[proofplan]
The product $M_n = \prod_{i=1}^n X_i$ is a non-negative martingale, so the [Almost Sure Martingale Convergence Theorem](/theorems/1157) gives $M_n \to M_\infty$ a.s. The dichotomy is controlled by the auxiliary martingale $N_n = \sqrt{M_n} / \prod_{i=1}^n a_i$, where $a_i = \mathbb{E}[\sqrt{X_i}]$. The factorisation $M_n = N_n^2 (\prod_{i=1}^n a_i)^2$ separates random and deterministic contributions. When $\prod_n a_n > 0$, the process $(N_n)$ is $L^2$-bounded, and Doob's $L^p$ inequality yields an integrable majorant for $(M_n)$, establishing uniform integrability and hence $L^1$ convergence. When $\prod_n a_n = 0$, the deterministic factor vanishes while $N_\infty$ remains a.s. finite, forcing $M_\infty = 0$.
[/proofplan]
[step:Verify that $M$ is a non-negative martingale and establish a.s. convergence]
[Set](/page/Set) $\mathcal{F}_n = \sigma(X_1, \ldots, X_n)$ for $n \geq 1$ and $\mathcal{F}_0 = \{\varnothing, \Omega\}$. Since each $X_n \geq 0$, the product $M_n = \prod_{i=1}^n X_i$ is non-negative and $\mathcal{F}_n$-measurable. By independence,
\begin{align*}
\mathbb{E}[M_n] = \mathbb{E}\Bigl[\prod_{i=1}^n X_i\Bigr] = \prod_{i=1}^n \mathbb{E}[X_i] = 1.
\end{align*}
For the conditional expectation, since $M_n$ is $\mathcal{F}_n$-measurable and $X_{n+1}$ is independent of $\mathcal{F}_n$,
\begin{align*}
\mathbb{E}[M_{n+1} \mid \mathcal{F}_n] = \mathbb{E}[M_n X_{n+1} \mid \mathcal{F}_n] = M_n \, \mathbb{E}[X_{n+1}] = M_n.
\end{align*}
Thus $(M_n)_{n \geq 0}$ is a non-negative martingale with respect to $(\mathcal{F}_n)$. By the [Almost Sure Martingale Convergence Theorem](/theorems/1157), every non-negative supermartingale converges a.s. to a finite [limit](/page/Limit). Therefore $M_n \to M_\infty$ a.s. for some $M_\infty \geq 0$.
[guided]
We verify the three conditions for the [Almost Sure Martingale Convergence Theorem](/theorems/1157): adaptedness, integrability, and the conditional expectation identity.
**Adaptedness.** Each $X_i$ is $\mathcal{F}_i$-measurable, so $M_n = \prod_{i=1}^n X_i$ is $\mathcal{F}_n$-measurable.
**Integrability.** Since $X_1, \ldots, X_n$ are independent and non-negative, the expectation of the product equals the product of the expectations:
\begin{align*}
\mathbb{E}[M_n] = \prod_{i=1}^n \mathbb{E}[X_i] = 1.
\end{align*}
In particular, $\sup_n \mathbb{E}[|M_n|] = 1 < \infty$, so $(M_n)$ is bounded in $L^1$.
**Conditional expectation.** We factor $M_{n+1} = M_n \cdot X_{n+1}$. Since $M_n$ is $\mathcal{F}_n$-measurable and $X_{n+1}$ is independent of $\mathcal{F}_n$ (and integrable), we may extract $M_n$ from the conditional expectation and replace $\mathbb{E}[X_{n+1} \mid \mathcal{F}_n]$ by $\mathbb{E}[X_{n+1}] = 1$.
A.s. convergence alone does not determine whether $\mathbb{E}[M_\infty] = 1$ or $M_\infty = 0$ a.s. Fatou's lemma gives only the one-sided bound $\mathbb{E}[M_\infty] \leq \liminf_n \mathbb{E}[M_n] = 1$. The dichotomy between parts (i) and (ii) addresses which alternative holds.
