Let $\mathcal A$ be an algebra on a set $X$, $\mu_0$ a premeasure on $\mathcal A$,
and $\mathcal M$ the $\sigma$-algebra generated by $\mathcal A$.
Let $\mu^*$ be the Caratheodory outer measure induced by $\mu_0$, $\mu$ the restriction of $\mu^*$ on $\mathcal M$.
Let $\mathcal C$ be the set of all $\mu^*$-measurable sets, $\bar{\mu}$ the restriction of $\mu^*$ on $\mathcal C$.
Suppose $\mu_0$ is $\sigma$-finite, i.e. there exists a sequence $E_n, n = 1, 2, \cdots$ of members of $\mathcal A$ such that $X = \cup_{n=1}^{\infty} E_n$, $\mu_0(E_n) \lt \infty$ for all $n$. I claim that $(X, \mathcal C, \bar{\mu})$ is the completion of $(X, \mathcal M, \mu)$.
Let $\bar{\mathcal M}$ be the completion of $\mathcal M$ with respect to $\mu$.
Since $\mathcal M\subset \mathcal C$, $\bar{\mathcal M}\subset \mathcal C$.
Hence it suffices to prove that $\mathcal C \subset \bar{\mathcal M}$.
Since $\mu_0$ is $\sigma$-finite, it suffices to prove that if $E\in \mathcal C$ and $\bar{\mu}(E) \lt \infty$, then $E\in \bar{\mathcal M}$.
By the definition of $\bar{\mu}$, for each integer $n \ge 1$, there is a sequence $A_j, j= 1, 2, \cdots$ such that $\sum_{j=1}^{\infty} \mu_0(A_j) \lt \bar{\mu}(E) + 1/n$, where $A_j \in \mathcal{A}$, $E \subset \bigcup_{j=1}^\infty A_j.$ Let $F_n = \bigcup_{j=1}^{\infty} A_j$. Then $E \subset F_n$ and $\mu(F_n) \le \sum_{j=1}^{\infty} \mu_0(A_j) \lt \bar{\mu}(E) + 1/n$.
Let $F = \cap_{n=1}^{\infty} F_n$. Then $F\in \mathcal M$, $E \subset F$, and $\bar{\mu}(E) \le \mu(F) \le \mu(F_n) \lt \bar{\mu}(E) + 1/n$ for all $n\ge 1$. Hence $\bar{\mu}(E) = \mu(F)$.
Similarly there exists $G\in \mathcal M$ such that $F - E \subset G$ and $\mu(G) = \bar{\mu}(F - E)$ = 0.
Then $E = (F - G) \cup (E\cap G)$, $F - G \in \mathcal M$, and $E\cap G$ is a subset of the $\mu$-null set $G$. Hence $E \in \bar{\mathcal M}$. This completes the proof.
Now consider the family $\mathcal A$ of finite disjoint unions of intervals of the form $[a, b)$ or $(-\infty, c)$, where $-\infty \lt a\lt b\le \infty$ and $c\in \mathbb R$.
It is elementary and well known that $\mathcal A$ is an algebra on $\mathbb R$ and there is a unique premeasure $\mu_0$ on $\mathcal A$ such that $\mu_0([a, b)) = b - a$ whenever $a$ and $b$ are finite.
Then $\mathcal M$ and $\mathcal C$ defined above are the families of Borel sets and Lebesgue measurable sets in $\mathbb R$ respectively and you get the picture.
Define the premeasure $\mu_f$ on the algebra of half-open intervals $(a,b]$ by $\mu_f((a,b])=f(b)-f(a).$
Now, $\mu_f$ extends $uniquely$ to a measure $\mu$ on $\mathscr B(\mathbb R).$ Next, define, for each $A\in \mathscr B(\mathbb R)$, $\nu (A)=\int_A f'd\lambda. \ $ Since $\nu$ also extends $\mu_f,\ $ we have by uniqueness that $\nu=\mu$ and absolute continuity follows immediately, and clearly $\frac{d\mu}{d\lambda}=f'$.
edit: I think I have a proof from scratch:
The Lebesgue-Stieljes measure $\mu$ is the one that extends $\mu_f$. You want to prove from scratch (without the Monotone Class Theorem or similar) that
$\mu (A)=\int _Af'd\lambda.\ $
Now, clearly $\nu $ defined by $\nu(A)=\int _Af'd\lambda\ $ is a positive measure since $f$ is increasing. Therefore
$A\subseteq B\Rightarrow \int _Af'd\lambda)\le \int _Af'd\lambda.$
Note that $\mu_f(I)=f(b)-f(a)\ $ for $any$ interval with endpoints $a,b$ and that the result is clearly true if $A=(a,b]\ $ or if $\mu(A)=\infty.$
If $A$ is Borel such that $\mu(A)<\infty,\ $ there is a sequence of disjoint intervals $\left \{ (a_i,b_i) \right \}_{i\in \mathbb N}$ such that $A\subseteq \bigcup_i(a_i,b_i)\ $ and
$\mu(A)>\mu \left ( \bigcup_i (a_i,b_i) \right )-\epsilon=\sum_i (f(b_i)-f(a_i))-\epsilon=\int _{\cup_i (a_i,b_i)}f'd\lambda-\epsilon>\int_Af'd\lambda-\epsilon,\ $ so
$\mu (A)\ge \int_Af'd\lambda.$
Similarly, there is a sequence of disjoint intervals $\left \{ (c_i,d_i) \right \}_{i\in \mathbb N}$ such that $A\subseteq \bigcup_i(c_i,d_i)\ $ and
$\nu(A)=\int_Af'd\lambda >\int _{\bigcup _i(c_i,d_i)}f'd\lambda -\epsilon=\sum_i(f(d_i)-f(c_i))-\epsilon =\mu (\bigcup _i(c_i,d_i))-\epsilon>\mu(A)-\epsilon,$ so
$\mu (A)\le \int_Af'd\lambda.$
The result follows.
Best Answer
The algebra $\Sigma_0$ is defined in the following way. $F \in \Sigma_0$ if $F$ can be written as $F = (a_1,b_1] \cap ... \cap (a_r,b_r].$ $\Sigma_0$ defined like that is an algebra on (0,1] and this allows to use later the theorem 1.7 to extend $(\Sigma_0,\mu_0)$ to $(\Sigma,\mu).$
Sets of the type $F = [a_1,b_1] \cap ... \cap [a_r,b_r]$ don't form an algebra on $[0,1]$. For example, if we have $E = [0,1/2],$ then $E^c = (1/2,1] \notin \Sigma_0.$ So you will not be able to use theorem 1.7 and define $\mu$ on (the $\sigma$-algebra) $\Sigma,$ which is the goal.