Proof of The Space of Piecewise Constant Paths in a Finite Set: Measurability of Evaluations, Restrictions, Left Limits, and Occupation Integrals

Throughout, $\mathsf{Path}=\mathsf{Path}(E,T)$, and for $p\in\mathsf{Path}$ we call the intervals $[0,t_1)$, $[t_j,t_{j+1})$ ($1\le j<k$) and $[t_k,T]$ of the definition (the single interval $[0,T]$ when $k=0$) the constancy intervals of $p$; they partition $[0,T]$, and each of them, except possibly the last, has its right endpoint strictly larger than its left endpoint and excluded, while the last one is $[t_k,T]$ or $[0,T]$. We use that the family of subsets of a set whose preimage under a given map lies in a given $\sigma$-algebra is itself a $\sigma$-algebra (preimages commute with complements and countable unions), so that a map into a generated $\sigma$-algebra is measurable as soon as the preimages of the generators are measurable. Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions is used for constants, indicators, linear combinations, products and pointwise limits of measurable real-valued maps.

Claim 1. Every subset $S$ of the finite set $E$ is the finite union of the singletons $\{y\}$, $y\in S$, so the preimage of $S$ under $p\mapsto p(u)$ is the finite union of the generators $\{p:p(u)=y\}$, $y\in S$, a member of $\mathcal{C}_T$; thus the evaluation map is measurable. Since $\phi(p(u))=\sum_{y\in E}\phi(y)\mathbf{1}\{p(u)=y\}$, the map $p\mapsto\phi(p(u))$ is a linear combination of indicators of members of $\mathcal{C}_T$, hence measurable. Now let $0<T'\le T$ and $p\in\mathsf{Path}$ with times $t_1<\dots<t_k$; let $k'$ be the number of indices $j$ with $t_j\le T'$. The restriction $p|_{[0,T']}$ is constant on $[0,t_1)$ if $k'\ge1$, on $[t_j,t_{j+1})$ for $j<k'$, and on $[t_{k'},T']$, the last interval being contained in $[t_{k'},t_{k'+1})$ if $k'<k$ and in $[t_k,T]$ if $k'=k$; if $k'=0$ it is constant on $[0,T']$, which is contained in $[0,t_1)$ (or in $[0,T]$ when $k=0$). Hence $p|_{[0,T']}\in\mathsf{Path}(E,T')$ with the times $t_1,\dots,t_{k'}$. The preimage under the restriction map of a generator $\{p':p'(u)=y\}$ of $\mathcal{C}_{T'}$ ($u\in[0,T']$) is the generator $\{p:p(u)=y\}$ of $\mathcal{C}_T$, so the restriction map is measurable.

Claim 2. Let $p\in\mathsf{Path}$ with times $t_1<\dots<t_k$ and let $t\in(0,T]$. If $k=0$ or $t\le t_1$, then $p$ is constant on $[0,t)$, and $\delta=t$ together with the constant value works. Otherwise let $j$ be the largest index with $t_j<t$; then $p$ is constant on $[t_j,t_{j+1})\supseteq[t_j,t)$ if $j<k$ (as $t\le t_{j+1}$), and on $[t_k,T]\supseteq[t_k,t)$ if $j=k$; so $\delta=t-t_j$ and the constant value work. If $y$ and $y'$ both have the property with $\delta$ and $\delta'$, then $y=p(s)=y'$ for $s=t-\min(\delta,\delta')\in[0,t)$; so $p(t-)$ is unique. For $u\in[0,T)$, the constancy interval containing $u$ has right endpoint strictly larger than $u$ (if it is the last one, its right endpoint is $T>u$ and $\delta=T-u$ works; otherwise it is of the form $[a,b)$ with $u\in[a,b)$, so $b>u$ and it contains $[u,u+\delta)$ for $\delta=b-u$), which gives the right-continuity assertion. For measurability, fix $t\in(0,T]$ and $y\in E$ and let $n_0$ be a natural number with $1/n_0\le t$. We claim

If $p(t-)=y$ with the $\delta$ above, then for every $m\ge\max(n_0,1/\delta)$ the point $t-1/m$ lies in $[t-\delta,t)$, so $p(t-1/m)=y$, and $p$ belongs to the right side. Conversely, if $p(t-1/m)=y$ for all $m\ge n$, then, with $\delta$ as in the existence part, $p(t-1/m)=p(t-)$ for all $m\ge\max(n,n_0,1/\delta)$, so $p(t-)=y$. The right side is a countable union of countable intersections of generators, hence a member of $\mathcal{C}_T$; as in claim 1 this gives the measurability of $p\mapsto p(t-)$.

