Proof of Progressive Measurability: Sections, Right-Continuous Adapted Processes, Arithmetic, and Indefinite Time Integrals

Throughout, $(\Omega,\mathcal{F},P)$, $T$, and the filtration $(\mathcal{F}_t)_{t\in[0,T]}$ are as in the statement, $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra on the real line, and for $t\in[0,T]$ the $\sigma$-algebra $\mathcal{B}_{[0,t]}$ on $[0,t]$ is as in the definition of progressive measurability. Measurable, for a real-valued map, means measurable with respect to the named $\sigma$-algebra and $\mathcal{B}(\mathbb{R})$; sums, products, constants, indicators, absolute values, and pointwise limits of measurable real-valued maps on a fixed measurable space are handled by the measurable-arithmetic lemma, cited by claim below. We first record three principles.

Composition. If $(U,\mathcal{U})$, $(V,\mathcal{V})$, and $(W,\mathcal{W})$ are measurable spaces and $\varphi:U\to V$ and $\psi:V\to W$ are measurable, then $\psi\circ\varphi$ is measurable, since $(\psi\circ\varphi)^{-1}(C)=\varphi^{-1}\big(\psi^{-1}(C)\big)\in\mathcal{U}$ for every $C\in\mathcal{W}$.

Insertion and projection. Let $(U,\mathcal{U})$ and $(V,\mathcal{V})$ be measurable spaces and let $\mathcal{U}\otimes\mathcal{V}$ be their product $\sigma$-algebra, generated by the measurable rectangles $A\times B$. For a fixed $v_0\in V$, the insertion $u\mapsto(u,v_0)$ is measurable for $\mathcal{U}$ and $\mathcal{U}\otimes\mathcal{V}$: the preimage of a rectangle $A\times B$ is $A$ if $v_0\in B$ and $\emptyset$ otherwise, a member of $\mathcal{U}$ either way, so claim 2 of the generator criterion applies. Symmetrically $v\mapsto(u_0,v)$ is measurable for a fixed $u_0\in U$. The projections $(u,v)\mapsto u$ and $(u,v)\mapsto v$ are measurable for $\mathcal{U}\otimes\mathcal{V}$ and $\mathcal{U}$, respectively $\mathcal{V}$, the preimages of $A$ and of $B$ being the rectangles $A\times V$ and $U\times B$.

Enlargement. If $\mathcal{V}\subseteq\mathcal{V}'$ are $\sigma$-algebras on $V$, then $\mathcal{U}\otimes\mathcal{V}\subseteq\mathcal{U}\otimes\mathcal{V}'$: every measurable rectangle of the former is one of the latter, so the generated $\sigma$-algebra $\mathcal{U}\otimes\mathcal{V}$ is contained in $\mathcal{U}\otimes\mathcal{V}'$ by minimality. Consequently a map measurable for $\mathcal{U}\otimes\mathcal{V}$ is measurable for $\mathcal{U}\otimes\mathcal{V}'$.

Step 1: claim 1. Let $X$ be progressively measurable and fix $t\in[0,T]$. The restriction $R_t$ of $(s,\omega)\mapsto X_s(\omega)$ to $[0,t]\times\Omega$ is measurable for $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$. Composing with the insertion $\omega\mapsto(t,\omega)$, measurable for $\mathcal{F}_t$ and $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$, exhibits $X_t=R_t\circ(\omega\mapsto(t,\omega))$ as $\mathcal{F}_t$-measurable; since $\mathcal{F}_t\subseteq\mathcal{F}$, $X_t$ is in particular a random variable, so $X$ is adapted. Likewise, for fixed $\omega$, the path section $s\mapsto X_s(\omega)$ on $[0,t]$ is $R_t\circ(s\mapsto(s,\omega))$, measurable for $\mathcal{B}_{[0,t]}$ by the insertion and composition principles. Finally, the case $t=T$ of progressive measurability says precisely that $(s,\omega)\mapsto X_s(\omega)$ on $[0,T]\times\Omega$ is measurable for $\mathcal{B}_{[0,T]}\otimes\mathcal{F}_T$, and the enlargement principle with $\mathcal{F}_T\subseteq\mathcal{F}$ gives measurability for $\mathcal{B}_{[0,T]}\otimes\mathcal{F}$.

