Let $f: \mathbb{R}^n \to \mathbb{R}$ be right-continuous ($f(x) = \lim\limits_{h\to 0^+} f(x+h) \,\forall x \in \mathbb{R}^n$) and $n$-increasing (i.e. $\Delta_{(a,b]} F \geq 0\,\forall a \leq b$ (here $\leq $ is understood componentwise), where the volume of a multivariate interval under $f$ is $\Delta_{(a,b]}F := \sum_{i\in \{0,1\}^n} (-1)^{\sum_{j=1}^n i_j} F(a_1^{i_1}b_1^{1-i_1},\cdots, a_n^{i_n}b_n^{1-i_n}).$ Show that there is a unique Borel measure $\mu_F$ such that $\mu_F((a,b]) = \Delta_{(a,b]} F, a\leq b.$

I think I should use Caratheodory's extension theorem. I'm thinking of using the ring $R := \{(a,b] \in \mathbb{R}^n : a\leq b\} = \mathcal{B}(\mathbb{R}^n),$ where $\mathcal{B}(\mathbb{R}^n)$ denotes the Boreal $\sigma$-algebra on $\mathbb{R}^n.$ I need to define a premeasure $\mu_0$ on $R,$ say $\mu_0(\cup_{i=1}^\infty (a_i, b_i]) = \sum_{i=1}^\infty \Delta_{(a_i,b_i]} F.$ Then if I can verify this to be a premeasure, one would clearly have that $\mu_F(A) = \mu_0(A)$ for every non-null set of $R,$ so by Caratheodory's extension theorem, $\mu_F$ is unique.

Is this incorrect?


1 Answer 1


As you mentioned in your posting, this may be obtained by application of Caratheorody's extension theorem, although it maybe more convenient to consider semirings (see Kallenberg's Foundations of probability)

Consider measures on the Borel space $(\mathbb{R}^d,\mathscr{B}(\mathbb{R}^d))$. For ${\bf x},\,{\bf y}\in\,\mathbb{R}^d$, we use the notation ${\bf x}\leq {\bf y}$, and ${\bf x}<{\bf y}$ to indicate that $x_k\leq y_k$ and $x_k<y_k$ respectively and let us also denote ${\bf e}=(1,\ldots,1)^\top$. Finally, consider the collection $\mathscr{E}$ of all $d$--dimensional intervals $\prod^d_{k=1}(a_k,b_k]=({\bf a},{\bf b}]$ with $a_k\leq b_k$, which is clearly a semiring.

For any function $F:\mathbb{R}^d\longrightarrow\mathbb{R}$ define the $j$-increment operator $$\Delta_j(a,b)F({\bf s})= F(s_1,\ldots,s_{j-1},b,s_{j+1},\ldots,s_d)-F(s_1,\ldots,s_{j-1},a, s_{j+1},\ldots,s_d)$$ and the obvious convention for $j=1$ and $j=d$. Clearly $\Delta_j(a, b)$ is linear.

For pairs $\{(a_j,b_j):j=1,\ldots,n\}$ with $a_j<b_j$, We use $\prod^n_{j=1}\Delta_j(a_j,b_j)$ the composition $\Delta_n(a_n,b_n)\circ\prod^{n-1}_{j=1}\Delta_j(a_j,b_j)$, where $\prod^1_{j=1}\Delta_j(a_j,b_j)=\Delta_1(a_1,b_1)$.

Let $F:\mathbb{R}^d\longrightarrow\mathbb{R}$ be right--continuous, i.e., $\lim_{{\bf x}\searrow{\bf a}}F({\bf x})=F({\bf a})$. For $a\leq b$ and $1<j<d$ denote by

Theorem: Suppose that $F$ is right--continuous and has nonnegative increments i.e., $\mu(({\bf a},{\bf b}]):=\prod^d_{j=1}\Delta_j(a_j,b_j)F\geq0$ for any $d$--dimensional interval $({\bf a},{\bf b}]$. Then $\mu$ admits an extension to a measure on a $\sigma$--algebra $\mathcal{M}_\mu\supset\mathscr{B}(\mathbb{R}^d)$.

Sketch of the proof:

Clearly $\mu(\emptyset)=0$ and $\mu$ is finitely additive on $\mathscr{E}$. We now prove that $\mu$ is countably subadditive on $\mathscr{E}$. If $({\bf a},{\bf b}]=\bigcup^\infty_{m=1} ({\bf a}(m),{\bf b}(m)]$, the right--continuity and positivity of the increments of $F$ imply that for any $\varepsilon>0$, there are ${\bf a}_\varepsilon$ and ${\bf b}_\varepsilon(j)$ such that $$ \mu(({\bf a},{\bf b}])< \mu(({\bf a}_\varepsilon,{\bf b}])+\tfrac{\varepsilon}{2}; \quad \mu(({\bf a}(m),{\bf b}_\varepsilon(m)])< \mu(({\bf a}(m),{\bf b}(m)])+\tfrac{\varepsilon}{2^{m+1}} $$ Since the close box $[{\bf a}_\varepsilon,{\bf b}]$ is compact and $$\begin{align}\label{compactbox} [{\bf a}_\varepsilon,{\bf b}] \subset ({\bf a},{\bf b}]\subset \bigcup^\infty_{m=1} ({\bf a}(m),{\bf b}(m)]\subset \bigcup^\infty_{m=1} ({\bf a}(m),{\bf b}_\varepsilon(m))\tag{1}label{compactbox} \end{align} $$ there is $N_0\in\mathbb{N}$ such that $({\bf a}_\varepsilon,{\bf b}]\subset[{\bf a}_\varepsilon,{\bf b}]\subset \bigcup^{N_0}_{m=1} ({\bf a}(m),{\bf b}_\varepsilon(m))$. Finite additivity implies finite subadditivity on the semiring $\mathscr{E}$, so $$\begin{align} \mu(({\bf a},{\bf b}])&<\mu(({\bf a}_\varepsilon,{\bf b}]) +\tfrac{\varepsilon}{2} \leq \sum^{N_0}_{m=1}\mu(({\bf a}(m),{\bf b}_\varepsilon(m)]) + \tfrac{\varepsilon}{2}\\ &\leq \sum^\infty_{m=1}\mu(({\bf a}(m),{\bf b}(m)])+\varepsilon \end{align} $$ Countably subadditivity of $\mu$ on $\mathscr{E}$ follows by letting $\varepsilon\searrow0$. The conclusion follows from Carath'eodory's extension theorem.


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