# What is the measure space a right continuous, non-decreasing function defines?

(Dealing with the real line $\mathbb{R}$) A right continuous, non-decreasing function $g$ defines a pre-measure on the algebra of half-open intervals: $$\mu _g (a,b] = g(b)-g(a)$$

This pre-measure can be extended to a measure using the Caratheodory process - first defining an outer measure (which turns out to be metric outer measure, which guarantees it is defined on the borel sets). This process provides us with a complete measure space.

How can we calculate the $\sigma$-algebra corresponding to this measure space?

Examples:

1. The function $x\mapsto x$ - Noting the borel measure it defines is also invariant under translations - we deduce it is "the" Borel measure on the real line and so (with a little bit more arguing) we understand the complete $\sigma$-algebra the Caratheodory process produces is the Lebesgue $\sigma$-algebra.
2. The function $x\mapsto 0$ - This produces the $0$-measure, and so every subset of $\mathbb{R}$ is measurable.
• These are called Lebesgue–Stieltjes measures en.wikipedia.org/wiki/Lebesgue%E2%80%93Stieltjes_integration but there is an ambiguity with "calculate". We can not "calculate" the algebra even for the usual Lebesgue measure, under some replacements of the axiom of choice all sets are Lebesgue measurable, and anything using it hardly counts as "calculating". As I recall, if one measure has density relative to the other its algebra is no less, so you can get some information that way, this is related to the Radon-Nicodym theorem en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem – Conifold Oct 5 '16 at 20:12
• Thank you for the links. I am asking this question from within the ZF realm :). Yes you are right, but the problem is that some Lebesgue-Stieltjes measures are singular to Lebesgue measure, and so, they have no Radom-Nikodym derivative (For example, the delta distribution, coming from the right continuous, non-decreasing Heaviside function). – Ranc Oct 5 '16 at 20:24
• For singular measures you may want to take a look at Hausdorff measures and geometric measure theory. They are fractional analogs of the Lebesgue measure, and I imagine their algebras would be describable in a way similar to Borel's, but with fractals and Cantorian sets in the mix. In particular, all Borel subsets are still Hausdorff measurable due to the metric outer measure property en.wikipedia.org/wiki/Metric_outer_measure – Conifold Oct 5 '16 at 20:37
• The process you (somewhat implicitly) describe is by far the best to describe the $\sigma$-algebra: take the completion of the Borel $\sigma$-algebra wrt the measure. – user138530 Oct 18 '16 at 6:08

## 1 Answer

Basic analysis of tractable cases:

Language: We denote $\mathfrak{B}(\cdot)$ , $\mathfrak{L}(\cdot)$, $\mathbb{P}(\cdot)$ , $(\cdot)^*$ the borel sets, the collection of lebesgue measurable sets , the powerset, the cylinderical $\sigma$-algebra of $(\cdot)$.

Also note that every borel measure $\lambda$ has a decomposition $\lambda_{\text{ac}} + \lambda_{\text{d}} + \lambda_{\text{sc}} = \lambda$ where the LHS corresponds to the decomposition into absolutely continuous, discrete and singular continuous parts of the measure $\lambda$ (with respect to the Lebesgue measure).

$g$ is absolutely continuous ($\lambda_{\text{ac}}$ case):
Note that the derivative is guaranteed to exist almost-everywhere, and so this sets cover $\mathbb{R}$ up to a lebesgue null-set. Because this case is not tractable yet, we further assume $g$ is $C^0$, and in such cases we divide the real line into 2 sets: $\{g'>0 \}\cup \{g'\leq0\}$.
On the first set, $g$ has an absolutely continuous inverse. Since both $g$ and its inverse have Luzin's property-N ; we realize $N$ is a null set iff $g(N)$ is, and so the completion of the borel sigma algebra $\mathfrak{B}\left(\{g'>0 \} \right)$ is simply $\mathfrak{L}(\{g'>0 \})$.
On the other set; since $g$ is non-decreasing we have $\{g'\leq0 \}=\{g'=0\}$ and so $g$ is constant there. Every set has measure $0$ and so it is measureable.
Conclusion: the $\sigma$-algebra is $$\big[\mathfrak{L}(\{g'>0 \}) \cup \mathbb{P}(\{g'=0 \}) \big]^*$$

$g$ is discrete ($\lambda_{\text{d}}$ case):
This is also easy and corresponds to the case where $g$ is a non-decreasing step function. Every subset of $\mathbb{R}$ is measurable.

$g$ is continuous but not absultely $(\lambda_{\text{sc}}$ case):
This happens when $g$ is of the form of Cantor-Lebesgue function. Since it is not absolutely continuous, it is singular with respect to the lebesgue measure and we have already dealt with the case of discrete measures, so we are left to understand the continuous case.
This case is the less tractable case, and it is shown by an example that we cannot further analyse it. Denote $c(x)$ the Cantor-Lebesgue function and notice it is continuous (but not AC) and non-decreasing. The image of the (null-set) cantor-set under $c$ has positive lebesgue measure and so the cantor set has positive measure under the Lebesgue-Stieltjes measure generated with $c$. Also note $c$ is constant at some parts and so we conclude there is non obvious inclusion between this $\sigma$-algebra and $\mathfrak{L}$.

Back to my question. We investigate $g=1_{[0,\infty)} \sqrt x$ and are trying to realize the $\sigma$-algebra, $\mathfrak{A}$, the construction generates. By the first case we have: $$\mathfrak{A} = \big[\mathfrak{L}\big([0,\infty)\big) \ \cup \ \mathbb{P}\big((-\infty,0]\big) \big]^*$$

That is, the cylindrical $\sigma$-algebra of the lebesgue measurable sets in $[0,\infty)$ and the power set of $(-\infty,0]$.