must converge $P$-almost surely to some non-negative limit $M_\infty$. Particularly interesting are those martingales whose limit is not almost surely $0$.

When investigating whether the limit of $M$ is $P$-almost surely $0$ or not we shall make use of the following result commonly known as the spine decomposition.

\[spine\_decomp\] Let $(M_t)_{t \geq 0}$ be as above. Then $$\label{eq_spine_decomp}
\tilde{Q} \Big( M_t \big\vert \tilde{\mathcal{G}}_\infty \Big) = 
spine(t) + \sum_{n=1}^{n_t} (A_n - 1) spine(S_n),$$ where $$spine(t) = \tilde{M}^{(1)}_t \mathrm{e}^{- \hat{\beta}t - \hat{\beta}_0 \tilde{L}_t}.$$

Note that we have used $\tilde{Q}$ to denote the expectation corresponding to probability measure $\tilde{Q}$ which is a common practice.

We also recall from standard measure theory that if $A \in \mathcal{F}_t$ for some $t \geq 0$ then $$Q(A) = \int_A M_t \mathrm{d}P$$ whereas if $A \in \mathcal{F}_\infty$ then $$\label{durrett}
Q(A) = \int_A M_\infty \mathrm{d}P + Q \big( A \cap \{ \limsup_{t \to \infty} M_t = \infty\} \big).$$ The latter identity can be found in [@Durrett], p.241.

\[M\_pm\] Let $$\label{eq_M_pm}
M_t^\pm := \sum_{u \in N_t} \mathrm{e}^{- \hat{\beta}_0 |X^u_t| - \frac{1}{2} \hat{\beta}_0^2 t - \hat{\beta} t} \text{ , } \qquad t \geq 0$$ be the $P$-martingale derived through the procedure described above by taking $$\label{M_one}
\tilde{M}^{(1)}_t = \mathrm{e}^{- \hat{\beta}_0 |\xi_t| 