$T<\infty$ almost surely but it is not bounded. If you want the equallity for a fixed $t$ it is not possible as the gaussian law $\mathcal{N}(x,t)$ and $\mathcal{N}(y,t)$ are different.

This is called coalescent brownian motion. Set $T=\inf{t:S_t=B_t}$ and set $W_t = 1_{t\leq T}S+1_{t>T}B_t$.

Just that the vector space generated by the variable $\mathcal{F}_i$ measurable are in direct sum : $\forall \xi_1,\cdots,\xi_n\in L^2(\Omega,\mathcal{F_1},\mu)\times...\times L^2(\Omega,\mathcal{F_n},\mu)$. $\xi_1+\cdots +\xi_n=0\Rightarrow \xi_1=0,\cdots,\xi_n=0$.

If one restrict to $\xi$ such that $\mathbb{E}(\xi)=0$. Is your IMP equivalent to $$ L^2(\Omega,\mathcal{F}_1,\mu)\oplus L^2(\Omega,\mathcal{F}_2,\mu)\oplus\cdots \oplus L^2(\Omega,\mathcal{F}_n,\mu)$$?

@losif Pinelis : The second point you mention is not true for all t but almost all $t$. Calling $\Omega$ the probability space of the brownian motion then $\int_{-L}^L\int_{\Omega} 1_{U(t)=1}dt d\mu=2L=\int_{\Omega} \int_{-L}^L1_{U(t)=1}dt d\mu$. So with probability 1, for almost all $t\in [-L,L]$ $U(t)=1$

@losif Pinelis: The Brownian motion is scale invariant ($f^{(r)}(t):=\frac{1}{\sqrt{r}}W_{rt}$ is also a brownian motion). So $K(0,t)$ have the same law as $K(0,r^{-1}t)$. So $\mathbb{P}(\frac{1}{2r}\int K(0,t)dt\geq a)$ is independant of $r$.

Yes "continuous-time", I edited the question.

Otherwise you can redo the CLT proof with the characteristic function $\mathbb{E}(e^{i\alpha (S_k-\mu N_k)/\sigma\sqrt{N_k}})=\mathbb{E}(\mathbb{E}(e^{i\alpha (X-\mu )/\sigma\sqrt{N_k}})^{N_k})\approx \mathbb{E}((1-\frac{\alpha^2}{2N_k})^{N_k})\approx e^{-\alpha^2/2}$.