Step 2: claim 2. Let $X$ be adapted with right-continuous paths and fix $t\in[0,T]$. Suppose first $t=0$. The restriction is $(0,\omega)\mapsto X_0(\omega)$, the composition of the projection $(0,\omega)\mapsto\omega$, measurable for $\mathcal{B}_{[0,0]}\otimes\mathcal{F}_0$ and $\mathcal{F}_0$, with the $\mathcal{F}_0$-measurable $X_0$; the composition principle applies.

Now let $t>0$ and fix a natural number $n$. Define $g_n:[0,t]\to[0,t]$ by $g_n(0)=0$ and $g_n(s)=kt/n$ for $s\in\big((k-1)t/n,\,kt/n\big]$, $k\in\{1,\dots,n\}$, and put $X^n(s,\omega)=X_{g_n(s)}(\omega)$ on $[0,t]\times\Omega$. Writing $\pi(s,\omega)=\omega$ for the projection and $J_k=\big((k-1)t/n,\,kt/n\big]$, we have, pointwise on $[0,t]\times\Omega$,

Each $J_k$ and the singleton $\{0\}$ belong to $\mathcal{B}_{[0,t]}$: rays $(a,\infty)$ belong to $\mathcal{B}(\mathbb{R})$ by claim 2 of the Borel generator lemma, so every interval $(a,b]=(a,\infty)\setminus(b,\infty)$ is Borel, as is $\{0\}=\big(\bigcap_{j\in\mathbb{N}}(-\tfrac1j,\infty)\big)\setminus(0,\infty)$, and $J_k\subseteq[0,t]$ and $\{0\}\subseteq[0,t]$ are then of the form (Borel set) $\cap\,[0,t]$. Hence the indicators $\mathbf{1}_{J_k\times\Omega}$ and $\mathbf{1}_{\{0\}\times\Omega}$ are measurable for $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$ (indicators of measurable rectangles; claim 1 of the arithmetic lemma), and each $X_{kt/n}\circ\pi$ is measurable for $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$ by adaptedness ($X_{kt/n}$ is $\mathcal{F}_{kt/n}$-measurable, and $\mathcal{F}_{kt/n}\subseteq\mathcal{F}_t$ by the filtration property) together with the projection and composition principles; likewise $X_0\circ\pi$. By claims 2 and 3 of the arithmetic lemma, $X^n$ is measurable for $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$.

At every $(s,\omega)\in[0,t]\times\Omega$, $X^n(s,\omega)\to X_s(\omega)$ as $n\to\infty$. Indeed, for $s=0$ every term equals $X_0(\omega)$. For $s>0$ and each $n$ there is exactly one $k$ with $s\in J_k$, and then $s\le g_n(s)=kt/n<s+t/n$, so $|g_n(s)-s|\le t/n$; given $\varepsilon>0$, claim 3 of the Archimedean property yields a natural $n_0$ with $\tfrac1{n_0}<\varepsilon/t$, and $t/n\le t/n_0<\varepsilon$ for $n\ge n_0$, so $t/n\to0$ and $g_n(s)\to s$ by domination (claim 3 of Order Properties of Limits of Real Sequences). The sequence $(g_n(s))_{n\in\mathbb{N}}$ lies in $[s,t]\subseteq[s,T]$, so $X^n(s,\omega)=X_{g_n(s)}(\omega)\to X_s(\omega)$ by right-continuity of the path at $s$. By claim 5 of the arithmetic lemma (pointwise limits), the restriction of $(s,\omega)\mapsto X_s(\omega)$ to $[0,t]\times\Omega$ is measurable for $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$. As $t$ was arbitrary, $X$ is progressively measurable.

