I am going through the paper by Longstaff and Schwartz (2001) on American-options pricing, and something got me confused.
There, in equation $(1)$ the continuation value at time $t_k$, $F(\omega; t_k)$, is defined as follows $$F(\omega; t_k) = E_Q\left[ \sum_{j = k+1}^{K} \exp \left( - \int_{t_k}^{t_j} r(\omega, s ) \, ds \right) C(\omega, t_j ; t_k, T) \; \Bigg\vert \; \mathcal{F}_k \right].$$
However, I think that it should instead be rewritten as $$F(\omega; t_k) = \sum_{j = k+1}^{K} D(t_k, t_j) \cdot E_Q\left[ C(\omega, t_j ; t_k, T) \; \Bigg\vert \; \mathcal{F}_k \right],$$ where $D(t_k, t_j) $ are the different discount factors at $t_k$, which are $\mathcal{F}_k$-measurable. My idea is nothing different to the martingale condition under the risk-free probability measure $Q$.
So my question is, is there something that I am missing here?
I appreciate every comment or discussion on the subject.
Edit
What I really mean is, isn't the price of a derivative at time $t$, chosen a numerary $\mathcal{N}$, given by
$$\dfrac{V(t,T)}{\mathcal{N}(t,T)} = E_{Q} \Bigg[ \dfrac{V(T,T)}{\mathcal{N}(T,T)} \Bigg\vert \mathcal{F}_t \Bigg]$$
where if $\mathcal{N}$ is chosen to be a zero coupon bond, then $\mathcal{N}(t,T) = D(t,T)$ and $\mathcal{N}(T,T) = 1$.
Longstaff, Schwartz - Valuing American Options by Simulation: A Simple Least-Squares Approach (2001)