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I am self-learning basic stochastic calculus. In my book, the author first defines the Ito integral for simple step adapted processes and then extends it to a larger class $\mathcal{L}_{c}^{2}(T)$ of integrands. This class has processes $(X_t:t\geq 0)$ satisfying:

(1) $X_t$ is adapted.

(2) The norm of $X_t$:

$$\int_{0}^{T}E[X_t^2]dt < \infty$$

(3) $(X_t)$ is almost surely continuous.

We proceed by verifying the below claim.

Approximation Lemma. Let $X\in \mathcal{L}_{c}^{2}(T)$. Then, there exists a sequence $(X^{(n)})$ of simple step adapted processes in $S(T)$, such that:

$$\lim_{n \to \infty} \int_{0}^{T} \mathbf{E}[(X_{t}^{(n)} - X_t)^2]dt = 0$$

My book is terse, so I want to expand on the steps.

(1) Does an $\epsilon-\delta$ style proof for step $1$ make sense, and is it rigorous? (2) How do I apply DCT in step $2$?

Proof.

(1) For a given $n$, consider the partition $\{\frac{jT}{2^{n}},\frac{(j+1)T}{2^{n}}\}$ and the simple step adapted process given by :

$$X_{t}^{(n)}=\sum_{j=0}^{n}X_{t_{j}}\mathbf{1}_{(t_{j},t_{j+1}]}(t)$$

In other words, we give the constant value $X_{t_{j}}$ on the whole interval $(t_{j},t_{j+1}]$. By continuity of paths of $X$, it is clear that $X_{t}^{(n)}(\omega)\to X_{t}(\omega)$ at any $t\leq T$ and for any $\omega$.

$(\star)$ Justification.

Let $A$ be the set of all $\omega$, such that $\lim_{t\to s}X(t,\omega)=X(s,\omega)$.Then, $\mathbb{P}(A)=1$.

Pick an arbitrary $\epsilon>0$ and fix a point $s\in[0,T]$. By definition of continuity, $(\forall\omega\in A)$, $(\exists\delta>0)$ such that $|t-s|<\delta$ implies $|X_{t}-X_{s}|<\epsilon$. There exists $N\in\mathbf{N}$, such that $\frac{1}{2^{N}}<\delta$.

Consider $(X^{(n)}:n\geq N)$. There exists a sequence of dyadic intervals $I_{N}\supset I_{N+1}\supset\ldots$ containing the point $s$.

Thus, the process $X^{(n)}$ takes the (random) but constant value $X_{\frac{jT}{2^{n}}}^{(n)}$ on the interval $\frac{jT}{2^{n}}<t\leq\frac{(j+1)T}{2^{n}}$.

For all $n\geq N$, since $l(I_{n})<\delta$, it follows that $|X_{\frac{jT}{2^{n}}}-X_{s}|<\epsilon$. But, $X^{(n)}$ takes the value $X_{\frac{jT}{2^{n}}}$ over $\left(\frac{jT}{2^{n}},\frac{(j+1)T}{2^{n}}\right]$. So, for all $n\geq N$, $X_{s}^{(n)}=X_{\frac{jT}{2^{n}}}$. Consequently, for all $n\geq N$, $\left|X_{s}^{(n)}-X_{s}\right|<\epsilon$.

This is true for all $\omega\in A$. Thus, $X_{t}^{(n)}\xrightarrow{a.s.}X_t$. $\blacksquare$

(2) Therefore, by the dominated convergence theorem, we have:

$$\lim_{n \to \infty} \int_{0}^{T} \mathbf{E}[(X_t^{(n)} - X_t)^2] dt = 0$$

($\star$) Justification.

I am not sure how to go about this. DCT has a certain bound condition that needs to be fulfilled, before we can write:

$$\lim_{n \to \infty} \int_{0}^{T} \mathbf{E}[(X_{t}^{(n)} - X_t)^2]dt = 0$$

So, any inputs would help.

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  • $\begingroup$ This question has been answered on math.stackexchange. $\endgroup$
    – mark leeds
    Jul 29, 2023 at 2:11

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