Suppose $X$ is a subordinator (an increasing Levy process) with Laplace exponent $\Phi$, i.e. $$ \exp(-\Phi(\lambda)) = E(\exp(-\lambda X_1)). $$ Let $\mathcal{F} = (\mathcal{F}_t)$ denote the natural filtration of $X$. Then it's easy to check that $$ M_t := \exp(-\lambda X_t + \Phi(\lambda) t) $$ is a unit mean martingale.
I have often seen the following written: "We can define a new probability measure $Q$ on $\mathcal{F}_{\infty} := \sigma (\cup \{ \mathcal{F}_t : t \geq 0\})$ by
$$ \frac{dQ}{dP}\bigg\vert_{\mathcal{F}_t} = M_t." $$
My question is this: Certainly, a probability measure $Q^t$ can be defined on $\mathcal{F}_t$ by replacing $Q$ with $Q^t$ in the equation above. However, the claim above is that there exists a probability measure $Q$ on $\mathcal{F}_{\infty}$ such that its restriction to $\mathcal{F}_t$ has density $M_t$ with respect to $P$. This is true in case the martingale $M$ is uniformly integrable (then, with $M_{\infty}$ for its limit, we can write $E(M_\infty|\mathcal{F}_t) = M_t$ and the claim follows). But what if $M$ isn't uniformly integrable? Does the required $Q$ always exist in this context (i.e. where $M$ is derived from a subordinator as above)?
Many thanks.
You are right to be suspicious. The recipe you describe specifies a finitely additive set function $Q$ on $\cup_t\mathcal F_t$ such that $Q|_{\mathcal F_t} = M_t\cdot P|_{\mathcal F_t}$ for each $t\ge 0$. Conditions are known (much milder than the uniform integrability of $M$) that guarantee that $Q$ admits a $\sigma$-finite extension to $\sigma(\cup_t\mathcal F_t)$. These conditions are met by the usual path spaces (continuous paths, or cadlag paths). See, for example, Parthasarathy's book Probability Measures on Metric Spaces, section 4 of Chapter V.
For your subordinator example, the martingale $M$ is not uniformly integrable; in fact, $M_\infty=0$ a.s. In effect, because the process $X^*_t:=X_{t+1}-X_1$, $t\ge 0$, is a subordinator with the same distribution as $X$, you have $M_{t+1}=M_1\cdot M^*_t$, where $M^*$ is defined in terms of $X^*$ just as $M$ was defined in terms of $X$, and $M_1$ and $M^*$ are independent. Consequently, $M_\infty=M_1\cdot M^*_\infty$, a.s., the factors on the right being independent. Unless $X$ is degenerate, you have $E\sqrt{M_1}<\sqrt{E(M_1)}=1$, and upon taking square roots and then expectations in the equality $M_\infty=M_1\cdot M^*_\infty$ you are forced to conclude that $E(M_\infty)=0$.