# Construction of Ito Integral

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}}.

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.

• This question has been answered on math.stackexchange. Jul 29, 2023 at 2:11