Claim 3. Fix $y\in E$. For a natural number $n$ and $i\in\{1,\dots,n\}$ put $I_{n,i}=((i-1)T/n,\,iT/n]$, a member of $\mathcal{B}_{[0,T]}$, and define $g_n:[0,T]\to[0,T]$ by $g_n(0)=0$ and $g_n(u)=iT/n$ for $u\in I_{n,i}$; the sets $\{0\}$ and $I_{n,1},\dots,I_{n,n}$ partition $[0,T]$, and $u\le g_n(u)<u+T/n$ for every $u$. Then

where $\mathbf{1}_I$ denotes the indicator of a set $I$; each summand is the indicator of a measurable rectangle of $\mathcal{B}_{[0,T]}\otimes\mathcal{C}_T$, so the left side is $\mathcal{B}_{[0,T]}\otimes\mathcal{C}_T$-measurable as a function of $(u,p)$. For every $(u,p)$ the sequence $\mathbf{1}\{p(g_n(u))=y\}$ converges to $\mathbf{1}\{p(u)=y\}$: if $u=T$ then $g_n(T)=T$ for every $n$; if $u<T$, claim 2 gives $\delta>0$ with $p=p(u)$ on $[u,u+\delta)\cap[0,T]$, and for $n>T/\delta$ we have $g_n(u)\in[u,u+\delta)$, so $p(g_n(u))=p(u)$. Being a pointwise limit of measurable maps, $(u,p)\mapsto\mathbf{1}\{p(u)=y\}$ is $\mathcal{B}_{[0,T]}\otimes\mathcal{C}_T$-measurable.

Now let $\phi:E\to\mathbb{R}$ and put $M=\max_{y\in E}|\phi(y)|$. The map $(u,p)\mapsto\phi(p(u))=\sum_{y\in E}\phi(y)\mathbf{1}\{p(u)=y\}$ is $\mathcal{B}_{[0,T]}\otimes\mathcal{C}_T$-measurable and bounded by $M$ in absolute value. For fixed $p$ its section $u\mapsto\phi(p(u))$ is $\mathcal{B}_{[0,T]}$-measurable by claim 3 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable, and, being bounded, it is integrable over $[0,T]$: by claim 4 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (applied with the second function equal to the constant $1$), a $\mathcal{B}_{[0,T]}$-measurable $f$ with $\int_{[0,T]}f^2\,du<\infty$ has integrable absolute value, and here $f^2\le M^2$ gives $\int_{[0,T]}f^2\,du\le M^2\int_{[0,T]}1\,du=M^2T$ by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, the integral of the constant $1$ being $\lambda_{[0,T]}([0,T])=T$ by Simple Function and Its Integral and claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval; so $f$, and in particular $u\mapsto\phi(p(u))$, is integrable. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, $|\int_{[0,T]}\phi(p(u))\,du|\le\int_{[0,T]}|\phi(p(u))|\,du\le\int_{[0,T]}M\,du=MT$, and, the indicators $u\mapsto\mathbf{1}\{p(u)=y\}$ being integrable by the same argument,

It remains to see that $p\mapsto\int_{[0,T]}\mathbf{1}\{p(u)=y\}\,du$ is $\mathcal{C}_T$-measurable for each $y$. For each $n$ and each $p$, the map $u\mapsto\mathbf{1}\{p(g_n(u))=y\}$ is the nonnegative simple function $\mathbf{1}_{\{0\}}(u)\mathbf{1}\{p(0)=y\}+\sum_{i=1}^{n}\mathbf{1}_{I_{n,i}}(u)\mathbf{1}\{p(iT/n)=y\}$ on $[0,T]$, whose integral is $\sum_{i=1}^{n}(T/n)\mathbf{1}\{p(iT/n)=y\}$ by additivity (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) and Simple Function and Its Integral, the integral of $c\,\mathbf{1}_I$ being $c\,\lambda_{[0,T]}(I)$, with $\lambda_{[0,T]}(\{0\})=0$ and $\lambda_{[0,T]}(I_{n,i})=T/n$ by claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval and Existence of Lebesgue Measure on the Real Line; this is a $\mathcal{C}_T$-measurable function of $p$, being a linear combination of indicators of generators. For fixed $p$, the maps $u\mapsto\mathbf{1}\{p(g_n(u))=y\}$ are $\mathcal{B}_{[0,T]}$-measurable, converge pointwise on $[0,T]$ to $u\mapsto\mathbf{1}\{p(u)=y\}$ as shown above, and are dominated by the constant $1$, which is integrable over $[0,T]$ since $\lambda_{[0,T]}([0,T])=T<\infty$; by the dominated convergence theorem their integrals converge to $\int_{[0,T]}\mathbf{1}\{p(u)=y\}\,du$. Hence $p\mapsto\int_{[0,T]}\mathbf{1}\{p(u)=y\}\,du$ is a pointwise limit of $\mathcal{C}_T$-measurable maps, so it is $\mathcal{C}_T$-measurable, and the display shows that $p\mapsto\int_{[0,T]}\phi(p(u))\,du$ is a linear combination of measurable maps. This completes the proof.