= Solution
A convenient precise version is the <almost sure submartingale convergence theorem>: if $(S_n)$ is a real <submartingale> and $\sup_n\mathbb E S_n^+<\infty$, then $S_n$ converges <almost surely> to a finite <integrable random variable>. In particular the <martingale convergence theorem> applies to any <martingale> with $\sup_n\mathbb E|S_n|<\infty$. It asserts neither <convergence in L1> nor preservation of the initial <expected value> without additional <uniform integrability>.
Here is a proof including the crossing estimate. Fix real $a<b$, and let $U_n[a,b]$ count completed <upcrossings> by time $n$. Use a predictable indicator $H_k$ which holds one unit during each crossing attempt: enter when $S_k\leq a$, exit when $S_k\geq b$. With
$$
G_n=\sum_{k=0}^{n-1}H_k(S_{k+1}-S_k),
$$
each completed trade gains at least $b-a$, while the possible unfinished trade loses no more than $(S_n-a)^-$. Hence $G_n\geq(b-a)U_n[a,b]-(S_n-a)^-$. The <submartingale> property and $0\leq H_k\leq1$ give $\mathbb E G_n\leq\mathbb E(S_n-S_0)$: the complementary predictable gains have nonnegative <expected value>. Therefore
$$
(b-a)\mathbb E U_n[a,b]\leq\mathbb E(S_n-S_0)+\mathbb E(S_n-a)^-=\mathbb E(S_n-a)^++a-\mathbb E S_0.
$$
The right side is uniformly bounded, since $(S_n-a)^+\leq S_n^++|a|$. By the <monotone convergence theorem>, every rational interval has finitely many <upcrossings> <almost surely>. A real sequence whose lower and upper limits differ crosses some rational interval infinitely often, so the <countable> intersection of these events gives an extended-real limit. Furthermore
$$
\mathbb E S_n^- =\mathbb E S_n^+-\mathbb E S_n\leq\sup_m\mathbb E S_m^+-\mathbb E S_0,
$$
so $\sup_n\mathbb E|S_n|<\infty$. The <Fatou lemma> makes the limit finite and <integrable>, completing the proof. Applying this theorem to $-Z_n$ gives the <almost sure supermartingale convergence theorem> for nonnegative <supermartingales>.
For the given adapted multiplicative drift, put
$$
P_0=1,\qquad P_n=\prod_{k=0}^{n-1}(1+Y_k),\qquad Z_n=\frac{X_n}{P_n}.
$$
The next denominator $P_{n+1}$ is $\mathcal F_n$-<measurable>, and $P_n\geq1$ makes $Z_n$ <integrable>. Thus
$$
\mathbb E[Z_{n+1}\mid\mathcal F_n]=\frac{\mathbb E[X_{n+1}\mid\mathcal F_n]}{P_{n+1}}\leq\frac{(1+Y_n)X_n}{P_{n+1}}=Z_n.
$$
The nonnegative <supermartingale> $Z_n$ has a finite <almost sure convergence> limit $Z_\infty$. On the probability-one event where the specified series is finite,
$$
0\leq\log P_n=\sum_{k<n}\log(1+Y_k)\leq\sum_{k\geq0}Y_k<\infty.
$$
Consequently $P_n\uparrow P_\infty\in[1,\infty)$ and the <multiplicative correction for summable adapted drift> gives
$$
\boxed{X_n\longrightarrow P_\infty Z_\infty<\infty\quad\text{almost surely}.}
$$
The argument does not require a finite <expected value> for the infinite product or for the total drift.
Back to article page
