= Solution
The <almost sure supermartingale convergence theorem> applies to the $L^1$-bounded <supermartingale> $X$. Hence $X_n\to X_\infty$ <almost surely>, and the <Fatou lemma> gives $\mathbb E|X_\infty|\leq\sup_n\mathbb E|X_n|<\infty$. Take the version of $X_\infty$ obtained from the sequence, so it is measurable for $\mathcal F_\infty=\sigma(\bigcup_n\mathcal F_n)$. Set
$$
M_n=\mathbb E[X_\infty\mid\mathcal F_n],\qquad Y_n=X_n-M_n.
$$
The <tower property of conditional expectation> makes $M$ a <martingale>, and the <uniform integrability of conditional expectations> makes it <uniformly integrable>. Explicitly, with $Z=X_\infty$, $C=\mathbb E|Z|$, and $R,K>0$, we have $|M_n|\leq R+\mathbb E[|Z|\mathbf1_{\{|Z|>R\}}\mid\mathcal F_n]$, so
$$
\mathbb E[|M_n|\mathbf1_{\{|M_n|>K\}}]\leq\frac{RC}{K}+\mathbb E[|Z|\mathbf1_{\{|Z|>R\}}].
$$
First choose $R$ large, then $K$ large. This proves <uniform integrability> uniformly in $n$.
The <uniformly integrable martingale convergence theorem> gives a limit $M_\infty$ both <almost surely> and in $L^1$. For $A\in\mathcal F_j$ and $n\geq j$, $\mathbb E[M_n\mathbf1_A]=\mathbb E[X_\infty\mathbf1_A]$. Passing to the $L^1$ limit and then extending from the algebra $\bigcup_j\mathcal F_j$ to $\mathcal F_\infty$ identifies $M_\infty=X_\infty$ <almost surely>. Finally,
$$
\mathbb E[Y_{n+1}\mid\mathcal F_n]=\mathbb E[X_{n+1}\mid\mathcal F_n]-M_n\leq Y_n,
\qquad Y_n=X_n-M_n\longrightarrow0\quad\text{a.s.}
$$
Thus \b[$X=M+Y$ with $M$ uniformly integrable and $Y$ a supermartingale tending almost surely to zero]. The <terminal decomposition of an L1-bounded supermartingale> does not require $Y\geq0$ or $Y_n\to0$ in $L^1$. Indeed, for $X$ equal to the negative of the <fair-coin doubling martingale>, $X_\infty=0$, $M=0$, and $Y=X$ is negative and has constant absolute mean one.
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