Proof of Adapted Mean-Square Continuous Processes are Ito Integrable

Write $\lVert\cdot\rVert_2$ for the mean-square norm of Square-Integrable Random Variables and the Mean-Square Inner Product, and for $t\in(0,T]$ and $k\ge1$ let $\tau_k(t)=iT/k$ for the unique $i$ with $t\in(iT/k,(i+1)T/k]$, so $H^k_t=H_{\tau_k(t)}$ and $0\le t-\tau_k(t)\le T/k$.

Step 1 (Simple adapted). For fixed $k$, the data $\bigl((iT/k)_{i=0}^{k},(H_{iT/k})_{i=0}^{k-1}\bigr)$ is a representation in the sense of Simple Adapted Process: the coefficients $H_{iT/k}$ are square-integrable and $\mathcal{F}_{iT/k}$-measurable by hypothesis. So $H^k$ is a simple adapted process on $(0,T]$.

Step 2 (Condition (b)). Fix $t\in(0,T]$. Since $\tau_k(t)\to t$ and $(H_u)$ is mean-square continuous at $t$, we get $\lVert H^k_t-H_t\rVert_2=\lVert H_{\tau_k(t)}-H_t\rVert_2\to0$ as $k\to\infty$.

Step 3 (Condition (a)). Let $R=\int_{\mathbb{R}}\mathbf{1}_{(0,T]}\rho\,d\lambda$, which is finite by clause (iv) of Ito Integrator of Intensity Type; if $R=0$ condition (a) is trivial (every integral there vanishes), so assume $R>0$. Let $\varepsilon>0$. By Uniform Mean-Square Continuity on a Compact Interval there is $\delta>0$ such that $\lVert H_u-H_v\rVert_2^{2}<\varepsilon/(2R)$ whenever $u,v\in[0,T]$ with $|u-v|<\delta$. Choose $K$ with $2T/K<\delta$. For $j,k\ge K$ and every $t\in(0,T]$, both $\tau_j(t)$ and $\tau_k(t)$ lie within $T/K$ of $t$, so $|\tau_j(t)-\tau_k(t)|<\delta$ and

By monotonicity of the Lebesgue integral (Linearity and Monotonicity of the Lebesgue Integral),

Thus $(H^k)$ is an approximating sequence for $(H_t)_{t\in(0,T]}$, which is therefore It^{o} integrable on $(0,T]$. This proves claim 1.

Step 4 (Explicit isometry). By the final clause of Uniform Mean-Square Continuity on a Compact Interval (with $0<T$), $e(t)=\mathbb{E}[H_t^{2}]$ is continuous on $[0,T]$, both $[0,T]$ and the codomain $\mathbb{R}$ carrying the metric of the real line. Write $\bar e$ for the extension of $e$ to $\mathbb{R}$ by $0$. Then $\bar e$ is measurable. Indeed, let $a\ge0$; since $\bar e$ vanishes off $[0,T]$, its superlevel set is $E_a=\{t\in[0,T]:e(t)>a\}$. If $t_0\in E_a$, continuity of $e$ at $t_0$ applied with $\varepsilon=e(t_0)-a>0$ supplies $\delta_{t_0}>0$ such that $e(t)>a$ for every $t\in[0,T]$ with $|t-t_0|<\delta_{t_0}$; hence $E_a=O_a\cap[0,T]$ with $O_a=\bigcup_{t_0\in E_a}(t_0-\delta_{t_0},t_0+\delta_{t_0})$ open, and $[0,T]$ is a Borel set, so $E_a$ is Borel. For $a<0$ the superlevel set is all of $\mathbb{R}$. The generator criterion of Measurable Function and Real-Valued Measurable Function therefore applies. The same argument applied to $\mathbf{1}_{(0,t]}\bar e$, whose superlevel sets for $a\ge0$ are $E_a\cap(0,t]$, shows that $\mathbf{1}_{(0,t]}\bar e$ is measurable for every $t\in(0,T]$. Now fix $t\in(0,T]$. The family $(H_u)_{u\in[0,t]}$ satisfies the hypotheses of the present lemma with $T$ replaced by $t$, so by claim 1 (applied on $(0,t]$) it is It^{o} integrable there with its own approximating sequence; by claim 2 of Existence and Uniqueness of the Mean-Square Extension of the Elementary Stochastic Integral, the resulting It^{o} integral agrees almost surely with $\int_0^t H_u\,dM_u$ in the sense of Ito Integrable Process and the Ito Integral (which is built from the restricted approximating sequence for the same family), and in particular the two have the same second moment. By claim 3 of Properties of the Ito Integral: Linearity, Isometry, Martingale Property, and Mean-Square Continuity applied on $(0,t]$ with the measurable function $\mathbf{1}_{(0,t]}e$,

which is claim 2.

Step 5 (Deterministic integrands). Let $f:[0,T]\to\mathbb{R}$ be continuous on $[0,T]$, with $[0,T]$ and the codomain $\mathbb{R}$ both carrying the metric of the real line, and set $H_t=f(t)$, the constant random variable. Constants are measurable with respect to every $\sigma$-algebra (preimages are $\emptyset$ or $\Omega$) and square-integrable, and $\lVert H_u-H_v\rVert_2=|f(u)-f(v)|$ (the expectation of the constant $(f(u)-f(v))^{2}$), so mean-square continuity of $(H_t)$ follows from continuity of $f$. Also $\mathbb{E}[H_u^{2}]=f(u)^{2}$. Claim 3 is now the specialization of claims 1 and 2. $\square$