Really struggling with this lemma.
Not sure about the general structure of the proof. Why have we chosen g to be orthogonal to all functions of the form 4.3.1?
Why should $G(\lambda)=0$, does it have something to do with the inner product on $L^2$.
What gives us the justification of the analytic extension and why would we want to do this?
I cannot find any justification in my book that if $G=0$ on $\mathbb{R^n}$ that as $G$ is analytic $G$ would be equal to $0$ on $\mathbb{C^n}$.
Next I cannot understand why we are using the fourier inversion.
Finally I cannot see how the last paragraph justifies that the subset is dense. Why is $g=0$ significant? I think this may be help me understand the general structure of the proof.
The proof is based on this standard fact from functional analysis:
So for this proof, $H = L^2(\mathcal{F}_T, P)$ and $E$ is the linear span described in the statement. We choose an arbitrary $g \in E^\perp$ and show we must have $g=0$. This will show that $E^\perp = \{0\}$ and hence $E$ is dense.
Next verify that if $h \in L^2([0,T])$ is a step function, having a jump of size $\lambda_i$ at the point $t_i$, $i=1,\dots,n$, and constant in between, then $\int_0^T h(t)\,dB_t = \lambda_1 B_{t_1} + \dots + \lambda_n B_{t_n}$. (This is immediate from the definition of the stochastic integral.) Then $$\exp\left(\int_0^T h(t)\,dB_t - \frac{1}{2} \int_0^t h(t)^2\,dt\right) = \exp\left(-\frac{1}{2} \int_0^T h(t)^2\,dt\right) \cdot \exp(\lambda_1 B_{t_1} + \dots + \lambda_n B_{t_n})$$ which is a scalar multiple of $\exp(\lambda_1 B_{t_1} + \dots + \lambda_n B_{t_n})$, which is therefore in $E$. Since $g$ was assumed to be in $E^\perp$, we have $$\int_\Omega \exp(\lambda_1 B_{t_1} + \dots + \lambda_n B_{t_n}) g\,dP = 0$$ which is (4.3.2).
The fact that $G$ has a complex analytic extension should be explained in Exercise 2.8(b). But rather than using facts about real analytic functions, it is probably easier to verify explicitly that $G$ is complex differentiable by differentiating under the integral sign (hopefully familiar from a measure theory course). You will need some estimates to justify this which are presumably the same ones from Exercise 2.8(b). Alternatively, use differentiation under the integral sign to show that $G$ satisfies the Cauchy-Riemann equations.
Now it is a standard fact from complex analysis that if $G$ is analytic on $\mathbb{C}^n$ and vanishes on $\mathbb{R}^n$, then $G=0$. When $n=1$ you can say more: if the zeros of $G$ have a limit point (which $\mathbb{R}^1$ certainly does) then $G=0$. You can find this as Theorem 6.9 in Bak and Newman. Our statement for general $n$ can then be proved by induction. Suppose it holds for $n-1$, and suppose $G : \mathbb{C}^n \to \mathbb{C}$ is analytic and vanishes on $\mathbb{R}^n$. Fix $w \in \mathbb{R}$ and consider the function $G_w(z_1, \dots, z_{n-1}) = G(z_1, \dots, z_{n-1}, w)$. Then $G_w : \mathbb{C}^{n-1} \to \mathbb{C}$ is analytic and vanishes on $\mathbb{R}^{n-1}$ so by the inductive hypothesis $G_w = 0$. Next fix $z_1, \dots, z_{n-1} \in \mathbb{C}$ and let $f(u) = G(z_1, \dots, z_{n-1}, u)$, so that $f : \mathbb{C} \to \mathbb{C}$ is analytic. For $u \in \mathbb{R}$ we have $f(u) = G_u(z_1, \dots, z_{n-1}) = 0$, so $f$ vanishes on $\mathbb{R}$ and by the $n=1$ case we have $f = 0$. Thus $G(z_1, \dots, z_{n-1}, u) = 0$ for all $u \in \mathbb{C}$. But $z_1, \dots, z_{n-1}$ were arbitrary so we must have $G=0$.
Finally, why do we use the Fourier transform? Well, because it works. Once we have the complex extension of $G$, we know that for any $y_1, \dots, y_n$ we get $\int_\Omega \exp\left(i y_1 B_{t_1} + \dots + i y_n B_{t_n}\right) g\,dP = 0$. But, roughly speaking, the Fourier inversion formula says that any reasonable function $\phi$ of $B_{t_1}, \dots, B_{t_n}$ is an "infinite linear combination" of functions of the form $\exp(\lambda_1 B_{t_1} + \dots + \lambda_n B_{t_n})$ (think of the integral in the inverse Fourier transform as a sort of infinite sum; in fact, it really is the limit of finite sums). So it is not surprising that we then get $\int_\Omega \phi\left(B_{t_1},\dots, B_{t_n}\right) g\,dP = 0$.
On the other hand, it had previously been shown (presumably Lemma 4.3.1) that the subspace $F$ spanned by functions of the form $\phi\left(B_{t_1},\dots, B_{t_n}\right)$, where $n$ ranges over all integers, was dense in $L^2(\mathcal{F}_T, P)$, and the conclusion of my previous paragraph says that $g \in F^\perp$. By the other direction of the theorem at the top of this answer, we conclude that $g=0$, which is what we needed.