I am following the proof of the martingale representation theorem in the book of Schilling and Partzsch and there is a part in it that I don't fully grasp.
Nomenclature: $B$ is Brownian motion and $X \in L^2_{\mathcal{P}}(\lambda_T \times P)$ means $X$ is a predictable process and $E[\int_0^T\lvert X_s \rvert^2\,ds]<\infty$.
Define: $$\mathcal{H}_T^2 = \left\{M_T : M_T = x + \int_0^TX_s\,dB_s \qquad x \in \mathbb{R}, \quad X \in L^2_{\mathcal{P}}(\lambda_T \times P)\right\}$$ Fact: $\mathcal{H}_T^2$ is a closed linear subspace of $L^2(P)$. Consider this already known. Also note that $e^{i\theta B_T} \in \mathcal{H}_T^2$ for every $\theta \in \mathbb{R}$.
Claim: $\mathcal{H}_T^2 = L^2(P)$.
The proof of this claim begins by saying that given the fact above we only need to show that $\mathcal{H}_T^2$ is dense in $L^2(P)$, which makes sense. But then the authors say it is sufficient to show that for any $Y \in L^2(P)$ $$Y \perp \mathcal{H}_T^2 \implies Y = 0$$ This is also fine but what does that have to do with density? The "latter" approach uses the projection theorem as far as I can tell. When someone says they will show $A$ is dense in $B$ I assume they will construct some approximating sequence in $A$ of an arbitrary element in $B$.
Moving on, the authors show that
$$Y \perp e^{i\theta B_T} \implies Y= 0$$
and conclude the proof of the claim without any further remarks. Obviously, there are a lot more elements in $\mathcal{H}_T^2$ so what is it about $x \mapsto e^{i\theta x}$ that makes it sufficient for this purpose? I know that the family of functions generated by considering integer values of $\theta$ forms an orthonormal basis for square integrable functions on the interval $[0,1]$ but I don't see how that helps here.