Let $P(t,T)$ denote the time $t$ price of a zero-coupon bond maturing at time $T$ and $\mathbb{Q}_T$ be the associated equivalent martingale measure which uses $P(t,T)$ as numeraire. Then, for any $\mathcal{F}_T$-measurable payoff $\xi$, the time $t$ value of $\xi$ is given by $$V_t=P(t,T)\cdot\mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t].$$ The undiscounted time $t$ price is given by $$\tilde{V}_t = \frac{V_t}{P(t,T)} = \mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t].$$ And indeed, $(\tilde{V}_t)$ is a $\mathbb{Q}_T$-martingale. Assuming integrability and adaptness (trivial), we need to show the martingale property. To this end, let $0\leq s<t\leq T$. Then, by the tower law,
\begin{align*}
\mathbb{E}^{\mathbb{Q}_T}[\tilde{V}_t\mid\mathcal{F}_s] &= \mathbb{E}^{\mathbb{Q}_T}\left[\mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t]\bigg|\mathcal{F}_s\right] \\
&= \mathbb{E}^{\mathbb{Q}_T}[\xi\mid\mathcal{F}_s] \\
&= \tilde{V}_s.
\end{align*}
Please note the following:
- This result is completely independent of the Bachelier model and equally applies to the Black-Scholes model, the Heston model and others.
- If interest rates are deterministic, so are bond prices and back accounts. Thus, the forward measure $\mathbb{Q}_T$ coincides with the ``standard'' risk-neutral measure $\mathbb{Q}$ which uses a risk-free bank account $B_t=e^{\int_0^t r(s)\mathrm{d}s}$ as numeraire.
