I am having difficulty following the proof of Lemma 7 from the following post : https://almostsuremath.com/2010/03/03/existence-of-the-stochastic-integral/#scn_intexists_thm2
The goal here is to extend the stochastic integral operator $\mu$ with respect to a cadlag adapted process $X$, given by $\mu(\xi):=\int_0^t \xi dX$ for a fixed time $t$, from the elementary predictable processes $\xi$ to the bounded predictable processes $b\mathcal{P}$.
In doing so, we take $\mathcal{A}$ to be the set of uniformly bounded, continuous and adapted processes, which forms a subalgebra of $b\mathcal{P}$ and generates $b\mathcal{P}$.
The rest of the proof is straightforward analysis, however, I am not able to see some facts as stated here.
Elementary predicatble processes are of the form
$$\xi_t = Z_0 1_{t=0}+\sum_{k=1}^n Z_k 1_{s_k<t\le t_k}$$for an $\mathscr{F}_0$-measurable random variable $Z_0$, real numbers $s_k, t_k \ge 0$ and $\mathscr{F}_{s_k}$-measurable random variables $Z_k$. The integral with respect to any other process $X$ up to time $t$ can be written out explicitly as,
$$\int_0^t \xi dX= \sum_{k=1}^n Z_k (X_{t_k \wedge t}-X_{s_k \wedge t}).$$
My questions are as follows:
- So choosing $\epsilon>0$, we can smooth out any elementary predictable process $\xi$ can be smoothed out as a continuous and adapted process,
$$\zeta_s^{\epsilon} := \int_0^1 \xi_{(s-u\epsilon)\vee 0}du.$$
Why is this process continuous and adapted? Is it continuous because it is just a Riemann integral of a bounded function(for each fixed $\omega$)? And then adaptedness follows as it is a limit of Riemann sums of functions before time $s$?
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Why is the Riemann approximation to the above integral given by the elementary processes $n^{-1}\sum_{k=1}^n \xi_{(s-k\epsilon/n\vee 0}$ a uniform approximation?
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Given $X_s^n := n^{-1} \sum_{k=1}^n X_{(s+k\epsilon/n)\wedge t}$, how do we get the identity $$n^{-1}\int_0^t \sum_{k=1}^n \xi_{(s-k\epsilon/n)\wedge 0} dX_s = \int_0^t \xi dX^n + \xi_0(X_0^n -X_0)?$$
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It is clear that $X_s^n \to \tilde{X_s^\epsilon}$ as $n\to \infty$, but how can we pass the limit inside the integral here to get $$\lim_{n\to \infty} \int_0^t \xi dX^n = \int_0^t \xi d\tilde{X^\epsilon}?$$
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Finally, how are we able to justify $\zeta^\epsilon \to \xi$ and $\tilde{X^\epsilon}\to X$ here as we need to pass to the limits inside the integral again.
I would greatly appreciate some detailed explanation on these analysis proofs.
Best Answer
You are correct.
Denote $A_{n,k}(s):=n^{-1}\xi_{(s-k\epsilon/n)\lor 0}$, $B_{n,k}(s):=\int_{(k-1)/n}^{k/n}\xi_{(s-u\epsilon)\lor 0}du$. Then the Riemann sum is $$\frac{1}{n}\sum_{k=1}^{n}\xi_{s-(k\epsilon/n)\lor0}=\sum_{k=1}^{n}A_{n,k}(s),$$ and the integral is $$\int_{0}^{1}\xi_{(s-u\epsilon)\lor0}du=\sum_{k=1}^{n}B_{n,k}(s).$$ Since $\xi$ is an elementary process,
if $[t_{\ell}\leq s-(k-1)\epsilon/n<s-k\epsilon/n\leq s_{\ell}]$ for some $\ell$, then $A_{n,k}(s)=B_{n,k}(s)$.
If $[s-\epsilon(k-1)/n,s-\epsilon k/n]$ contains at least one discontinuity point of $\xi$, then $A_{n,k}$ can be different with $B_{n,k}$. We claim that for $s\in[0,t]$, there are at most $m(\xi,t)$ of such terms, since the number of discontinuity points on $[0,t]$ is at most $m(\xi,t)$. Also note that $\xi$ is uniformly bounded, say $\sup_{s\in[0,t]}|\xi|\leq K$. We further have $|A_{n,k}(s)-B_{n,k}(s)|\leq 2K/n$.
