If you know all about stochastic integration please just skip to my question; the first part is just fixing notation and background.
Let $\Omega, \mathcal{A},\mu$ be a probability space. And let $W:\Omega \times [0,T]\rightarrow \mathbb{R}$ be a Brownian motion adapted to a filtration $\mathcal{F}_t\subset \mathcal{A}, t\in [0,T]$. I just read how one may define a stochastic integral $\int_0^TXdW$ where $$X: \Omega \times [0,T]\rightarrow \mathbb{R}$$ is $ \mathcal{A}\otimes\mathcal{B}([0,T])$-measurable, X is $\mathcal{F}_t$-adapted, and where $X\in L^2(\Omega \times [0,T])$). The proof is based on the fact that the $L^2$ condition implies that there is a sequence of adapted $L^2$ "simple processes", $$\sigma_n(\omega,t)=\sum_{i=0}^{k}f_i(\omega)\cdot \mathbb{1}_{(t_i,t_{i+1}]}(t), ~~~(f_i\in L^2(\Omega,~\mathcal{F}_{t_i}),~t_0=0,~t_{k_n+1}=T)$$ converging to $X$ in $L^2(\Omega \times [0,T]).$ For $t\in (t_j,t_{j+1}]$ one defines $$(\int_0^T\sigma_ndW)(\omega,t)=f_j\cdot(W_{t}-W_{t_j}) +\sum_{i=0}^{j-1}f_i\cdot(W_{t_{i+1}}-W_{t_i}),$$and then one shows that $t\mapsto (\int_0^T\sigma_ndW)(\omega,t)$ is continuous for all $\omega,n$, and that for almost every $\omega$, the functions $t\mapsto (\int_0^T\sigma_ndW)(\omega,t)$ satisfy the Weierstrass M test, and thus converge to a continuous function $t\mapsto Y(\omega,t)$. We define $\int_0^TXdW:=Y$.
That is all clear.
Now the book I'm reading (Intro to Stochastic Calculus Applied to Finance by Lamberton and Lapeyre) says we need to extend this integral by relaxing the $X\in L^2(\Omega \times [0,T])$ condition. Namely the claim is that if $$\mathcal{H}= \{ X: \Omega \times [0,T]\rightarrow \mathbb{R}~~~|~~X ~is~\mathcal{F}_t- adapted, ~ and~for~almost~all~\omega~we~have~\int_0^T|X(\omega,t)|^2dt<\infty \}$$
and if $\mathcal{C}$ is the space of adapted and almost surely continuous processes on $\Omega\times [0,T]$, then there exists a unique linear mapping $$\mathcal{H}\rightarrow\mathcal{C},~X\mapsto \int_0^T XdW$$
satisfying the following two properties
This integral agrees with the previous definition for simple processes (the third centered line in this post).
If $H_n$ is a sequence of processes in $\mathcal{H}$ such that the sequence of function $\bigg(\omega\mapsto\int_0^TH_n^2(\omega,t)dt\bigg)_{n\in \mathbb{N}}$ converges in probability to $0$, then the sequence $$\bigg(\omega\mapsto sup_{t\leq T}|(\int_0^T H_ndW)(\omega,t)|\bigg)_{n\in \mathbb{N}}$$ also converges to $0$ in probability
Finally here's my question:
The book says that it is clear that conditions 1 and 2 imply that the new integral agrees with the old (almost surely for all $t\in [0,T]$) whenever the integrand $X$ is in $L^2(\Omega \times [0,T])$. Why?? The only thing I can think to do is to again take a sequence of simple processes, $\sigma_n$, converging to $X$ in $L^2(\Omega \times [0,T])$. Then clearly the sequence $$\bigg(\omega\mapsto\int_0^T|\sigma_n(s,\omega) ds - X(s,\omega)|^2 ds\bigg)_{n \in \mathbb{N}}$$ converges to $0$ in $L^1(\Omega)$ and therefore converges to $0$ in probability. Thus by 2, we have that $$\bigg(\omega \mapsto sup_{0\leq t \leq T}\bigg|(\int_0^T \sigma_ndW)(\omega, t)-(\int_0^TXdW)(\omega, t)\bigg|\bigg)_{n\in \mathbb{N}}$$ converges to zero in probability. Now what we want is that for almost all $\omega$, for all $t$ we have $$(\int_0^T \sigma_ndW)(\omega, t)-(\int_0^TXdW)(\omega, t)\rightarrow 0.$$ This would suffice because we know $\int_0^T \sigma_ndW$ agrees with the previously defined integral by property 1, and the sequence $\int_0^T \sigma_ndW$ converges for almost all $\omega$ for all $t$ to the previously defined integral of $X$. But as far as I can tell, convergence in probability doesn't guarantee this.