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From David William's book Probability with Martingales.

Doob's Optional stopping theorem:

Let $T$ be a stopping time and $X$ be a supermartingale. Then $X_T$ is integrable and $E(X_T)\le E(X_0)$ in each of the following situations

(i) $T$ is bounded (for some $N \in \mathbb{N}$, $T(\omega)\le N$);

(ii) $X$ is bounded (for some $K \in \mathbb{R}^{+}$, $|X(\omega)|\le K$ for every $n$ and every $\omega$) and $T$ is a.s finite.

(There are more conditions but I'm not too worried about them)

My question is that what is the difference between (i) and (ii)? I understand the proof behind them in the sense that I can justify all the steps in the proof but I believe I'm missing some subtlety not required in the proof because intuitively $I$ thought the the conditions placed on T were the same. Proof 1: choose $n=N$, proof 2 uses Dominated Convergence Theorem.

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If $T \sim Geo(p)$, $T$ is a.s. finite but there is no $N$ such that $T(w) \leq N a.s.$ – madprob Nov 11 '12 at 2:49
have you solved this question? or are you still looking for an answer? – Conrado Costa Jun 30 '15 at 20:12
@ConradoC madprob's counter example was exactly what I was looking for but I couldn't flag it as an answer. I could just tag your answer as well if you provide one. – Pk.yd Jul 23 '15 at 10:51
@Pk.yd That is not needed, It's all clarified. – Conrado Costa Jul 23 '15 at 12:59

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