Let $(\Omega, \mathscr{F}, \{\mathscr{F_n}\}_{n \in \mathbb{N}}, \mathbb{P})$ be a filtered probability space, and let $Y = ({Y_n})_{n \in \mathbb{N}}$ be a/an $(\{\mathscr{F_n}\}_{n \in \mathbb{N}}, \mathbb{P})$-supermartingale.
Prove $Y_S$ is integrable if $Y$ is a bounded supermartingale and $S$ is an a.s. finite stopping time.
So far all I was able to show is that $Y_{S \wedge n}$ is integrable and $E[Y_S] \le E[Y_0]$.
Hints pls?
As we have already discussed this depends on your definition of a "bounded process".
If you mean that $Y$ is (uniformly) bounded, i.e. there exists $C>0$ such that $|Y_n| \leq C$ for all $n \in \mathbb{N}$, i.e. $$|Y_n(\omega)| \leq C \qquad \text{for all $\omega \in \Omega$ and $n \in \mathbb{N}$} \tag{1},$$ then the claim does hold true. Indeed: Fix $\omega \in \Omega$. Then $$|Y_{S}(\omega)| \stackrel{\text{def}}{=} |Y_{S(\omega)}(\omega)| \stackrel{(1)}{\leq} C.$$ This shows that $Y_S$ is bounded; hence in particular integrable.
If you mean that each $Y_n$ is bounded, i.e. there exists $C_n>0$ such that $|Y_n| \leq C_n$, then the claim does, in general, not hold true as the following counterexample shows.
Consider $\Omega := \mathbb{N}$ endowed with the counting measure
$$\mathbb{P}(d\omega) := c \sum_{n \in \mathbb{N}} \frac{1}{n^2} \delta_n(d\omega)$$
where the constant $c$ is chosen such that $\mathbb{P}(\mathbb{N})=1$. Define a (non-negative) random variable by
$$X(\omega) := \frac{1}{\omega}, \qquad \omega \in \Omega.$$
Then $Y_k :=-k^2 X$ is a supermartingale (see the lemma below), each $Y_k$ is bounded and
$$S := \inf\{k \in \mathbb{N}; Y_k = -k\}$$
defines a (finite) stopping time. By the very definition of $(Y_k)_{k \in \mathbb{N}}$ and $S$, we have $$\{S=k\} = \left\{\omega \in \Omega; -k^2 \frac{1}{\omega} = -k \right\} = \{k\}.$$
Consequently,
$$\mathbb{E}(|Y_S|) = \sum_{k=1}^{\infty} |k| \cdot \mathbb{P}(S=k) =c \sum_{k=1}^{\infty} \frac{k}{k^2} = \infty,$$
i.e. $Y_S$ is not integrable.
The following lemma (which we used for the above counterexample) is not difficult to prove:
Lemma: Let $X \geq 0$ be an integrable random variable on a probability space $(\Omega,\mathcal{A},\mathbb{P})$ and $(a_k)_{k \in \mathbb{N}} \subseteq \mathbb{R} \backslash \{0\}$ a sequence of non-increasing real numbers. Then $Y_k := a_k X$ defines a supermartingale (with respect to the canonical filtration).