Step 3: claim 3. Fix $t\in[0,T]$. The restrictions to $[0,t]\times\Omega$ of the families $cX$, $X+Y$, and $XY$ are the scalar multiple, sum, and product of the restrictions of $X$ and of $Y$, each measurable for $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$; claims 2 and 3 of the arithmetic lemma give their measurability. For the deterministic family: if $t>0$, the restriction $u|_{[0,t]}$ is continuous on $[0,t]$ by claim 1 of the restriction lemma, hence measurable for $\mathcal{B}_{[0,t]}$ and $\mathcal{B}(\mathbb{R})$ by claim 4 of the measurable-limits toolkit applied on the interval $[0,t]$; the map $(s,\omega)\mapsto u(s)$ on $[0,t]\times\Omega$ is its composition with the projection $(s,\omega)\mapsto s$, measurable by the projection and composition principles. If $t=0$ the restriction is the constant $(0,\omega)\mapsto u(0)$, measurable by claim 1 of the arithmetic lemma. Hence the deterministic family is progressively measurable, and the weighted family $(u(t)X_t)_{t\in[0,T]}$ is progressively measurable as the product of two progressively measurable families.

Step 4: claim 4. Existence and bound. Fix $\omega\in\Omega$ and $t\in(0,T]$ (for $t=0$ the integral is $0$ by the convention of the statement). By claim 1 the path section $x(s)=X_s(\omega)$, $s\in[0,t]$, is measurable for $\mathcal{B}_{[0,t]}$, with $|x|\le K$ everywhere. Its positive and negative parts $x^{\pm}$, as in the definition of the Lebesgue integral, are measurable, nonnegative, and bounded by $K$, hence also measurable in the ray sense of Lebesgue Integral of a Nonnegative Measurable Function, which for nonnegative real-valued maps agrees with Borel measurability as recorded in that definition. The constant function $K$ on $[0,t]$ is $K\,\mathbf{1}_{[0,t]}$, whose integral is $K\,\lambda_{[0,t]}([0,t])=K\,t$ by the indicator-integral lemma, claim 1 of linearity and monotonicity of the Lebesgue integral, and claim 1 of the interval toolkit; so by the monotonicity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral,

Hence $x$ is integrable with respect to $\lambda_{[0,t]}$, $Y_t(\omega)=\int_{[0,t]}x\,d\lambda_{[0,t]}$ is defined, and $|Y_t(\omega)|\le\int_{[0,t]}|x|\,d\lambda_{[0,t]}\le K\,t$ by claim 2 of Linearity and Monotonicity of the Lebesgue Integral (the modulus bound and monotonicity, with $|x|\le K$).

Increments. Let $0\le r\le t\le T$ and fix $\omega$. If $r=0$ the bound $|Y_t-Y_0|=|Y_t|\le Kt$ was just proved (and for $t=0$ there is nothing to prove); so let $0<r<t$ (the case $r=t$ is trivial). Write $\tilde{h}$ for the zero extension to $\mathbb{R}$ of a function $h$ on a compact interval, as in claim 2 of the interval toolkit. With $x$ the path section on $[0,t]$ as above and $x_r$ its restriction to $[0,r]$, claim 2 of the toolkit gives $Y_t(\omega)=\int_{\mathbb{R}}\tilde{x}\,d\lambda$ and $Y_r(\omega)=\int_{\mathbb{R}}\widetilde{x_r}\,d\lambda$, with $\lambda$ Lebesgue measure. Pointwise on $\mathbb{R}$, $\tilde{x}=\widetilde{x_r}+\tilde{x}\,\mathbf{1}_{(r,t]}$: the two summands agree with $\tilde{x}$ on $[0,r]$ and on $(r,t]$ respectively and vanish elsewhere. All three functions are integrable with respect to $\lambda$: the first two by claim 2 of the toolkit and the integrability already proved, and the third because it is measurable (the interval $(r,t]$ is Borel as in Step 2, and the product with its indicator is measurable by claims 1 and 3 of the arithmetic lemma) with $|\tilde{x}\,\mathbf{1}_{(r,t]}|\le K\,\mathbf{1}_{(r,t]}$, whose integral is $K\,\lambda\big((r,t]\big)\le K\,\lambda\big([r,t]\big)=K\,(t-r)$ by the indicator-integral lemma, monotonicity of a measure, and claim 1 of the interval toolkit (whose restricted measure agrees with $\lambda$ on $\mathcal{B}_{[r,t]}$, so $\lambda([r,t])=t-r$). By linearity (claim 2 of Linearity and Monotonicity of the Lebesgue Integral),

by the modulus bound and monotonicity of claim 2 of Linearity and Monotonicity of the Lebesgue Integral applied to $\pm\,\tilde{x}\,\mathbf{1}_{(r,t]}\le K\,\mathbf{1}_{(r,t]}$.

