I encountered such an exercise:
Let $H=\chi_{[0,+\infty)}$ be the Heaviside function, and $\{r_n\}_{n=1}^\infty$ is a given real sequence. Let $$\mathcal{R}:=\left\{\bigcup_{i=1}^k (a_i,b_i] : k \in \mathbb N; a_i, b_i \in \mathbb R, \forall i=1,\ldots , k\right\}$$ be a ring on $\mathbb R$. For $x\in \mathbb R$, we define $f(x)= \sum_{n=1}^\infty \frac{1}{2^n} H(x-r_n)$. Define a set function $\mu$ on $\mathcal{R}$ by $$\mu((a,b])= f(b)-f(a) , \quad \forall (a,b]\in \mathcal{R}. $$ Then $\mu$ is a measure on $\mathcal{R}$. Determine the Carathéodory outer measure $\mu^*$ induced by $\mu$ and the $\sigma$-algebra $\mathfrak{M}_{\mu^*}$ of $\mu^*$-measurable sets.
Some related definitions.
Definition (ring). A ring on a set $\Omega$ is a subset $\mathcal{R}$ of the power set $2^\Omega$ with the following properties:
- $\varnothing \in \mathcal{R}$.
- $A, B \in \mathcal{R} \Longrightarrow A \cap B \in \mathcal{R}$.
- $A, B \in \mathcal{R} \Longrightarrow A \setminus B \in \mathcal{R}$.
Definition (measure). Let $\mathcal{A} \subset 2^\Omega$ with $\varnothing ∈ \mathcal{A}$. We call measure on $\mathcal{A}$, any map $μ : \mathcal A \to [0, +∞]$ with the following properties:
- $μ(\varnothing) =0$.
- (countably additive) $A ∈ \mathcal{A}$, $A_n ∈ \mathcal{A}$ with $A_n$ mutually disjoint and $A= \bigcup_{n=1}^\infty A_n$ $\Longrightarrow$ $\mu (A)= \sum_{n=1}^\infty \mu (A_n)$.
Definition (outer measure). We define an outer measure on $Ω$ as being any map $μ^∗ : 2^\Omega → [0, +∞]$ with the following properties:
- $μ^∗(\varnothing) = 0$.
- $A \subset B \Longrightarrow μ^∗(A) \leq μ^∗(B)$.
- $\mu^* \left(\bigcup_{n=1}^\infty A_n \right)\leq \sum_{n=1}^\infty \mu (A_n)$.
Definition ($\sigma$-algebra of $\mu^*$-measurable sets). Let $\mu^* :2^\Omega \to [0,+\infty]$ be an outer measure on $\Omega$. If $A\subset \Omega$ satisfies: $$ μ^* (T ) = μ^*(T \cap A) + μ^∗(T \cap A^c) ,\quad \forall T \subset Ω ,$$ then we say $A$ is $\mu^*$-measurable. Denote by $\mathfrak{M}_{\mu^*}$ the collection of all the $\mu^*$-measurable subsets of $\Omega$. We can prove that $\mathfrak{M}_{\mu^*}$ is indeed a $\sigma$-algebra, and the restriction $\mu^*|_{\mathfrak{M}_{\mu^*}}:\mathfrak{M}_{\mu^*}\to [0,+\infty]$ of $\mu^*$ on the $\sigma$-algebra $\mathfrak{M}_{\mu^*}$ of $\mu^*$-measurable sets becomes a measure, i.e., $(\Omega, \mathfrak{M}_{\mu^*}, \mu^*|_{\mathfrak{M}_{\mu^*}})$ is a measure space.
Definition (Carathéodory outer measure). Let $\mathcal{R}$ be a ring on $Ω$ and $μ : \mathcal{R} → [0, +∞]$ be a measure on $\mathcal{R}$. For all $T \subset Ω$, we define: $$μ^*(T) : = \inf\left\{\sum_{n=1}^\infty \mu (A_n) : \text{$\{A_n\}_{n=1}^\infty$ is an $\mathcal{R}$-cover of $T$} \right\},$$ where an $\mathcal{R}$-cover of $T$ is defined as any sequence $\{A_n\}_{n=1}^\infty$ of elements of $\mathcal{R}$ such that $T \subset \bigcup_{n=1}^\infty A_n$. And by convention $\inf \varnothing= +\infty$. $\mu^*$ is called the Carathéodory outer measure induced by $\mu$ (of course, it is an outer measure, by definition).
My attempt.
Clearly, $f$ is a non-decreasing, right-continuous funtion on $\mathbb R$ with $f(-\infty)=0$ and $f(+\infty)=1$, and $\mu$ is a measure on the ring $\mathcal{R}$. Moreover, I found that if we restrict the measure $$\nu:= \sum_{n=1}^\infty \frac{1}{2^n} \delta_{r_n} \quad (\text{where $\delta_{r_n}$ is the Dirac measure on the point $r_n$})$$ on the ring $\mathcal{R}$, then it is consistent with $\mu$. Thus I guess the Carathéodory outer measure $\mu^*$ induced by $\mu$ is exactly $\nu$ (then we can deduce that all the subsets of $\mathbb R$ are $\mu^*$-measurable), but I cannot explicitly prove my claim, I wonder whether we need to use the regularity of the measure $\nu$.