My understanding:
If a function $f$ is continuous and $\lim_{n\to \infty}x_n=x$, then $f(x)=f(\lim_{n\to \infty}x_n)=\lim_{n\to \infty}f(x_n)$. That is, you can take the limit out. Similarly, $\lim_{n\to \infty}\cup_{k=1}^nA_k=\cup_{k=1}^{\infty}A_k$, the "continuity" property of measure implies
$$
m(\cup_{k=1}^{\infty}A_k)=m(\lim_{n\to \infty}\cup_{k=1}^nA_k)=\lim_{n\to \infty}m(\cup_{k=1}^nA_k)=\lim_{n\to \infty}A_n
$$
Note $\{A_n\}_{n=1}^{\infty}$ is an ascending collection of measurable sets, so $A_n=\cup_{k=1}^nA_k$.
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
We shall be using the following result:
Lemma. if $E$ is Lebesgue measurable and $m(E)>0$, then for every $\alpha\in (0, 1)$, there is an interval $I$ such that $m(E \cap I) > \alpha \, m(I)$.
Proof. It is a consequence of the Lebesgue Density Theorem, according to which, if $m(E)>0$, then for almost every point $x \in E$ $$ \lim_{\varepsilon\to 0} \frac{m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)}{2\varepsilon} = 1. $$ Hence for such an $x$ we can choose an $\varepsilon$, such that $$ \frac{m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)}{2\varepsilon} > \alpha $$ or equivalently $$ m\big(E \cap (x-\varepsilon,x+\varepsilon)\big)> \alpha\,m\big((x-\varepsilon,x+\varepsilon)\big)\big). \tag*{$\Box$} $$
In our case, as $m(E)>0$, let an interval $I=[a,b]$, such that $m(E\cap I)>\frac{4}{5}m(I)$, and set $t=m(I)/5>0$. Clearly, writing $E_t=E+t$, $$ m\big(E_t\cap [a+t,b]\big)=m\big(E\cap [a,b-t]\big) \ge m(E\cap I)-m\big([b-t,b]\big)\ge \tfrac{3}{5}m(I), $$ and similarly $$ m\big(E\cap [a+t,b-t]\big),\,m\big(E_t\cap [a+t,b-t]\big)\ge\frac{2}{5}m(I), $$ while $m\big([a+t,b-t]\big)=\frac{3}{5}m(I)$. Therefore $$ m(E\cap E_t)\ge m\big(E\cap E_t\cap [a+t,b-t]\big)\ge\frac{1}{5}m(I). $$