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.
It's clear that this definition is the same of lower integral for Riemann integral (...)
That is false. As a counter-example, the function $\mathbf{1}_{\mathbb{[0,1] \backslash Q}}: [0,1] \to \mathbb{R}$ has lower integral for Riemann integral equal to $0$, and "lower" integral according to the Lebesgue definition equal to $1$. The point, as mentioned by Ian at the comments, is that not every simple function is a step function: the function above being an example.
The function above is also an example of one which is Lebesgue integrable but not Riemann integrable, so the statement as it is in the title is not true.
However, the statement
$$\sup \left\{ \int \phi : \ \phi \leq f \right\} = \inf \left\{ \int \phi : \ \phi \geq f \right\}$$
is true, if $f$ is a non-negative bounded function which is not zero only on a finite measure set.* To see this, it suffices to show a sequence of simple functions $\phi_n \geq f$ such that $\lim \int \phi_n =\int f$.
Pick $M \mathbf{1}_E \geq f$. Now, we then have $M\mathbf{1}_E - f \geq 0.$ It follows that there is an increasing sequence $s_n$ of simple functions such that $s_n \to M\mathbf{1}_E-f$ and $s_n \leq M\mathbf{1}_E-f.$ By the monotone convergence theorem, $\int s_n \to \int M\mathbf{1}_E -\int f$.
We have that $f \leq M\mathbf{1}_E-s_n$, so that $\phi_n:=M\mathbf{1}_E-s_n$ is a sequence of simple functions satisfying what we want, since
$$\int \phi_n=\int M\mathbf{1}_E-\int s_n \to\int f.$$
*If $f$ doesn't satisfy those hypotheses (i.e., bounded and not zero only on a finite measure set), the right side is always infinity so the question is a little irrelevant.
Best Answer
Note that $\mathcal{U}(f,P) = \sum_{i=1}^n \sup_{A_i}f\cdot \mu(A_i) = \int\psi_P$ where $\psi_P = \sum_{i=1}^n \sup_{A_i}f\cdot \chi_{A_i}$ is a simple function.
Since $\psi_P \geqslant f$, we have $\inf \{\int\psi:\psi \geqslant f, \psi \text{ simple} \} \leqslant \mathcal{U}(f,P)$ and, it follows that taking the infimum over all partitions $P$, we have $\inf \{\int\psi:\psi \geqslant f, \psi \text{ simple} \} \leqslant \inf_P\mathcal{U}(f,P)$.
On the other hand, if $\psi = \sum_{j=1}^m b_j \chi_{B_j}$ is the canonical representation of a simple function $\psi \geqslant f$, then $b_j \geqslant \sup_{B_j} f$ and $\int \psi = \sum_{j=1}^m b_j \mu(B_j) \geqslant \sum_{j=1}^m \sup_{B_j} f\cdot \mu(B_j) = \mathcal {U}(f,P_\psi) $ for some partition $P_\psi$. This implies that $\inf \{\int\psi:\psi \geqslant f, \psi \text{ simple} \} \geqslant \inf_P \mathcal{U}(f,P).$
Hence, $\inf \{\int\psi:\psi \geqslant f, \psi \text{ simple} \} = \inf_P \mathcal{U}(f,P)$ and by a similar argument we also have $\sup \{\int\varphi:\varphi \leqslant f, \varphi \text{ simple} \} = \sup_P \mathcal{L}(f,P)$.
Since $f$ is bounded, for any $\epsilon>0$ there exists, by the simple approximation lemma, simple functions $\varphi_\epsilon$ and $\psi_\epsilon$ such that $\varphi_\epsilon \leqslant f \leqslant \psi_\epsilon$ and $\psi_\epsilon - \varphi_\epsilon < \epsilon$.
Thus,
$$\int \varphi_\epsilon \leqslant \sup \{\int\varphi:\varphi \leqslant f, \varphi \text{ simple} \} = \sup_P \mathcal{L}(f,P) \\ \leqslant \inf_P \mathcal{U}(f,P) = \inf \{\int\psi:\psi \geqslant f, \psi \text{ simple} \} \leqslant \int\psi_\epsilon,$$
and, for all $\epsilon > 0$,
$$0 \leqslant \inf_P \mathcal{U}(f,P) -\sup_P \mathcal{L}(f,P) \leqslant \int \psi_\epsilon - \int \varphi_\epsilon < \epsilon \mu(X)$$
Therefore,
$$\inf_P \mathcal{U}(f,P) = \sup_P \mathcal{L}(f,P) = \int f $$