Let $a, b \in \mathbb{R}$ with $a < b$, and $f: [a, b] \to \mathbb{R}$ be a monotonically increasing, right-continuous function. Show that there exists a monotonically increasing function $g: [a, b] \to \mathbb{R}$ and a Lebesgue-null set $N$ such that $f$ and $g$ are differentiable on $[a, b] \setminus N$, the derivative of $g$ is zero everywhere on $[a, b] \setminus N$, and the derivative $f'$ satisfies
$$f(x) - f(a) = \int_a^x f'(t) \,d\lambda(t) + g(x)$$
for all $x \in [a, b] \setminus N$.
My first approach:
Let $N$ be the set of points of discontinuity of $f$. Define $g: [a, b] \to \mathbb{R}$ as follows:
$$g(x) = \lim_{{t \to x^+}} f'(t)$$
Note that $g(x)$ is well-defined for each $x$ because $f$ is right-continuous. Furthermore, since $f$ is monotonically increasing, $g(x)$ is monotonically increasing as well. Both $f$ and $g$ are differentiable on the set $[a, b] \setminus N$, since we have basically cut out the "bad" points for $f$ and since $f$ is monotonically increasing on $[a,b]/N$, it is also differentiable. The differentialibility of $g$ follows from its definition as a limit. Which is the same reasion the derivative of $g$ is zero almost everywhere.
However, I am a little bit lost regarding this integral identity. How can I derive that one? I know the fundamental theorem of calculus gives me $$f(x) = f(a) + \int_{a}^{x} f'(t) d\lambda(t)$$ if $f$ is continuous. But how can I argue that adding the function $g$ on the right side "compensates" the missing condition of $f$ being continuous?
And how does my first approach look like in general? Are there any mistakes which I made? Could anybody give me feedback on that?