Limit of the expectation in Galton-Watson-process using a Martingale Let $(Z_n)_{n\in\mathbb{N}_0}$ be a Galton-Watson process
$$
Z_{n+1} = \sum^{Z_n}_{i=1}X_{n,i}  \text{ for }n \in \mathbb{N} 
$$
with $Z_0=1$ and $(X_{n,i})_{n\in\mathbb{N}_0,i\in\mathbb{N}}$ i.i.d., and let $$m:=\mathbb{E}[X_{1,1}]<\infty,$$ $$\sigma^2:=\text{Var}[X_{1,1}]\in(0,\infty)$$ and
$$W_n = m^{-n}Z_n$$
Lemma 11.18 in Klenke states that $W$ is a martingale and $\mathbb{E}[Z_n]=m^n$.
In the proof of Theorem 11.19 it says, that $\text{lim}_{n\rightarrow\infty}W_n =: W_\infty$ exists since $W\geq0$ is a martingale and for $m\leq1$ $(Z_n)$ converges a.s. to $Z_\infty$ = 0 since $\sigma^2>0$.
This seems clear for me, when $m<1$, since $Z_n=m^nW_n$. However, I do not understand why this is the case for $m=1$. Then we have $\mathbb{E}[Z_n] = 1, \forall n$. Blackwell-Gershwick gives $\text{Var}[Z_{n+1}] = \text{Var}[Z_n] + \sigma^2$, so that $\text{Var}[Z_n]\rightarrow\infty$. But how does this indicate that $Z_{\infty} = 0$?
 A: Since $\mathbb P(W_n\geqslant0)=1$ and $\sup_n\mathbb E[W_n]=1<\infty$, from Doob's martingale convergence theorem we have that $W_n\stackrel{n\to\infty}\longrightarrow W$ for some random variable $W$. Let $f(s) = \sum_{n=0}^\infty p_ns^n$ be the generating function of $X_{1,1}$. Set $\rho_i = \mathbb P(W=0\mid Z_1=i)$, then $\rho_i=\rho_1^i$ as conditioned on $Z_1=i$, the process $\{Z_n\}$ is equivalent to the sum of $i$ copies of $\{Z_n\}$ conditioned on $Z_1=1$. It suffices then to show that $\rho:=\rho_1 = 0$.
For $1\leqslant j\leqslant Z_1$, let $Z_n(j)$ denote the branching process obtained from the the $j^{\mathrm{th}}$ child of the ancestor $\varnothing$. Then
$$
Z_{n+1} = \sum_{j=1}^{Z_1}Z_n(j),
$$
so
$$
W_{n+1} = m^{-(n+1)}\sum_{j=1}^{Z_1}Z_n(j) = m^{-1}\sum_{j=1}^{Z_1}W_n(j)\stackrel{n\to\infty}\longrightarrow m^{-1}\sum_{j=1}^{Z_1}W(j).
$$
It follows that
\begin{align}
\rho &= \mathbb P\left(\bigcap_{j=1}^{Z_1}\{W(j)=0\}\right)\\
&= \sum_{n=0}^\infty \mathbb P(Z_1=n)\mathbb P\left(\bigcap_{j=1}^{n}\{W(j)=0\}\right)\\
&= \sum_{n=0}^\infty p_n\rho^n\\
&= f(\rho).
\end{align}
Recall that the extinction probability $\pi:=\lim_{n\to\infty}\mathbb P(Z_n=0)$ satisfies $\pi=f(\pi)$. Since $m\leqslant 1$, it follows that $\rho=1$.
A: If $m=1$, $Z_n$ is a critical Galton-Watson process. Therefore, the extinction event has probability $1$:
$$
\rho:=P(\exists N>0, \forall n \geq N, Z_n = 0) = 1.
$$
In this case, you have
$$
Z_n \xrightarrow[n\to \infty]{} 0,
$$
as surely.
