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Let $W$ be a P-Brownian motion on $[0,1]$ and tilt by the bounded constant drift integrand one, so $dQ=e^{W_1-1/2}dP$. Then $W_s^Q=W_s-s$ is Q-Brownian motion, either by Girsanov theorem or direct Gaussian exponential tilting. Set $a=1/2$,The density of $W_a$ under P is ${\pi }^{-1/2}{e}^{-x^2}$, so ${\mathbb{E}}_{P}H={\pi }^{-1/2}\int (1+x^2{)}^{-1}dx<\infty$. Independence of future Brownian increments makes $M$ a continuous true P-martingale. Under Q, that density is multiplied by $e^{x-1/4}$, so ${\mathbb{E}}_{Q}H=\infty$.

For $s>a$, its compensated local martingale isThe future Q-Brownian increment is independent of the positive variable $H$ and has positive finite mean absolute value. Hence the displayed variable has infinite expected absolute value under Q and is not a true martingale. This shows why the general Girsanov theorem promises a local martingale, while a true-martingale conclusion needs an additional integrability condition such as Girsanov compensation with square-integrable input.

The initial value can first be subtracted: it is ${\mathcal{F}}_0$-measurable and has the same integrability under both measures, since the density has conditional expectation one on ${\mathcal{F}}_0$. Put $N=M-⟨M,X⟩$. Since $dZ=ZdX$, the Itô product rule and the quadratic covariation identity $d⟨M,Z⟩=Zd⟨M,X⟩$ giveThis is a local P-martingale. Stop $N$ at increasing bounded levels; the resulting products with $Z$ are true martingales, since they are local martingales dominated at stopping times by a constant times the uniformly integrable density process. Part (c) then shows that the stopped $N$ processes are Q-martingales. ConsequentlyThe unqualified true-martingale assertion printed in this part needs an additional integrability hypothesis. One sufficient version is a square-integrable P-martingale $M$. Indeed $Z_t\in L^2(P)$ by boundedness of $Y$, and the covariation bound givesThe Doob L2 maximal inequality, ${\mathbb{E}}_P⟨M{⟩}_t<\infty$, and Cauchy-Schwarz inequality therefore give ${\mathbb{E}}_Q{\sup }_{s\le t}|N_s|<\infty$. Its local martingale identities pass to the limit, making it a true Q-martingale. This is Girsanov compensation with square-integrable input.

Here is a counterexample to the broader printed assertion. On $[0,1]$, take $X_s=W_s$, so $Y_s\equiv 1$ and $Z_s=e^{W_s-s/2}$. Put $a=1/2$ andUnder P, $W_a$ has density ${\pi }^{-1/2}{e}^{-x^2}$, so ${\mathbb{E}}_{P}H<\infty$. Independence of future Brownian increments proves $M$ is a continuous true P-martingale. Under Q, $W_s^Q=W_s-s$ is Brownian and the density of $W_a$ is multiplied by $e^{x-1/4}$, so ${\mathbb{E}}_{Q}H=\infty$. The compensated process after $a$ isThe future Q-Brownian increment is independent of $H$ and has positive finite expected absolute value. Thus ${\mathbb{E}}_Q|N_s|=\infty$ for every $s>a$, so it cannot be a true martingale. This proves that Bounded Girsanov drift does not preserve true martingales under compensation.

For the requested application, the Brownian driver does satisfy the needed conditions. To distinguish the state from the density-driving integral, call the state $U$. Set $Y_s=b(U_s)/a(U_s)$; the stated coefficient bounds make this integrand bounded. Applying the local result to $M=W$ givesIt is a continuous local Q-martingale with quadratic variation $s$. The Lévy characterization of Brownian motion therefore makes it a Q-Brownian motion. Substituting $d{W}_s=d{\tilde{W}}_s+[b(U_s)/a(U_s)]ds$ into the original state equation givesThe Brownian driver after the change is $\tilde{W}$; writing it as $W$ again in the last printed equation amounts to renaming that driver.