Is it sufficient to define a measure on half-open intervals? Suppose I want to define measure $\mu$ on the Borel sigma-algebra of $[0,1]$. Is it sufficient to define $\mu$ on the half-open intervals $[a,b)$ where $0\leq a<b\leq 1$?
The reason I think this is so is because the value of $\mu$ on other intervals can be computed by the properties of a measure. For example:
$$[a,b] = \bigcap_{n>0} \big[a,b+{1\over n}\big)$$
so:
$$\mu([a,b]) = \lim_{n\to\infty} \mu\bigg(\big[a,b+{1\over n}\big)\bigg)$$
and:
$$(a,b) = \bigcup_{n>0} \big[a+{1\over n},b\big)$$
so:
$$\mu((a,b)) = \lim_{n\to\infty} \mu\bigg(\big[a+{1\over n},b\big)\bigg)$$
Is this true?
 A: Let us recall the setting of the Hahn-Caratheodory Theorem: first, let $X$ be a set and $\mathcal{S}$ a collection of subsets of $X$.  We say $\mathcal{S}$ is a semi-ring if $\phi \in \mathcal{S}$, $\mathcal{S}$ is closed under finite intersections, and $\mathcal{S}$ has the property:
$$A, B \in \mathcal{S} \quad \implies \quad \exists E_{1},E_{2},\dots,E_{n} \in \mathcal{S} \, : \, A \setminus B = E_{1} \cup E_{2} \cup \dots \cup E_{n},$$
where $n \in \mathbb{N}$ and $E_{i} \cap E_{j} = \phi$ for each $i,j$.  We call $\nu : \mathcal{S} \to [0,\infty]$ a pre-measure if $\nu(\phi) = 0$, $\nu$ is (finitely) additive, and $\nu$ is countably monotone, i.e. if $\{E_{n}\}_{n \in \mathbb{N}}$ is a disjoint subfamily of $\mathcal{S}$ and $E = \bigcup_{n = 1}^{\infty} E_{n} \in \mathcal{S}$, then
$$\mu(E) \leq \sum_{n = 1}^{\infty} \mu(E_{n}).$$ 
The Hahn-Caratheodory Theorem states that if $\nu$ is a premeasure on $(X,\mathcal{S})$, then there is a measure $\mu$ on $\sigma(\mathcal{S})$ such that $\mu(E) = \nu(E)$ whenever $E \in \mathcal{S}$.  Moreover, $\mu$ is unique if $\nu$ is a $\sigma$-finite premeasure, i.e. there is a family $\{E_{n}\}_{n \in \mathbb{N}} \subseteq \mathcal{S}$ such that $X = \bigcup_{n = 1}^{\infty} E_{n}$ and $\nu(E_{n}) < \infty$ for each $n$.  
If we set $X = [0,1]$ and $\mathcal{S} = \{[a,b) \, \mid \, 0 \leq a \leq b \leq 1\}$, then $\mathcal{S}$ is a semi-ring on $X$.  Provided your function $\mu$ is a pre-measure, then there is an extension to $\sigma(\mathcal{S})$ (which we may as well also denote by $\mu$), which will be unique provided the pre-measure was $\sigma$-finite.  
However, this won't uniquely define $\mu$ on the Borel $\sigma$-algebra $\mathscr{B}_{[0,1]}$.  Indeed, if I let $d\mu_{1} = dx$ and $d \mu_{2} = dx + d\delta_{1}$, where $dx$ denotes the Lebesgue measure and $\delta_{1}$ is the Dirac mass at $1$, then $\mu_{1}([a,b)) = \mu_{2}([a,b))$ for each $[a,b) \in \mathcal{S}$.  Nonetheless, $\mu_{1} \neq \mu_{2}$.  The problem is $\sigma(\mathcal{S}) \subsetneq \mathscr{B}_{[0,1]}$.  
If you wanted to have a chance of defining $\mu$ uniquely on $\mathscr{B}_{[0,1]}$, then you should really consider 
$$\mathcal{S} = \{[a,b) \cap [0,1] \, \mid \, a, b \in \mathbb{R}\},$$
which includes sets like $[0,1]$ that will see the difference between $\mu_{1}$ and $\mu_{2}$, for example.  
A: You need left-continuity to establish the property
$$
\lim_{x\to b-0} \mu\big([a,x)\big) = \mu\big([a,b)\big).
$$
For every left-continuous, non-decreasing function $f:[0,1]\to[0,1]$ with $f(0)=0$, $f(1)=1$ there is a unique Borel-measure $\mu$ such that $\mu\big([a,b)\big)=f(b)-f(a)$ whenever $0\le a<b\le1$.
