# Prove that Lebesgue-Stieltjes outer measure is an outer measure.

For any non-decreasing $f: \mathbb{R} \to \mathbb{R}$ and $S \subseteq \mathbb{R}$, The Lebesgue-Stieltjes outer measure associated with $f$ is $$\lambda^*_f(S) = \inf\left\{\sum_{j=0}^\infty (\lim_{x \uparrow b_j}f(x) - \lim_{x \uparrow a_j}f(x)) \mid a_j, b_j \in \mathbb{R} \land S \subseteq \bigcup_{j=0}^\infty [a, b[ \right\}: \mathcal{P}(\mathbb{R}) \to [0,\infty]$$

Theorem:

Let $S_0, S_1, \dots \subseteq \mathbb{R}$. Then $$\lambda^*_f\left(\bigcup_{j=0}^\infty S_j\right) \leq \sum_{j=0}^\infty \lambda^*_f(S_j)$$

My attempt: Construct a two dimensional array of half-open intervals $I_{j,k}$ such that($\epsilon$ is arbitrary) $$S_j \subseteq \bigcup_{k=0}^\infty I_{j, k} \land \sum_{k=0}^\infty \lambda_f(I_{j, k}) \leq \lambda^*_f(S_j) + 2^{-j} \epsilon$$ and consequently prove $$\lambda^*_f\left(\bigcup_{j=0}^\infty S_j\right) \leq \sum_{j=0}^\infty \lambda^*_f(S_j) + \epsilon$$ However, this argument is not supposed to work when I replace the $\lim_{x\uparrow b} f(x)$ in the definition of L-S outer measure with $f(b)$. What is wrong here?

## 1 Answer

(Before I begin: The question uses the function $\lambda_f$ but doesn't define it. According to Fremlin's Measure Theory, which OP identified in the bounty comment as the source of the question, it's a set function on the set of half-open intervals, such that $\lambda_f(\varnothing) = 0$, and such that $\lambda_f \, [a_j, b_j[$ is the summand in the definition of $\lambda_f^*$ for $a<b$.)

There isn't anything wrong. The argument continues to work because the most important ingredient isn't any minutiae of $\lambda_f$ but rather the way we defined $\lambda_f^*$ from $\lambda_f$ as the infimum of sums of outputs of $\lambda_f$ over covers. The only property of $\lambda_f$ that we needed was its non-negativity, which we have regardless of whether $\lambda_f \left[a,b\right[$ is $\lim_{x \uparrow b} f(x) - \lim_{x \uparrow a} f(x)$, $f(b) - \lim_{x \uparrow a} f(x)$, or $f(b) - f(a)$.

It really doesn't take much to get an outer measure through this method: The same argument shows that given any triplet $(X, \mathcal C, \mu)$, where $X$ is a set, $\mathcal C \subseteq \mathcal P(X)$, and $\mu \colon \mathcal C \to [0, \infty]$ a set function(!) such that $\inf \mu(\mathcal C) = 0$, the set function \begin{align*} \mu^*\colon \mathcal P(X) &\to [0, \infty] \\ S &\mapsto \inf \left\{ \sum_{j=0}^\infty \mu(A_j) \;\middle|\; A_j \in \mathcal C,\, S \subseteq \bigcup_{j=0}^\infty A_j\right\} \end{align*} is an outer measure. For a quick reference, Wikipedia states a slightly less general version (they require $\varnothing \in \mathcal C$ and $\mu(\varnothing) = 0$) attributed to M. E. Munroe's Introduction to Measure and Integration.

The belief that "this argument is not supposed to work" might come from a misinterpretation of Fremlin. In Exercise 114X(a), he first defines $\lambda_g$ and $\theta_g$ ($\equiv\lambda_g^*$) (here $g$ is the non-decreasing function), and then invites the reader to do as follows:

Show that $\theta_g$ is an outer measure on $\mathbb R$. Let $\mu_g$ be the measure defined from $\theta_g$ by Carathéodory's method; show that $\mu_g I$ is defined and equal to $\lambda_g I$ for every half-open interval $I \subseteq \mathbb R$, and that every Borel subset of $\mathbb R$ is in the domain of $\mu_g$.

Subsequently in Exercise 114X(b), he asks

At which point would the argument of 114Xa break down if we wrote $\lambda_g \left[a, b \right[ = g(b) - g(a)$ instead of using [the given formula $$\lambda_g \left[a, b\right[ = \lim\nolimits_{x\uparrow b} g(x) - \lim\nolimits_{x \uparrow a} g(x)$$ ]?

We've seen that $\theta_g$ is an outer measure even with the alternate $\lambda_g$. The "breakdown" instead happens somewhere in the second sentence of the passage I quoted from 114X(a).