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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?

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up vote 3 down vote accepted

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.

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The general idea makes a lot of sense, but I think I spot two small mistakes. 1, $f$ has to be continuous to some side or necessarily to the right? 2, isn't $\bigcap_{n\in\mathbb N}[a, b + h_n) = [a, b]$, since $b\in[a, b + h_n)$ for every $h_n$? – Luke Sep 18 '11 at 21:47
@Luke: if you use left-closed intervals then $f$ has to be right-continuous. You can do a similar construction using right-closed intervals, in which case $f$ would need to be left continuous. In the end this doesn't really matter. The crux of the matter is that if $f$ is discontinuous at $x$ then the resulting measure $\mu_f$ will have an atom at $x$ (that is, the singleton {x} will have positive mass). As for your second point, you're right. I'll edit accordingly. – Mark Sep 19 '11 at 12:08
Um… You didn't change the type of bracket. I just did (feel free to roll back, even though I made pretty small changes). – Luke Sep 19 '11 at 19:25

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