Therefore, we have $$\sup_{s\in[0,t]}\biggl|\sum_{k=1}^{n}A_{n,k}(s)-\sum_{k=1}^{n}B_{n,k}(s)\biggr|\leq \sup_{s\in[0,t]}\sum_{k=1}^{n}|A_{n,k}(s)-B_{n,k}(s)|\leq \frac{2Km(\xi,t)}{n}\to0.$$
For each fixed $k$, a change of variable $u\mapsto (s-k\epsilon/n)$ gives $$\begin{align*}\int_{0}^{t}\xi_{(s-k\epsilon/n)\lor0}dX_{s}&=\int_{k\epsilon/n}^{t}\xi_{s-k\epsilon/n}dX_{s}+\int_{0}^{k\epsilon/n}\xi_{0}dX_{s}\\ &=\int_{0}^{t-k\epsilon/n}\xi_{u}dX_{u+k\epsilon/n}+\xi_{0}(X_{k\epsilon/n}-X_{0})\\ &=\int_{0}^{t}\xi_{u}dX_{(u+k\epsilon/n)\land t}+\xi_{0}(X_{(0+k\epsilon/n)\land t}-X_{0}).\end{align*}$$ Taking summation from $k=1$ to $n$ yields the desired result.
I was confused looking at this too because of the misleading notation, as there are two pairs of $k$ and $n$ show up here. The $k$ and $n$ in the definition of $\xi=Z_{0}1_{\{t=0\}}+\sum_{k=1}^{n}Z_{k}1_{\{s_{k}<t\leq t_{k}\}}$ is fixed (even won't change with $\omega$) once $\xi$ has been chosen. As a result, for any fixed time $t$, in the definition of integral $\int_{0}^{t}\xi dX= \sum_{k=1}^n Z_k (X_{t_k \land t}-X_{s_k \land t})$, the summation $n$ can be seen as finite. To make it more clear, for each $\omega$, we have $$\begin{align*} \lim_{n\to\infty}\int_{0}^{t}\xi dX^{n}&:=\lim_{n\to\infty}\sum_{i=1}^{m(\xi,t)}Z_{i}(X_{t_{i}\land t}^{n}-X_{s_{i}\land t}^{n})\\ &=\sum_{i=1}^{m(\xi,t)}\lim_{n\to\infty}Z_{i}(X_{t_{i}\land t}^{n}-X_{s_{i}\land t}^{n})\\ &=\sum_{i=1}^{m(\xi,t)}Z_{i}(\tilde{X}_{t_{i}\land t}^{\epsilon}-\tilde{X}_{s_{i}\land t}^{\epsilon})=:\int_{0}^{t}\xi d\tilde{X}^{\epsilon}, \end{align*}$$ where the change of limit and summation is valid since $m(\xi,t)$ is finite.
By definition, $\zeta^{\epsilon}_{0}=\xi_{0}$, and for any $s>0$, a change of variable $v\mapsto (s-u\epsilon)$ gives $$\begin{align*}\lim_{\epsilon\downarrow 0}\zeta_{s}^{\epsilon}&=\lim_{\epsilon\downarrow0}\int_{0}^{1}\xi_{(s-u\epsilon)\lor 0}du=\lim_{\epsilon\downarrow0}\int_{0}^{1}\xi_{s-u\epsilon}du=\lim_{\epsilon\downarrow0}\frac{\int^{s}_{s-\epsilon}\xi_{v}dv}{\epsilon}.\end{align*}$$ Since $\xi$ is predictable hence left-continuous. We have $\lim_{\epsilon\downarrow0}\sup_{v\in[s-\epsilon,s]}\xi_{v}=\lim_{\epsilon\downarrow0}\inf_{v\in[s-\epsilon,s]}\xi_{v}=\xi_{s}$, the desired convergence holds. Similar result holds for $\tilde{X}^{\epsilon}$ using cadlag of $X$. The change of limit and integration follows from the same trick as in 4 (for $\tilde{X}^{\epsilon}$ part) and the bounded convergence assumption of this integral that author mentioned in the post. Since $\xi$ is uniformly bounded, i.e. $\sup_{s\in[0,t]}|\xi|\leq K$, $$\sup_{s\in[0,t]}|\zeta^{\epsilon}|=\sup_{s\in[0,t]}\biggl|\int_{0}^{1}\xi_{(s-u\epsilon)\lor0}du\biggr|\leq \sup_{s\in[0,t]}\int_{0}^{1}|\xi_{(s-u\epsilon)\lor0}|du\leq\int_{0}^{1}Kdu=K.$$