Question about Rene Schilling's construction of Brownian Motion using complete ONS: Taking a $L^2$ limit out of the exponential This is part of the proof of Lemma 3.1 from Rene Schilling's Brownian Motion.(The full proof is attached at the bottom of my question.)
Consider the Hilbert space $L^2(dt)=L^2([0,1],dt)$ with scalar product $\langle f,g \rangle_{L^2} = \int_0^1 f(t)g(t)dt$, and assume that $(\phi_n)_{n \ge 0}$ is any complete ONS and let $(G_n)_{n \ge 0}$ be a sequence of real-valued iid Gaussian $N(0,1)$ random variables on the probability space $(\Omega, \mathscr{A},P)$. Set 
$$W_N(t) := \sum_{n=0}^{N-1} G_n \langle 1_{[0,t)}, \phi_n \rangle_{L^2} = \sum_{n=0}^{N-1} G_n \int_0^t \phi_n(s) ds.$$ 
Then the limit $W(t):= \lim_{N \to \infty} W_N(t) $ exists for every $t \in [0,1]$ in $L^2(P)$ and the process $W(t)$ satisfies the properties of Brownian Motion.
Proof. 
The proof first shows that using the independence  of $G_n$ and Parseval's identity wee get for every $t \in [0,1]$ $E[W_N(t)]^2 = t$ and $W(t) = L^2-\lim_N W_N(t)$ exists. 
An analogous calculation yields for $s<t $ and $u<v$ 
$$E(W(t)-W(s))(W(v)-W(u)) = \sum_{n=0}^\infty \langle 1_{[0,t)} - 1_{[0,s)}, \phi_n \rangle_{L^2} \langle 1_{[0,v)} - 1_{[0,u)}, \phi_n \rangle_{L^2} = \langle 1_{[s,t)} , 1_{[u,v)}\rangle_{L^2},$$ and we see that $E(W(t)-W(s))(W(v)-W(u)) = (v \wedge t - u \vee s)^+$, so $0$ if $[s,t) \cap [u,v) = \emptyset$. 
Question.
I have a question about the next line of the proof. It says in the text that: 
With this calculation we find for all $0 \le s < t \le u < v$ and $\xi , \eta \in \mathbb{R}$ 
$$E[\exp(i \xi ( W(t)-W(s)) + i \eta (W(v)-W(u)))] = \lim_N E[\exp(i \sum_{n=0}^{N-1} (\xi \langle 1_{[s,t)}, \phi_n \rangle + \eta 1_{[u,v)}, \phi_n \rangle ) G_n)].$$
I can't figure out how the above calculation is used to get this identity. What exactly allows us to take the limit outside of the exponent and the expectation when we have a $L^2$ limit? 
An argument I came up with is we can consider $g$ to be the bounded continuous function $g(x) = \exp(i ( \xi f(x) + \eta h(x)))$, where $f_n \to f$ in $L^2$ and $h_n \to h$ in $L^2$ ( Take $f_n = W_n(t) - W_n(s)$ and $h_n = W_n(v)-W_n(u)$. ) Then by Vitali's generalized dominated convergence theorem, we would get $\lim_n \exp(i(\xi f_n(x)+\eta h_n(x)))=g(x)$, which gives the identity above. 
However, this argument does not use the calculation  $E(W(t)-W(s))(W(v)-W(u)) = (v \wedge t - u \vee s)^+$. So I don't think this is what the author intended. 
I would greatly appreciate a justification of this limiting argument.
I attach below the full proof.


 A: Here is how I understand it, though I am not exactly sure what the author actually intended. First, for every $t \in [0,1]$, the sequence $(W_N(t))$ converges to $W(t)$ in $L^2(\mathbb{P})$. Therefore, for $0 \leq s < t \leq u < v$, the sequence $(W_N(t)-W_N(s),W_N(v)-W_N(u))$ converges to $(W_N(t)-W_N(s),W_N(v)-W_N(u))$ in $L^2(\mathbb{P})$, which means the characteristic functions converge: for any $\xi, \eta \in \mathbb{R}$,
$$
\mathbb{E} \left [ e^{i \xi (W_N(t) - W_N(s)) + i \eta (W_N(v) - W_N(u))} \right ] \to \mathbb{E} \left [ e^{i \xi (W(t) - W(s)) + i \eta (W(v) - W(u))}\right ].
$$
The rest of the computation shows that the left-hand side is
$$
\exp \left [ - \frac12 \sum_{n=0}^{N-1} \left ( \xi^2 \langle 1_{[s,t)}, \phi_n  \rangle^2 + \eta^2 \langle 1_{[u,v)}, \phi_n  \rangle^2 \right ) \right ].
$$
But
$$
\sum_{n=0}^{N-1} \langle 1_{[s,t)}, \phi_n  \rangle^2 \to \langle 1_{[s,t)}, 1_{[s,t)} \rangle = t - s
$$
in $L^2([0,1])$ since $(\phi_n)$ is a complete ONS. Similarly for the other part, and as $L^2$ convergence is preserved by bounded continuous functions, this yields that
$$
\exp \left [ - \frac12 \sum_{n=0}^{N-1} \left ( \xi^2 \langle 1_{[s,t)}, \phi_n  \rangle^2 + \eta^2 \langle 1_{[u,v)}, \phi_n  \rangle^2 \right ) \right ] \to \exp \left [ - \frac12  \xi^2 (t-s) + \eta^2 (v-u) \right ],
$$
and thus finally
$$
\mathbb{E} \left [ e^{i \xi (W(t) - W(s)) + i \eta (W(v) - W(u))}\right ] = e^{- \frac12  \xi^2 (t-s) + \eta^2 (v-u)}.
$$
But indeed, this does not use the computation of the second moment, which is definitely not enough to compute characteristic functions.
