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
Another approach is possible: since $f$ is continuous and nondecreasing we have: $$ \lambda_{f}((x,y])=f(y)-f(x)$$ Since $\lambda_{f} << \lambda$, from the differentiation theorem for monotone functions (they are a.e differentiable) and the theorem about measure derivation (see below), we have: $$ \frac{d\lambda_{f}}{d \lambda}=\lim_{r \rightarrow 0} \frac{\lambda_{f}(B(x,r))}{\lambda(B(x,r))}=\lim_{y \rightarrow x} \frac{\lambda_{f}((x,y])}{y-x}=f'(x) \ \ a.e $$ The claim follows by the Radon-Nikodym theorem.
For completeness, I post the theorems mentioned above:
Derivation of measures and Radon derivative
We define the derivative of the measure $\mu$ with respect to another measure $\lambda$ as: $$D_{\lambda}(\mu)(x)=\lim_{r \rightarrow 0} \frac{\mu(B(x,r))}{\lambda(B(x,r))}$$ (to be more precise we should define the upper and lower derivatives first, then the derivative would be defined only if lower and upper ones agree). Let now $\lambda$ be the lebesgue measure and $\mu$ a $\sigma$-finite measure absolutely continuous with respect to $\lambda$. Then the Radon-Nikodym derivative of $\mu$ wrt $\lambda$ and $D_{\lambda}(\mu)$ agree almost everywhere, that is: $$D_{\lambda}(\mu)(x)=\lim_{r \rightarrow 0} \frac{\mu(B(x,r))}{\lambda(B(x,r))}=\frac{d \mu}{d \lambda} \qquad a.e $$
This theorem can be proved (at least if $\lambda$ is the Lebesgue measure) with:
Lebesgue differentiation theorem
If $f \in L^1_{loc}(\mathbb{R})$, then $$ \lim_{r \rightarrow 0} \frac{1}{\lambda(B(x,r))} \int_{B(x,r)} f(t) \, d\lambda(t) = f(x) \qquad a.e $$
Note that this result is almost trivial is $f$ is continuous!