[/guided]
[/step]
[step:Bound $a_n = \mathbb{E}[\sqrt{X_n}]$ using [Jensen's inequality](/theorems/9)]
The [function](/page/Function) $\varphi: [0,\infty) \to \mathbb{R}$, $x \mapsto \sqrt{x}$ is concave. Since $X_n \geq 0$ and $\mathbb{E}[X_n] = 1 < \infty$, applying [Jensen's Inequality](/theorems/9) to the concave function $\varphi$ (reversing the convex inequality) gives
\begin{align*}
a_n = \mathbb{E}[\sqrt{X_n}] \leq \sqrt{\mathbb{E}[X_n]} = 1.
\end{align*}
Moreover, $a_n > 0$: since $\mathbb{E}[X_n] = 1 > 0$ and $X_n \geq 0$, we have $\mathbb{P}(X_n > 0) > 0$, so $\mathbb{E}[\sqrt{X_n}] > 0$. Thus $a_n \in (0, 1]$ for every $n \geq 1$.
[/step]
[step:Construct the auxiliary martingale $N_n$ and verify its properties]
Define
\begin{align*}
N_n: \Omega &\to [0, \infty) \\
\omega &\mapsto \frac{\sqrt{X_1(\omega) \cdots X_n(\omega)}}{a_1 \cdots a_n} = \frac{\sqrt{M_n(\omega)}}{\prod_{i=1}^n a_i}
\end{align*}
with $N_0 = 1$. By independence and the definition of $a_i$,
\begin{align*}
\mathbb{E}[N_n] = \frac{1}{\prod_{i=1}^n a_i} \, \mathbb{E}\Bigl[\prod_{i=1}^n \sqrt{X_i}\Bigr] = \frac{\prod_{i=1}^n a_i}{\prod_{i=1}^n a_i} = 1.
\end{align*}
Since $N_{n+1} = N_n \cdot \sqrt{X_{n+1}} / a_{n+1}$ with $N_n$ being $\mathcal{F}_n$-measurable and $X_{n+1}$ independent of $\mathcal{F}_n$,
\begin{align*}
\mathbb{E}[N_{n+1} \mid \mathcal{F}_n] = \frac{N_n}{a_{n+1}} \, \mathbb{E}[\sqrt{X_{n+1}}] = \frac{N_n}{a_{n+1}} \cdot a_{n+1} = N_n.
\end{align*}
Thus $(N_n)$ is a non-negative martingale. By the [Almost Sure Martingale Convergence Theorem](/theorems/1157), $N_n \to N_\infty$ a.s. for some finite $N_\infty \geq 0$.
The key algebraic identity linking $M$ and $N$ is
\begin{align*}
M_n = N_n^2 \Bigl(\prod_{i=1}^n a_i\Bigr)^{\!2}.
\end{align*}
Since $a_i \leq 1$ for all $i$, this gives the pointwise bound $M_n \leq N_n^2$.
[guided]
The idea behind $N_n$ is to "take the square root" of the product martingale and renormalise. The product $\sqrt{X_1 \cdots X_n}$ has expectation $\prod_{i=1}^n a_i$ (by independence), which depends on $n$ and may tend to zero. Dividing by $\prod_{i=1}^n a_i$ restores the unit-mean property, making $N$ a martingale.
Why is this construction useful? The second moment of $N_n$ is
\begin{align*}
\mathbb{E}[N_n^2] = \frac{\mathbb{E}[X_1 \cdots X_n]}{(\prod_{i=1}^n a_i)^2} = \frac{1}{(\prod_{i=1}^n a_i)^2}.
\end{align*}
This quantity is finite for each $n$, but remains bounded as $n \to \infty$ if and only if $\prod_{n=1}^\infty a_n > 0$. The factorisation $M_n = N_n^2 (\prod_{i=1}^n a_i)^2$ separates $M_n$ into a "random" factor $N_n^2$ (controlled by martingale theory) and a "deterministic" factor $(\prod_{i=1}^n a_i)^2$ (controlled by the product $\prod a_n$). This is the mechanism that drives the dichotomy.
