Let $(\Omega,\mathcal{A},\mathbb{P})$ a probability space. Let $(X_t)_{t \in [0,T]}$, $(Y_t)_{t \in [0,T]}$ (real-valued) centered Gaussian processes such that the finite dimensional distributions coincide, i.e.
$$\mathbb{E}X_t = \mathbb{E}Y_t = 0 \qquad \qquad C(s,t) := \mathbb{E}(X_s \cdot X_t) = \mathbb{E}(Y_s \cdot Y_t)$$
Moreover, assume that the trajectories $t \mapsto X(t,\omega)$, $t \mapsto Y(t,\omega)$ are in $L^2([0,T])$ for all $w \in \Omega$.
Does the equality
$$\mathbb{P}(\|X-\varphi\|_{L^2}<\delta) = \mathbb{P}(\|Y-\varphi\|_{L^2}<\delta) := \mathbb{P} \left\{\omega; \left(\int_0^T |Y(t,\omega)-\varphi(t)|^2 \, dt \right)^{\frac{1}{2}} < \delta \right\}$$ hold for $\varphi \in L^2([0,T]), \delta>0$? Equivalently: Do the distributions of the mappings
$$\Omega \ni \omega \mapsto X(\cdot,\omega) \in L^2([0,T]) \qquad \Omega \ni \omega \mapsto Y(\cdot,\omega) \in L^2([0,T]) \tag{1}$$
on $L^2([0,T])$, endowed with the Borel-$\sigma$-algebra generated by the $L^2$-norm, coincide?
Using the definition of the finite dimensional distributions (and Kolmogorov's extension theorem), it's not difficult to see that the distributions of the mappings in $(1)$ coincide with respect to the product-$\sigma$-algebra on $L^2([0,T])$.
Right from the intuition, I would say, that the open balls $B(\varphi,\delta)$ are not contained in the product-$\sigma$-algebra, because I need the information about the whole path to decide whether it is an element of this ball - consequently, this approach does not work out.
I would appreciate any hints & suggestions. Thanks!