= Solution
{c}
The products are nonnegative and measurable. <independence> and the unit means give $\mathbb EY_n=\prod_{k\leq n}\mathbb E\xi_k=1$, so they are integrable.
Let $\mathcal H_n=\sigma(Y_0,\ldots,Y_n)$ be the <natural filtration>. It is contained in $\sigma(\xi_1,\ldots,\xi_n)$, so $\xi_{n+1}$ is independent of $\mathcal H_n$ even if some products have vanished. Using the bounded or integrable nonnegative conditional-product identity,
$$
\mathbb E(Y_{n+1}\mid\mathcal H_n)
=Y_n\mathbb E(\xi_{n+1}\mid\mathcal H_n)
=Y_n.
$$
Thus \b[$Y$ is a <martingale> in its <natural filtration>].
Doob's convergence theorem in the form needed here states that a nonnegative <supermartingale> converges almost surely to a finite integrable limit; more generally a <submartingale> with uniformly bounded first absolute moments has such an <almost sure convergence>. Applying it to $Y$ gives
$$
\boxed{Y_n\to Y_\infty\text{ almost surely},\qquad
\mathbb E Y_\infty\leq1.}
$$
The <expected value> bound follows from <Fatou's lemma>. The theorem does not assert $L^1$ convergence or equality of <expected values> without a uniform-integrability hypothesis.
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