Equivalence of seprating measures and local martingale measure

Equivalence of seprating measures and local martingale measure

Let me first introduce some definitions. We are given a process
$X=\{X_t;0\le t\le T\}$, usually a semimartingale which is RCLL and
adapted to a filtration, which satisfies the usual condition. We call a
probability measure $Q$ a local martinagle measure, if $X$ is under $Q$ a
local martingale. $Q$ is called a separating measure if
$E_Q[G_T(\Theta)]\le 0$, where $G_T(\Theta)=\int_0^T\theta dX$ for
$\theta\in\Theta$ the space of predictable processes which are
$X$-integrable, i.e. the stochastic integral is well-defined, and
$\theta\cdot X:=\int\theta dX\ge -a$ for an $a\ge 0$.
The claim I want to prove is the following
Claim Let $Q$ be a separating measure for $X$. If $X$ is locally bounded,
i.e. there is a sequence of stopping time increasing to $T$ such that the
stopped process $X^{\tau_n}$ is bounded, then $Q$ a local martingale
measure for $X$.
I was able to prove the statement with bounded instead of locally bounded
an martingale measure instead of local martingale measure. Somehow, I have
trouble with the localization argument. For the bounded case, the
following works: let $\tau$ be a bounded stopping time. We want to prove
$E_Q[X_\tau]=E_Q[X_0]$, since this implies the martingale property of $X$
under $Q$. Choose $\theta=\mathbf1_{((0,\tau]]}\in\Theta$ since $X$ is
bounded. Then $G_T(\theta)=X_\tau-X_0$. By assumption $E_Q[G_T(\theta)]\le
0 \iff E_Q[X_\tau]\le E_Q[X_0]$. No do the same for
$\theta':=-\mathbf1_{((0,\tau]]}$ and we are done.
In the local part, let $\tau_n$ be this sequence. Can I argue like this:
Define $\theta:=\mathbf1_{((0,\tau_n\wedge \tau]]}\in \Theta$ since
$(\theta\cdot X)=(\mathbf1_{((0,\tau]]}\cdot
X^{\tau_n})=X^{\tau_n}_\tau-X^{\tau_n}_0$ and the latter is again bounded,
so we have $\theta\cdot X\ge -a$. Is my argument right? Thanks for you
help in advance.