Proof of The Pre-Stopping-Time Indicator and Stopped Time Integrals

Throughout, $\mathcal{B}_{[0,t]}$, $\mathcal{B}_{[0,T]}$, and the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ are as in the statement, and measurability is that of maps between measurable spaces. We use once and for all that a measurable real-valued function on a compact interval that is bounded in absolute value by a real $K\ge0$ is Lebesgue integrable there: its square is measurable with integral at most that of the constant $K^2$ by monotonicity of the nonnegative integral, the constant $K^2$ — a simple function — having integral $K^2\,\lambda_{[a,b]}([a,b])=(b-a)K^2<\infty$ directly from the definition of the nonnegative integral, with $\lambda_{[a,b]}([a,b])=b-a$ by claim 1 of the integral toolkit; so claim 4 of the same toolkit with $g=1$ shows the absolute value has finite integral, which is integrability. We call this the bounded-integrability remark.

Step 1 (claim 1). First let $t\in(0,T]$ and set

so that the restriction of $(s,\omega)\mapsto I_s(\omega)$ to $[0,t]\times\Omega$ is the indicator of $G_t$. We claim

where $\{u<\tau\}$ abbreviates $\{\omega\in\Omega:u<\tau(\omega)\}$, $[0,q)$ denotes $\{s\in[0,t]:s<q\}$, and $\mathbb{Q}$ is the set of rational numbers. For the inclusion of the right side in $G_t$: a point of $\{t\}\times\{t<\tau\}$ has $s=t<\tau(\omega)$, and a point of $[0,q)\times\{q<\tau\}$ has $s<q<\tau(\omega)$. Conversely, let $(s,\omega)\in G_t$. If $s=t$, then $(s,\omega)\in\{t\}\times\{t<\tau\}$. If $s<t$, then $s<\min(t,\tau(\omega))$, so by density of the rationals there is $q\in\mathbb{Q}$ with $s<q<\min(t,\tau(\omega))$; then $q\in\mathbb{Q}\cap[0,t)$ (as $0\le s<q<t$), $s\in[0,q)$, and $\omega\in\{q<\tau\}$.

Each set on the right belongs to the product $\sigma$-algebra $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$. Indeed $\{t\}=S_1\cap[0,t]$ and $[0,q)=S_2\cap[0,t]$ belong to $\mathcal{B}_{[0,t]}$, with $S_1=\{t\}$ closed and $S_2=(-1,q)$ open in the real line, hence Borel by claim 1 of the Borel toolkit together with its claim 2 identifying the metric Borel $\sigma$-algebra of the real line with $\mathcal{B}(\mathbb{R})$. By the definition of a stopping time and the closure of a $\sigma$-algebra under complements, $\{t<\tau\}=\Omega\setminus\{\tau\le t\}\in\mathcal{F}_t$, and $\{q<\tau\}=\Omega\setminus\{\tau\le q\}\in\mathcal{F}_q\subseteq\mathcal{F}_t$, the inclusion being the monotonicity of a filtration. Measurable rectangles belong to the product $\sigma$-algebra by its definition. Finally, $\mathbb{Q}$ is countable; fixing a surjection $n\mapsto q_n$ of $\mathbb{N}$ onto $\mathbb{Q}$ and setting $R_n=[0,q_n)\times\{q_n<\tau\}$ when $q_n\in[0,t)$ and $R_n=\emptyset$ otherwise, the union over $\mathbb{Q}\cap[0,t)$ equals $\bigcup_{n\in\mathbb{N}}R_n$, a countable union of members of $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$, hence a member. So $G_t\in\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$.

For $t=0$: $\mathcal{B}_{[0,0]}=\{\emptyset,\{0\}\}$ by the definition of progressive measurability, and $G_0=\{0\}\times\{0<\tau\}$ with $\{0<\tau\}=\Omega\setminus\{\tau\le0\}\in\mathcal{F}_0$, a measurable rectangle again.

Now fix $t\in[0,T]$ and a Borel set $S\subseteq\mathbb{R}$. The preimage of $S$ under the restriction of $I$ to $[0,t]\times\Omega$ is $\emptyset$, $G_t$, $([0,t]\times\Omega)\setminus G_t$, or $[0,t]\times\Omega$, according to which of the values $1$, $0$ lie in $S$; each of these belongs to $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$. Hence the restriction is measurable for every $t$, which is progressive measurability of $I$.