[/guided]
[/step]
[step:Part (i) — Establish $L^2$ boundedness of $(N_n)$ and dominate $\sup_n M_n$]
Assume $\prod_{n=1}^\infty a_n > 0$. Then
\begin{align*}
\sup_{n \geq 1} \mathbb{E}[N_n^2] = \sup_{n \geq 1} \frac{1}{(\prod_{i=1}^n a_i)^2} = \frac{1}{(\prod_{n=1}^\infty a_n)^2} < \infty,
\end{align*}
so $(N_n)$ is bounded in $L^2$. Since $N_n \geq 0$, the process $(N_n^2)_{n \geq 0}$ is a non-negative submartingale (by Jensen's inequality applied to the convex map $x \mapsto x^2$ and the martingale property of $N$). We apply [Doob's $L^p$ Inequality](/theorems/1159) with $p = 2$ to the martingale $(N_n)$. Setting $N_n^* = \sup_{0 \leq k \leq n} N_k$, the inequality gives
\begin{align*}
\|N_n^*\|_2^2 \leq \Bigl(\frac{p}{p-1}\Bigr)^{\!2} \|N_n\|_2^2 = 4 \, \mathbb{E}[N_n^2].
\end{align*}
Since $M_k \leq N_k^2$ for every $k$ (because $\prod_{i=1}^k a_i \leq 1$), and squaring is monotone on $[0,\infty)$,
\begin{align*}
\sup_{0 \leq k \leq n} M_k \leq \sup_{0 \leq k \leq n} N_k^2 = (N_n^*)^2.
\end{align*}
Therefore
\begin{align*}
\mathbb{E}\Bigl[\sup_{0 \leq k \leq n} M_k\Bigr] \leq \mathbb{E}[(N_n^*)^2] \leq 4 \, \mathbb{E}[N_n^2] \leq \frac{4}{(\prod_{n=1}^\infty a_n)^2}.
\end{align*}
The [sequence](/page/Sequence) $\sup_{0 \leq k \leq n} M_k$ is non-decreasing and non-negative. By the [Monotone Convergence Theorem](/theorems/509),
\begin{align*}
\mathbb{E}\Bigl[\sup_{n \geq 0} M_n\Bigr] = \lim_{n \to \infty} \mathbb{E}\Bigl[\sup_{0 \leq k \leq n} M_k\Bigr] \leq \frac{4}{(\prod_{n=1}^\infty a_n)^2} < \infty.
\end{align*}
[guided]
The goal is to establish uniform integrability of $(M_n)$. The standard route is to find an integrable majorant: if $|Y_n| \leq Z$ for all $n$ with $\mathbb{E}[Z] < \infty$, then $(Y_n)$ is uniformly integrable (since $\mathbb{E}[|Y_n| \, \mathbb{1}_{|Y_n| > \alpha}] \leq \mathbb{E}[Z \, \mathbb{1}_{Z > \alpha}] \to 0$ as $\alpha \to \infty$, uniformly in $n$).
The natural majorant is $Z = \sup_{n \geq 0} M_n$, and we need $\mathbb{E}[Z] < \infty$. The bound $M_k \leq N_k^2$ converts this into an $L^2$ maximal inequality for the martingale $(N_n)$.
[Doob's $L^p$ Inequality](/theorems/1159) requires that $(N_n)$ be a martingale (verified above) and $p > 1$. With $p = 2$, the constant is $p/(p-1) = 2$, so $\|N_n^*\|_2 \leq 2 \|N_n\|_2$. Squaring gives $\mathbb{E}[(N_n^*)^2] \leq 4 \, \mathbb{E}[N_n^2]$.
The [Monotone Convergence Theorem](/theorems/509) applies because $\sup_{0 \leq k \leq n} M_k$ is a non-decreasing sequence of non-negative [measurable functions](/page/Measurable%20Functions) converging pointwise to $\sup_{n \geq 0} M_n$.
