Removing a hypothesis when generalizing the Lebesgue measure Let $f:\mathbb R\to\mathbb R$ be a continuous increasing function. Define the (generalized) length of (finite) semiopen intervals,
$$
\begin{align}
\lambda_f:&\{[a,b):a,b\in\mathbb R\,;\;a\leq b\}\to[0,\infty),\\
&[a,b) \mapsto f(b)-f(a).
\end{align}
$$
Define also
$$
\begin{align}\theta^*_f:&\mathcal P(\mathbb R)\to[0,\infty],\\
&A\mapsto\inf\,\left\{\sum_{k\in\mathbb N}\,\lambda_f([a_k,b_k)):A\subset\bigcup_{k\in\mathbb N}[a_k,b_k)\right\},
\end{align}
$$
which can be shown to be an outer measure. (Therefore, with $\theta^*_f$, Carathéodory's method generates the measure $\mu_f$, the Lebesgue-Stieltjes measure.)
What changes in this rationale if we no longer assume $f$ has to be continuous?
 A: Note that, since $f$ is increasing, this already implies that it has finite one-sided limits in any real point. Of course, this does not imply continuity, but in order to eventually get a measure which is defined and finite on all intervals $[a,b)$, $f$ must at least be right-continuous. Indeed, if $\mu$ is a measure on the real line assigning finite value to every interval of this form, then $\mu ([a,b)) = \lim_{n \to \infty} \mu ([a,b+h_n))$ for any sequence $h_n > 0$ decreasing to $0$ (because $[a,b] = \bigcap_{n \in \mathbb{N}} [a,b+h_n)$ and $\mu(\{b\})=0\!$ ).
To apply Carathéodory's method, you need that the measure be a priori $\sigma$-additive on intervals of such form (in the case that their union is also of this form). The above illustrates that this won't be true if $f$ isn't right continuous. However, it turns out that right continuity (along with the monotonicity requirement) is enough in order to get a Borel measure using Carathéodory's method. This is simply because in this case we avoid the former obstruction and this is indeed the only one — Carathéodory's method then generally extends the measure to the smallest $\sigma$-algebra generated by such intervals, which is the Borel algebra.
