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Use $M_0=0$, as implicit in the specified initial value of $Y$. First let $[M{]}_t=t$. The Lévy characterization of Brownian motion makes $M$ Brownian motion relative to the given filtration. SetOn this event at time one, the drift has absolute size at most $C$, while $|Y_0|<1$, so $|Y_1|\ge |M_1|-|Y_0|-C>1$. Continuity forces exit by time one. ThereforeThe same estimate applies conditionally at each integer time without assuming anything about future drift independence. On $\{T>n\}$, $|Y_n|<1$; if $|M_{n+1}-M_n|>C+2$, the drift bound forces exit by $n+1$. That Brownian increment is independent of ${\mathcal{F}}_n$. HenceIn particular $T<\infty$ almost surely.

For general $Q$, write $K_t=[M{]}_t$ and $D_t={\int }_0^t{A}_{s}ds$. Continuity of $Q$ and monotonicity of its integral imply $Q_s\ge 0$ at all times outside one null set. The drift condition givesLet ${\tau }_u=\inf \{t:K_t>u\}$. The divergent clock makes ${\tau }_u$ finite for every $u$, and its continuity gives $K_{{\tau }_u}=u$. By the Dambis-Dubins-Schwarz theorem, $W_u=M_{{\tau }_u}$ is Brownian motion for ${\mathcal{G}}_u={\mathcal{F}}_{{\tau }_u}$. On intervals where $K$ is constant, both $M$ and $D$ are constant. Thus $d_u=D_{{\tau }_u}$ is a continuous adapted function of clock time satisfying $|d_v-d_u|\le C(v-u)$. The time-changed process isThe unit-clock argument uses only this increment bound on the drift, so its first exit time $S$ obeys $\mathbb{P}(S>n)\le (1-\epsilon {)}^n$. It is finite almost surely, and a corresponding finite ordinary time has $|Y|\ge 1$. This proves the exit with drift controlled by quadratic variation resultThis proof needs only a nonnegative measurable bracket density, a continuous divergent integral clock, and the drift increment bound; pointwise continuity of the density is unnecessary. This observation permits the application below with measurable SDE coefficients.

For the multidimensional SDE, put $a_{ij}(x)={\sum }_{k=1}^m{\sigma }_k^i(x){\sigma }_k^j(x)$. The generator of an SDE diffusion isFor every $f\in C_b^2({\mathbb{R}}^d)$, the Itô formula givesBounded derivatives and bounded noise fields make the right side square-integrable on each finite horizon, hence a true martingale. This is precisely the defining test-function property of an L-diffusion, in the martingale-problem sense.

Now fix $x$ and $\xi \ne 0$. Its scalar projection has martingale part $M_t^{\xi }={\sum }_k{\int }_0^t⟨\xi ,{\sigma }_k(X_s^x)⟩d{B}_s^k$ and drift $⟨\xi ,b(X_s^x)⟩$. The uniform ellipticity assumption givesLet $B_{\ast }={\sup }_z|b(z)|$ and take $R>|⟨\xi ,x⟩|$. Rescale the projection by $R$. Its bracket density is at least $\lambda |\xi {|}^2/R^2$, its absolute drift is at most $B_{\ast }|\xi |/R$, and hence that drift is at most $C_R$ times its bracket density, with $C_R=B_{\ast }R/(\lambda |\xi |)$. Its clock diverges, so the exit lemma forces it out of $(-1,1)$ in finite time. Equivalently it leaves the slab $|⟨\xi ,X_t^x⟩|<R$. Taking a countable sequence of integer $R$ tending to infinity proves the projection unboundedness of a uniformly elliptic diffusion:For the final uniqueness result, let ${\tau }_D=\inf \{t:X_t^x\notin D\}$, with $x\in D$. Since $D$ is bounded, it lies in a finite slab; the preceding finite slab-exit time bounds ${\tau }_D$ from above. Thus ${\tau }_D<\infty$ almost surely. Continuity gives $X_{{\tau }_D}^x\in \partial D$.

Set $h=u-v$. It is bounded, has bounded derivatives, is zero on the boundary, and satisfies $Lh=0$ in $D$. The stopped test-function identity says $h(X_{t\wedge {\tau }_D}^x)$ is a martingale; equivalently it is a bounded local martingale, which suffices. Taking expectations and then using bounded convergence as $t\to \infty$ givesTherefore the bounded-domain harmonic uniqueness for a diffusion conclusion isIn particular the argument supplies the representation $u(x)={\mathbb{E}}_{x}u(X_{{\tau }_D})$ and establishes uniqueness without selecting a unique diffusion law.