Path continuity. Fix $\omega$. For all $r,t\in[0,T]$, $|Y_t(\omega)-Y_r(\omega)|\le K|t-r|$ (the increment bound, in either order). Given $t\in[0,T]$ and $\varepsilon>0$, take $\delta=\varepsilon/(K+1)$: every $r\in[0,T]$ with $|r-t|<\delta$ has $|Y_r(\omega)-Y_t(\omega)|\le K|r-t|<\varepsilon$. This is continuity of the path at $t$ relative to $[0,T]$, both sides carrying the metric of the real line.

Adaptedness. $Y_0=0$ is constant, hence $\mathcal{F}_0$-measurable (claim 1 of the arithmetic lemma). Fix $t\in(0,T]$. The restriction $R_t$ of $X$ to $[0,t]\times\Omega$ is measurable for $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$; its pointwise positive and negative parts $R_t^{\pm}=\max(\pm R_t,0)$ are measurable by claims 1, 2, and 4 of the arithmetic lemma, nonnegative, bounded by $K$, and measurable in the sense of Lebesgue Integral of a Nonnegative Measurable Function as noted above. The triple $([0,t],\mathcal{B}_{[0,t]},\lambda_{[0,t]})$ is a finite measure space by claim 1 of the interval toolkit, and $(\Omega,\mathcal{F}_t,P_t)$, with $P_t$ the restriction of $P$ to $\mathcal{F}_t$, is a probability space: $\mathcal{F}_t$ is a $\sigma$-algebra on $\Omega$, and the defining properties of a probability measure for $P_t$ are instances of those of $P$, every set involved lying in $\mathcal{F}_t\subseteq\mathcal{F}$. Both spaces have finite total measure, hence are $\sigma$-finite (covered by a single set of finite measure). By the Tonelli clause of Tonelli and Fubini Theorems applied to $R_t^{+}$ and to $R_t^{-}$ on this product — their domain $\sigma$-algebra is exactly $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$ — the maps $\omega\mapsto\int_{[0,t]}R_t^{\pm}(s,\omega)\,\lambda_{[0,t]}(ds)$ are $\mathcal{F}_t$-measurable in the $[0,\infty]$-valued ray sense of Lebesgue Integral of a Nonnegative Measurable Function; their values are finite (at most $Kt$, as in the first part), so they are real-valued and $\mathcal{F}_t$-measurable in the Borel sense by the bridging sentence of that definition, and at every $\omega$,

by the definition of the Lebesgue integral applied to the section $x=R_t(\cdot\,,\omega)$, whose positive and negative parts are the sections of $R_t^{\pm}$. So $Y_t$ is $\mathcal{F}_t$-measurable by claim 2 of the arithmetic lemma, and $Y$ is adapted.

Progressive measurability. Every path of $Y$ is continuous on $[0,T]$, hence right-continuous in the sense of the statement: let $s\in[0,T]$ and let $(s_j)_{j\in\mathbb{N}}$ be a sequence in $[s,T]$ converging to $s$; given $\varepsilon>0$, continuity of the path at $s$ relative to $[0,T]$ provides $\delta>0$ with $|Y_{s'}(\omega)-Y_s(\omega)|<\varepsilon$ for all $s'\in[0,T]$ with $|s'-s|<\delta$, and the definition of the limit provides an index beyond which $|s_j-s|<\delta$; hence $Y_{s_j}(\omega)\to Y_s(\omega)$. By claim 2, $Y$ is progressively measurable. $\blacksquare$