Let $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0},\mathbb P)$ a filtered probability space satisfying the usual conditions (especially $\mathcal F_0$ contains all null sets) and $\mathbb Q$ a further probability measure with $\mathbb Q\ll\mathbb P$. If we now restict the measures for each $t$ to the sub-$\sigma$-algebras $\mathcal F_t$ of the filtration, the property $$\mathbb Q|_{\mathcal F_t}\ll\mathbb P|_{\mathcal F_t}$$ is conserved. Thus the Radon Nikodym derivatives
$$S_t:=\frac{\mathrm d\mathbb Q|_{\mathcal F_t}}{\mathrm d\mathbb P|_{\mathcal F_t}}=\mathbb E_{\mathbb P}\left(\left.\frac{\mathrm d\mathbb Q}{\mathrm d\mathbb P}\right|\mathcal F_t\right)
$$
exist and form a martingale. As the Radon Nikodym derivative is only unique up to a $\mathbb P$-null set, I'm asking myself:
1. Question: Can the stochastic process $S:=(S_t)_{t\geq0}$ be choosen monotonically increasing?
The background of this question is: As $S$ is a martingale converging a.s. and in $L^1$ to $S_\infty:=\frac{\mathrm d\mathbb Q}{\mathrm d\mathbb P}$, I want to show that for a monotonacally decreasing convex function $\phi$ the process $\phi(S)$ is a submartingale converging in $L^1$ to $\phi(S_\infty)$. But to deduce that $\phi(S)$ is a submartingale I need $\phi(S)$ to be integrable (as a requirement for the Jensen inequality).
2. Question: Can be shown without any further assumptions that $\phi(S)$ is integrable?
In the final step I want to use something like the monotone convergence theorem to deduce the $L^1$-convergence of $S$.
Best Answer
1. No, outside the trivial case $\Bbb Q=\Bbb P$. Because $S$ is a martingale, $\Bbb E_{\Bbb P}[S_t]=\Bbb E_{\Bbb P}[S_u]$ for all $0\le t<u$. If it were true that $S_t\le S_u$, then you would have $\Bbb P[S_t=S_u]=1$.
2. If $\phi$ is understood to be a decreasing convex function mapping $[0,\infty)$ to $\Bbb R$, then the integrability of each $\phi(S_t)$ is assured. On the one hand, $\phi(x)\le\phi(0)$ for each $x\ge 0$, so $\phi(S_t)$ is bounded above. In the other direction, by convexity there are reals $a$ and $b$ such that $\phi(x)\ge a+bx$ for all $x\ge 0$. Thus, $\phi(S_t)\ge a+bS_t$. Because $S_t$ is integrable, these two bounds ensure the integrability of $\phi(S_t)$, which is therefore a submartingale. In addition, you have the bounds $$ a+bS_t-\phi(0)\le\phi(S_t)-\phi(S_\infty)\le \phi(0)-a-bS_\infty, $$ which together with the uniform integrability of $(S_t)_{0\le t\le\infty}$ is enough to guarantee that $\phi(S_t)$ converges to $\phi(S_\infty)$ in $L^1$.