Limit value of the product martingale $\exp(uX_n - nu^2 \sigma^2 / 2)$

Let $(Y_i)$ be i.i.d. random variables with $\mathbb{P}(Y\geq 0)=1$ and $\mathbb{E}(Y)\leq 1$. The sequence of partial products $M_n:=\prod_{i=1}^n Y_i$ is a non-negative supermartingale and so converges almost surely to a limit random variable $M_\infty$.

Martingale Lemma. Either $\mathbb{P}(Y=1)=1$ or $\mathbb{P}(M_\infty=0)=1$.

Proof. We have $$\lim_{\varepsilon\downarrow 0} \mathbb{P}(|Y-1|>\varepsilon)=\mathbb{P}(|Y-1|>0)=\mathbb{P}(Y\neq 1).$$ If this probability is zero, we are in case 1.

Otherwise, we can choose $\varepsilon>0$ so that $\mathbb{P}(|Y-1|>\varepsilon)>0$. Then the Borel-Cantelli lemma shows that $\mathbb{P}(|Y_i-1|>\varepsilon \mbox{ infinitely often})=1$. This implies that $Y_i$ fails to converge to 1 pointwise, so $\mathbb{P}(Y_i\not\to 1)=1$.

If an infinite product converges to a non-zero limit, the factors must converge to 1. Therefore we finally get $$1=\mathbb{P}(M_n\to M_\infty)=\mathbb{P}(M_n\to M_\infty, Y_i\not\to 1)=\mathbb{P}(M_\infty=0),$$ which is case 2.

Here is an elementary computation of $Z^u_\infty$. Start with the fact that $X_n/n\to0$ almost surely by the law of large numbers hence, almost surely, $ uX_n\leqslant nu^2/4$ for every $n$ large enough. Assume that $u\ne0$. The preceding remark shows that $uX_n-nu^2/2\to-\infty$ almost surely when $n\to\infty$, thus $P(Z^u_\infty=0)=1$ and $E(Z^u_\infty\mid\mathscr F_n)=0$ almost surely, in particular $E(Z^u_\infty\mid\mathscr F_n)\ne Z^u_n$ almost surely.