+ \hat{\beta}_0 \tilde{L}_t - \frac{1}{2} \hat{\beta}_0^2 t} \qquad \text{ , } t \geq 0.$$ a) If condition is satsfied (the $X \log X$ condition on the offspring distribution) then $$M^\pm_\infty > 0 \qquad P \text{-a.s.}$$ b) If condition is not satsfied then $$M^\pm_\infty = 0 \qquad P \text{-a.s.}$$

Recall that martingale when used as the Radon-Nikodym derivative has the effect of putting constant drift of magnitude $\hat{\beta}_0$ towards the origin onto $(\xi_t)_{t \geq 0}$. More details on this can be found in [@BH14] or [@BS02] (“Brownian Motion with alternating drift", pp. 128-129).

Let $Q^\pm$ and $\tilde{Q}^\pm$ be the probability measures associated with martingales and as previously described in this subsection.

A standard argument which relies only on point-recurrence of $(\xi_t)_{t \geq 0}$ under $\tilde{P}$ (see e.g. [@JHH09]) tells us that $P(M^\pm_\infty > 0) \in \{0, 1\}$.

So if we could show that $Q^\pm(\limsup_{t \to \infty} M^\pm_t = \infty) = 1$ then by taking $A = \Omega$ in we would get $1 = E M^\pm_\infty + 1$ and hence $M^\pm_\infty = 0 \ P$-almost surely.

On the other hand, if we could show that $Q^\pm(\limsup_{t \to \infty} M^\pm_t = \infty) = 0$ then, again by taking $A = \Omega$ in , 