Also we get that as long as the pre-measure is $\sigma$-finite then this extension is unique. So $\lambda^n$ is uniquely defined on the Borelsets by this definition on half open intervals.
In fact the method by Carathéodory does not only give us a measure, but in fact what we call an outer measure, that is a measure on the whole powerset where we replace $\sigma$-additivity by monotonity and $\sigma$-subadditivity (measures of countable unions are at most the sum of the individual measures). This then gives you a new idea of measurable sets (basically a set is measurable if it supports $\sigma$-additivity). We then get a $\sigma$-Algebra of all Lebesgue-measurable sets that is larger than the Borel-$\sigma$-algebra. (Without the axiom of choice we cannot find sets that are not measurable).