What fails when $\prod_n a_n = 0$? The $L^2$ norm $\mathbb{E}[N_n^2] = (\prod_{i=1}^n a_i)^{-2} \to \infty$, so Doob's inequality provides no finite bound. This is consistent with $(M_n)$ failing to be uniformly integrable in case (ii).
[/guided]
[/step]
[step:Part (i) — Conclude $L^1$ convergence via uniform integrability]
Since $0 \leq M_n \leq \sup_{k \geq 0} M_k$ and $\mathbb{E}[\sup_{k \geq 0} M_k] < \infty$, the family $(M_n)_{n \geq 0}$ is uniformly integrable. By the [UI Martingale Convergence Theorem](/theorems/1163), the uniformly integrable martingale $(M_n)$ converges in $L^1$ to $M_\infty$, and
\begin{align*}
\mathbb{E}[M_\infty] = \lim_{n \to \infty} \mathbb{E}[M_n] = 1.
\end{align*}
[guided]
The [UI Martingale Convergence Theorem](/theorems/1163) states that for a martingale, uniform integrability is equivalent to $L^1$ convergence (and to closability). The key consequence is the preservation of expectation: $\mathbb{E}[M_\infty] = \lim_n \mathbb{E}[M_n]$. Without uniform integrability, Fatou's lemma would give only $\mathbb{E}[M_\infty] \leq 1$, and as case (ii) shows, strict inequality ($\mathbb{E}[M_\infty] = 0 < 1$) does occur.
The condition $\mathbb{E}[M_\infty] = 1$ certifies that no mass is lost in the limit — the martingale "closes." In the language of the theorem, $M_n = \mathbb{E}[M_\infty \mid \mathcal{F}_n]$ a.s.
[/guided]
[/step]
[step:Part (ii) — Deduce $M_\infty = 0$ a.s. from the vanishing deterministic factor]
Assume $\prod_{n=1}^\infty a_n = 0$. Since $a_n \in (0, 1]$, the partial products $\prod_{i=1}^n a_i$ form a non-increasing sequence in $(0, 1]$ with $\prod_{i=1}^n a_i \to 0$ as $n \to \infty$.
From the factorisation $M_n = N_n^2 (\prod_{i=1}^n a_i)^2$, we have for each $\omega \in \Omega$,
\begin{align*}
0 \leq M_n(\omega) = N_n(\omega)^2 \Bigl(\prod_{i=1}^n a_i\Bigr)^{\!2}.
\end{align*}
Since $N_n \to N_\infty$ a.s. with $N_\infty$ finite, the factor $N_n(\omega)^2$ remains bounded on the a.s. event $\{N_\infty < \infty\}$, while $(\prod_{i=1}^n a_i)^2 \to 0$. Therefore $M_n(\omega) \to 0$ a.s., and since $M_n \to M_\infty$ a.s., we conclude $M_\infty = 0$ a.s.
[guided]
The factorisation $M_n = N_n^2 (\prod_{i=1}^n a_i)^2$ separates the random and deterministic contributions. Even though $N_n^2$ may be unbounded in $L^2$ when $\prod a_n = 0$ (since $\mathbb{E}[N_n^2] = (\prod_{i=1}^n a_i)^{-2} \to \infty$), the a.s. limit $N_\infty$ is still finite. This is because the [Almost Sure Martingale Convergence Theorem](/theorems/1157) requires only $L^1$ boundedness ($\mathbb{E}[N_n] = 1$ for all $n$), not $L^2$ boundedness.
With $N_\infty(\omega) < \infty$ a.s., the product $N_n(\omega)^2 \cdot (\prod_{i=1}^n a_i)^2$ is a bounded sequence times a sequence converging to zero, giving $M_n(\omega) \to 0$.
This shows that when $\prod a_n = 0$, the martingale $(M_n)$ is **not** uniformly integrable: $\mathbb{E}[M_n] = 1$ for all $n$ but $\mathbb{E}[M_\infty] = 0$. The $L^1$ mass "escapes to infinity" — the martingale concentrates its mass on events of vanishing probability. This is the prototypical example of a non-closable martingale.
[/guided]
[/step]
