I have read the following theorem related to an $X \in L^2(\Omega)$ random variable's integral representation:
Let $w$ denotes a Wiener process with its natural filtration $\mathcal{F}$. Fix $T<\infty$. Let $h$ be an arbitrary $h\in L^{2}\left(\left[0,T\right]\right)$ deterministic function. The set $\mathcal{U}$ of processes that has the form
$$\left(\mathcal{E}\left(h\bullet w\right)\right)\left(T\right)\dot{=}\exp\left(\int_{0}^{T}h_{t}dw_{t}-\frac{1}{2}\int_{0}^{T}h_{t}^{2}d\left[w\right]_{t}\right)$$
are dense in $L^{2}\left(\Omega,\mathcal{F}_T,\mathbf{P}\right)$.
The proof says that it is enough to prove that for any arbitrary $g\in\mathcal{F}_T$, if $g$ is orthogonal to the set of $\mathcal{U}$, then $g\overset{a.s.}{=}0$. Other words, if $$\mathbb{E}\left(\left(\left(\mathcal{E}\left(h\bullet w\right)\right)\left(T\right)\right)\cdot g\right)=0\Longrightarrow g\overset{a.s.}{=}0.$$
From this point it proves that indeed $g\overset{a.s.}{=}0$. However, I don't really understand why it implicates that $\mathcal{U}$ is dense in $L^{2}\left(\Omega,\mathcal{F}_T,\mathbf{P}\right)$. So currently I'm only interested in why this statement is enough to prove in order to prove that $\mathcal{U}$ is indeed dense in $L^{2}\left(\Omega,\mathcal{F}_T,\mathbf{P}\right)$?
The set $\mathcal U$ is not dense in $L^2$ as it contains only non-negative random variables with unit expectation. However, this shows that $\text{span}(\mathcal U)$ is dense in $L^2$, or alternatively that $\overline{\text{span}(\mathcal U)} = L^2$.
To prove this, given any $g \in L^2$, we can use Hilbert space projection to write $g = u + v$ where $u \in \overline{\text{span}(\mathcal U)}$ and $v$ is orthogonal to $\overline{\text{span}(\mathcal U)}$. In particular, $v$ is orthogonal to $\mathcal U$, so $v = 0$ a.s. This shows $g = u$ a.s., so $g \in \overline{\text{span}(\mathcal U)}$. Since $g \in L^2$ was arbitrary, we conclude $\overline{\text{span}(\mathcal U)} = L^2$.