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 $|X_{t}-X_{s}|<\epsilon$. By the Archimedean property, 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 $I_{N}\subseteq I_{N+1}\subseteq\ldots$ 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.
For any $t \in [0,T]$, $X_{t}^{(n)} \to X_{t}$ almost surely. Since almost sure convergence implies $L^2-$convergence, it follows that for any $t \in [0,T]$,
$$\mathbf{E}[(X_t^{(n)} - X_t)^2] \to 0$$
I am not sure how to go from here. 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.
I decided to write an answer to more clearly explain what I was trying to say. As written, this does prove the assertion that bounded simple processes are dense in the space $\mathcal{L}_c^2(T)$, but not that the specific sequence of simple functions that the author proposes converges to $X_t$ in the $L^2$ sense.
We first prove the approximation lemma for $X_t$ that satisfies the above 3 conditions, and satisfies a boundedness condition: $$\exists M | \forall t \geq 0, a.s. \; \omega \in \Omega \quad |X_t(\omega)| \leq M.$$
In this case, $$X^{(n)}_t \xrightarrow{a.s} X_t$$ by the proof you gave above.
For verifying the second claim, note that under these assumptions $$(X_t^{(n)}(\omega) - X_t(\omega))^2 \leq (2M)^2$$ almost surely and that $\int_0^T \mathbb{E}[(2M)^2] dt < \infty.$ By dominated convergence, the second claim follows: for $|X_t(\omega)| \leq M$, $$\lim_{n \to \infty} \int_0^T \mathbb{E}[(X_t - X^{(n)})^2]dt = 0. \tag{(A)}.$$
For simplicity's sake, assume $X_t$ is a nonnegative process. To extend the result, note that almost surely for all $\omega$ and for all $t \geq 0$, $$\lim_{M \to \infty} M \wedge X_t(\omega) = X_t(\omega).$$ It's also straightforward to verify that $$|(X_t(\omega) - M \wedge X_t(\omega)|^2 \leq |X_t(\omega)|^2.$$ Further, since $\int_0^T \mathbb{E}[(X_t)^2] dt < \infty$, we will apply dominated convergence to conclude $$\lim_{M \to \infty} \int_0^T \mathbb{E}[(X_t - M \wedge X_t)^2] dt = 0 \tag{(B)}.$$
At this point, the book claims to prove the limit $\lim_{n \to \infty} \int_0^T \mathbb{E}[(X_t - X^{(n)})^2]dt = 0$ using the definition above. However, I will construct a different sequence of simple functions $Y^{(n)}_t$ that approximate $X_t$.
Fix $\epsilon_n= 2^{-n} > 0.$ From line $(B),$ there exists $M_n > 0$ large enough such that $$\int_0^T \mathbb{E}[(X_t - M \wedge X_t)^2]dt < \epsilon_n.$$
By line $(A)$, there exists an $N_n > 0$ such that $$\int_0^T \mathbb{E}[(M_n \wedge X_t - M_n \wedge X^{(N_n)}_t)^2] dt < \epsilon_n.$$
It follows that $Y^{(n)}_t := M_n \wedge X^{(N_m)}_t$ is close to $X_t$ in the $L_2$ sense, i.e.,
$$\int_0^T \mathbb{E}[(X_t - Y^{(n)}_t)^2]dt \leq 4\epsilon_n = 4\cdot 2^{-n}.$$
Note: the book proposes a different sequence of simple functions, but I don't see how they are bounded, let alone dominated above by $X_t$. If we just let $X_t(\omega) = 1-t$, it's clear that $X^{(n)}_t \geq X_t$.
If we know that $\mathbb{E}[X_t^2]$ was continuous (which is not a guarantee, see The continuity of the expectation of a continuous stochastic procees), we could perhaps conclude something, but I am not sure that the statement in the book holds without some argument more sophisticated than dominated convergence. Perhaps the author omitted an assumption.
If we assume a bound on the process like that of the Kolmogorov continuity criterion, (see Density of L2 simple processes) the simple functions given in this question will approximate the process. However, it's not a guarantee that our process is this nice.