Confusion with stochastic integral in likelihood ratio I am having some trouble understanding part of this paper.

We observe a stochastic process $Y$ on $I:=[0,1]$ where
  $$Y(t) = n^{1/2} \int_0^t f(x) \mathop{dx} + W(t)$$
  where $f \in L^1(I)$ is an unknown function and $W$ is standard Brownian motion.
  Suppose we want to perform a simple hypothesis test with null $f=0$ and alternative $f=g$ for some fixed $g \in L^2(I)$. Since $$\log(dP_g/dP_0)(Y) = n^{1/2} \int_I g \mathop{dY} - n \|g\|^2/2,$$ the Neyman-Pearson test rejects the null hypothesis at level $\alpha$ if the linear test statistic $\|g\|_{2,I}^{-1} \int_I g(x) \mathop{dY(x)}$ exceeds the $(1-\alpha)$-quantile of the standard Gaussian distribution. Note $\int_I g\mathop{dY} \sim \mathcal{N}\left(n^{1/2} \langle f,g\rangle_I, \|g\|_{2,I}^2\right)$.

I cannot derive the log likelihood ratio stated above. If I look only at $Y(1) \sim \mathcal{N}\left(n^{1/2} \int_I f(x) \mathop{dx}, 1\right)$, I can take the log ratio of the two densities and get
$$\log \frac{dP_g}{dP_0}(Y)
= -\frac{\left(Y-n^{1/2}\int_I g(x) \mathop{dx}\right)^2}{2} + \frac{Y^2}{2} = n^{1/2} Y \int_I g(x) \mathop{dx}-\frac{n}{2}\left(\int_I g(x) \mathop{dx}\right)^2.$$
This seems close to what they have, but I am unable to make the connection.
 A: Girsanov Theorem can be used to compute the likelihood for this problem. Under $\mathbb{P}^0$ the process $Y$ is an ordinary Brownian motion and under $\mathbb{P}^g$ the process is given by 
$$Y_t=n^{1/2}\int_0 ^t g(s)ds+W(t).$$
Consider the process 
$$X_t={\sqrt{n}}\int _0^t g(s)dW_s$$
and consider the measure change given by 
$$\left.\frac{d\mathbb{Q}}{d\mathbb{P^0}}\right\vert _{\mathcal{F_t}}:=\mathcal{E}(X)_t:=\exp\left(X_t-\frac{1}{2}[X]_t\right).$$
Then (if $\mathcal{E}$ is a positive martingale) Girsanov can be applied and implies that, under $\mathbb{Q}$
$$Y_t\overset{d}{=}W_t+[W,X]_t=W_t+{n}\int_0^t g(s)ds$$
and hence $\mathbb{Q}=\mathbb{P}^g$.
It follows that 
$$\frac{d\mathbb{P}^g}{d\mathbb{P}^0}=\exp\left(\sqrt{n}\int_0^T g(s) dW_s - \frac{n}{2}\int_0^T g(s)^2 ds\right).$$
The last part in the exponent equals the $L^2$-norm.
As to your approach: The first part of the likelihood you computed equals the integral $n^{1/2}\int _I g(s)dY$ but only for the case where $Y$ does not vary in time (as you implicitly assumed in your computation by only looking at $Y(1)$). For the correct likelihood however you have to take into account the entire trajectory. In other words you calculated the likelihood using an 'approximation' of the trajectory consisting of a constant (in time).
The second part is again equal to the norm, which was also obtained in the original text.
