I'm trying to prove that the stochastic integral defined for the set of square integrable local martingales is really an extension of ordinary stochastic integral. Define
$\mathcal{H}=\{(H_t)_{0\leq t\leq T}:H_t\,\, \text{adapted to}\,\,\mathcal{F}_t, E(\int_0^TH_s^2ds)<\infty\}$. And denote the stochastic integral by $J(H)_t$ for $H\in\mathcal{H}$.
Now $\tilde{\mathcal{H}}=\{(H_t)_{0\leq t\leq T}:H_t\,\, \text{adapted to}\,\,\mathcal{F}_t, \int_0^TH_s^2ds<\infty\,\,a.s.\}$.
Then my book says that there exists a unique linear map $\tilde{J}$ from $\tilde{\mathcal{H}}$ to the vector space of continuous processes defined on $[0,T]$ such that:
If $H$ is a simple process $\tilde{J}(H)_t=J(H)_t$ for all $t\in[0,T]$ and $P$-a.s.
If $(H_n)\subset \tilde{\mathcal{H}}$ such that $\int_0^T(H_s^n)^2ds$ converges to 0 in probability, then $\sup_{0\leq t\leq T}|\tilde{J}(H^n)(t)|$ converges to 0 in probability.
I want to show that for $H\in\mathcal H$, $\tilde{J}(H)_t=J(H)_t$ almost surely so that $\tilde{J}$ is really an extension. I tried to solve this by choosing the approximating simple process $H^n\in \mathcal{H}$, but I'm stuck in applying the second property of $\tilde{J}$.