Simplest version of Girsanov Theorem

Question

Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and the space of continuous functions $\Omega=C[0,T]$, equipped with the Borel $\sigma$-field $\mathcal{B}$. Let $B_t$ be the Brownian motion and let $X_t=B_t+\mu t$ be a Brownian motion with a drift. Let $P$ and $Q$ be the probabilities corresponding to $B_t$ and $X_t$ respectively, i.e. $P(A)=\mathbb{P}((t\mapsto B_t)\in A)$ and $Q(A)=\mathbb{P}((t\mapsto X_t)\in A)$. From the tilting identity $$E[f(X_{t_1} , ... , X_{t_n})] = E \left[f(B_{t_1} , ... , B_{t_n}) \underbrace{\exp \left(\mu B_T-\frac12\mu^2 T \right)}_{M_T}\right]$$ I have to deduce the "simple form" of the Girsanov theorem $E_Q(W)=E_P(WM_T)$. For this it suffices to show the result on indicator functions of cylindrical sets, i.e. $E_Q(1_A)=E_P(1_AM_T)$ where $A=\{\omega\in C[0,T]:\omega(t_1)\in[a_1,b_1],\ldots,\omega(t_n)\in[a_n,b_n]\}$. Apparently this follows directly from the tilting identity but I can't understand how. We have $$E_Q(1_A)=Q(A)=P((t\mapsto X_t)\in A)=...?...=E_P[1_AM_T]$$