Let $\mu$ be a probability measure on $(\mathbb{R},\mathcal{B}({\mathbb{R}}))$ and define
$F_\mu \colon x\mapsto \mu((-\infty,x])$.
Additionally, let the measure $\mu_F$ on $(\mathbb{R},\mathcal{B}({\mathbb{R}}))$ defined by
$\mu_F((a,b]) = F(b)-F(a)\quad \ a,b\in\mathbb{R},\text{ with } a<b$
where $F$ is a probability distribution, be called Stieltjes measure.
Now in my book it is stated that $\mu\mapsto F_\mu$ is a bijection. The implication that every probability distribution function can yield a probability measure is done via the measure extension theorem.
Why exactly is the measure extension theorem needed? What is the difficulty with extending the measure from $\mathcal{A}=\{(a,b]\colon a,b\in \mathbb{R}\}$? I could certainly define it on the ring of unions on $\mathcal{A}$ by just
$\mu(A)=\sum_{i=1}^n (a_i,b_i]$
where $A=\bigcup_{i=1}^n(a_i,b_i]$ for disjoint intervals $(a_i,b_i]$. This is easily seen to be a premeasure. So why wouldn't this work on a $\mathcal{B}(\mathbb{R})$ just fine?
And another question: My book states that we see from this bijection that every finite measure is a Stieltjes measure for some $F$ on the Borel algebra. How can I conclude this from the bijection theorem? I only know that for every probability measure I can construct a probability distribution function and for every probability distribution function a probability measure. But how do I know that every finite measure (or even just a probability measure) is of Stieljes form?