We fix some filtered probability space $(\Omega,\mathfrak{F},\{\mathfrak{F}_t\}_{t\in[0,T]},\mathbb{P})$. Let, for short, $L^2$ be the space of all progressively measurable processes in $L^2([0,T]\times\Omega)$ (we consider the product measure $\lambda\otimes\mathbb{P}$). We assume that it is known, that every continuous and bounded process $X\in L^2$ can be approximated in the $L^2$-sense by a sequence $(X^m)_{m\in\mathbb{N}}$ of simple processes.
Let $Y$ be a bounded (not necessarily continuous) process in $L^2$ for which it is known that there is some sequence of bounded, continuous processes $Y^m\in L^2$ converging in $L^2$-sense to $Y$. We want to prove that this sequence $(Y^m)$ can be replaced by a sequence of simple processes in $L^2$. In my lecture, it was suggested to approximate each $Y^m$ by a sequence $(Y^{m,n})_{n\in\mathbb{N}}$ of simple processes in $L^2$ and then to just take the "diagonal sequence" $(Y^{n,n})_{n\in\mathbb{N}}$ which then should converge to $Y$ in the $L^2$-sense. Does somebody know how to prove the latter convergence assertion?
Let $(Y^m)_{m \in \mathbb{N}}$ be a sequence of bounded continuous processes converging in $L^2$ to $Y$. By assumption, there exists a sequence $(Y^{m,n})_{n \in \mathbb{N}}$ of simple processes such that
$$Y^{m,n} \stackrel{n \to \infty}{\to} Y^m$$
for each fixed $m \in \mathbb{N}$. Taking a further subsequence (if necessary), we may assume that
$$\|Y^{m,n}-Y^m\|_{L^2} \leq \frac{1}{n} \qquad \text{for all $n \in \mathbb{N}$}. \tag{1}$$
Then, using $(1)$ with $m=n$, we get
$$\|Y^{n,n}-Y\| \leq \|Y^{n,n}-Y^n\| + \|Y^n-Y\| \leq \frac{1}{n} + \|Y^n-Y\|.$$
Since by assumption $Y^n \to Y$, we can choose $n \geq N$ sufficiently large such that
$$\|Y^{n,n}-Y\| \leq 2 \epsilon \qquad \text{for all $n \geq N$}.$$
This finishes the proof.