The remaining assertions of claim 1 follow: by claim 1 of the progressive measurability toolkit, $I$ is adapted — so $\{t<\tau\}$, the preimage of $\{1\}$ under the $\mathcal{F}_t$-measurable function $I_t$, belongs to $\mathcal{F}_t$ — and every path of $I$ is measurable on $[0,T]$ (the path section at $t=T$). That the path at $\omega$ equals $1$ on $[0,\tau(\omega))$ and $0$ on $[\tau(\omega),T]$ is immediate from the definition of $I$.

Step 2 (claim 2). Fix $\omega$, $K$, and $t$ as in the claim, and write $\rho=\min(t,\tau(\omega))\in[0,t]$. The function $s\mapsto I_s(\omega)X_s(\omega)$ on $[0,T]$ is measurable — a product of measurable real-valued functions is measurable by measurability of sequentially continuous functions of measurable maps, applied to $(x,y)\mapsto xy$ — and bounded by $K$; restrictions of measurable functions to compact subintervals are measurable (for Borel $S$, the preimage under the restriction is the intersection of the original preimage with the subinterval, and a trace of a trace is a trace); so by the bounded-integrability remark all integrals appearing below exist.

If $t=0$, both sides of the asserted identity are $0$ by the convention $\int_{[0,0]}\cdot\,ds=0$.

If $t>0$ and $\rho=0$ (that is, $\tau(\omega)=0$): the left side is $0$ by the convention, and the right side is $0$ because the integrand vanishes identically on $[0,t]$ — no $s\ge0$ satisfies $s<\tau(\omega)=0$ — and the integral of a function vanishing everywhere is $0$ by claim 1 of the null-set integral lemma (with the empty null set).

If $t>0$ and $\rho>0$: write $\mathbf{1}_{[0,\rho]}(s)$ for the function on $[0,t]$ equal to $1$ for $s\le\rho$ and $0$ otherwise (measurable, being the indicator of the trace of a closed set). The zero extension to $\mathbb{R}$ of the restriction of the path of $X$ to $[0,\rho]$ and the zero extension to $\mathbb{R}$ of $s\mapsto\mathbf{1}_{[0,\rho]}(s)X_s(\omega)$ on $[0,t]$ coincide: both equal $X_s(\omega)$ on $[0,\rho]$ and $0$ elsewhere. Applying claim 2 of the integral toolkit on $[0,\rho]$ and on $[0,t]$ therefore gives

Moreover, for $s\in[0,t]$ one has $\mathbf{1}_{[0,\rho]}(s)=1$ exactly when $s\le\tau(\omega)$ (given $s\le t$, the condition $s\le\min(t,\tau(\omega))$ reduces to $s\le\tau(\omega)$), while $I_s(\omega)=1$ exactly when $s<\tau(\omega)$; so the integrands $\mathbf{1}_{[0,\rho]}(s)X_s(\omega)$ and $I_s(\omega)X_s(\omega)$ agree for every $s\in[0,t]$ except possibly $s=\tau(\omega)$. The set $\{\tau(\omega)\}\cap[0,t]$ is either empty or the degenerate interval $[\tau(\omega),\tau(\omega)]$, whose restricted Lebesgue measure is its length $0$ by the definition of the restricted Lebesgue measure and the interval-length property of Lebesgue measure; so it is a null set, and by claim 2 of the null-set integral lemma (both integrands integrable and agreeing off it),

Combining the two displays proves claim 2.

Step 3 (claim 3). By claim 1 above and claim 3 of the progressive measurability toolkit, the product family $(I_tX_t)_{t\in[0,T]}$ is progressively measurable, and it is bounded by $K$ since $0\le I_t\le1$ everywhere. Claim 4 of the same toolkit then yields everything except the last identity: $J_t(\omega)$ is defined for every $t$ and $\omega$, $|J_t(\omega)-J_r(\omega)|\le K(t-r)$ for all $0\le r\le t\le T$ and $\omega$, every path of $J$ is continuous on $[0,T]$, and $J$ is adapted and progressively measurable. Finally, for every $\omega$ the path of $X$ is measurable on $[0,T]$ (claim 1 of the toolkit, $X$ being progressively measurable) and bounded by $K$, so claim 2 applies at every $\omega\in\Omega$ and every $t\in[0,T]$ and gives $J_t(\omega)=\int_{[0,\min(t,\tau(\omega))]}X_s(\omega)\,ds$. $\blacksquare$