= Bounded Girsanov drift does not preserve true martingales under compensation
{c}
{title2=$M\in\mathcal M(P)\not\Rightarrow M-\langle M,W\rangle\in\mathcal M(Q),\quad dQ=e^{W_1-1/2}dP$}
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 $a=1/2$,
$$
H=\frac{e^{W_a^2}}{1+W_a^2},\qquad M_s=H(W_s-W_a)\mathbf1_{\{s>a\}}.
$$
The density of $W_a$ under P is $\pi^{-1/2}e^{-x^2}$, so $\mathbb E_PH=\pi^{-1/2}\int(1+x^2)^{-1}dx<\infty$. Independence of future Brownian increments makes $M$ a continuous true P-martingale. Under Q, that density is multiplied by $e^{x-1/4}$, so $\mathbb E_QH=\infty$.
For $s>a$, its compensated local martingale is
$$
M_s-\langle M,W\rangle_s=H(W_s^Q-W_a^Q).
$$
The 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>.
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