Q) Let $X_n$ be a submartingale with $\sup X_n<\infty$. Let $\xi_n = X_n – X_{n-1}$ and let $E(\sup \xi_n^+)<\infty$. Show that $X_n$ converges a.s.
Martingale convergence theorem: If $X_n$ is a submartingale with $\sup EX_n^+<\infty$, then $X_n$ converges a.s.
If I take $T = \inf\{n:X_n\geq K\}$, then
$$\begin{align}
X_{n\wedge T}&\leq K + \sup\xi_n^+ \\
E(X_{n\wedge T})&\leq K + E(\sup\xi_n^+)<\infty \\
\implies \sup E(X_{n\wedge T}) &< \infty
\end{align}
$$
But how can I claim that $\sup E(X_{n\wedge T}^+) < \infty$, so that I can say $X_{n\wedge T}$ converges a.s. by Martingale convergence theorem?
If I have that, $X_n$ converges a.s. on $\{T=\infty\} = \{\sup X_n\leq K\}$ which would be true if I choose $K = \sup X_n +1 $.
Best Answer
Fix $K>0$. If $X_{n \wedge T}(\omega) \leq 0$, then $$X_{n \wedge T}^+(\omega)=0.$$ If $X_{n \wedge T}(\omega)>0$, then $$X_{n \wedge T}^+(\omega)=X_{n \wedge T}(\omega) \leq K + \sup_{n} \xi_n^+(\omega).$$ In summary, we get $$X_{n \wedge T}^+(\omega) \leq K+\sup_{n} \xi_n^+(\omega)$$ for all $\omega \in \Omega$. Hence, by the martingale convergence theorem, $X_n$ converges on $\{T=\infty\} \supseteq \{\sup_n X_n < K\}$.
Mind that it is not possible to chose $K=\sup_n X_n+1$. The constant $K$ in the above reasoning is a constant (not depending on $\omega$) whereas $\sup_n X_n+1$ is a random variable. It is, in general, not possible to find a deterministic constant $K$ such that $\sup_n X_n(\omega)\leq K$ for all $\omega \in \Omega$. We need to argue a bit differently:
From our earlier consideration we know that $X_n$ converges a.s. on $\{\sup_n X_n < K\}$. Since $K>0$ is arbitrary, this implies that $X_n$ converges a.s. on $$\left\{ \sup_n X_n < \infty\right\} = \bigcup_{K \in \mathbb{N}} \left\{ \{\sup_n X_n < K\right\}.$$ By assumption, $\{\sup_n X_n<\infty\}=\Omega$, and this finishes the proof.