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By the convergence theorem we know that given a L$^1$-bounded submartingale $X_t$, $t\geq1$, with $\sup_{t\geq0}\mathbb{E}[X_t^+]<\infty$, the sequence $X_t(\omega)$ converges $\mathbb{P}$-a.s. to some $X(\omega)$. This follows as an application of the upcrossing inequality. However, this theorem doesn't prove convergence in L$^1$. In fact, if we consider a martingale $M_t=(\frac{1-p}{p})^{S_t}$, where $S_t$ is an asymmetric random walk with $\mathbb{E}[X_i=1]=p, \mathbb{E}[X_i=-1]=1-p$, then $M_t\rightarrow0, \mathbb{P}$-a.s., but $\|M_t\|_1=\mathbb{E}[M_t]=\mathbb{E}[M_0]=1$, which contradicts L$^1$ convergence to $0.$

My question is, what are the assumptions required in order to have convergence in $L^1$ of $L^1$-bounded submartingales?

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You have that the following are equivalent for a martingale (EDIT:I read the question wrong):

$1)$ Convergence in $L^1$

$2)$ Uniform integrability

$3)$ The martingale being closed, i.e. $\exists Z $ s.t. $M_t=\mathbb{E}[Z|\mathcal{F_t}]$

provided you have right continuous sample paths.

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  • $\begingroup$ Does it also apply to submartingales? $\endgroup$
    – FunnyBuzer
    Jul 16, 2018 at 12:48
  • $\begingroup$ Oh, my bad, read martingale instead of sub-martingale. I don't think it does, sorry :( $\endgroup$
    – asdf
    Jul 16, 2018 at 12:49
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    $\begingroup$ Condition 3 obviously doesn't fit for submartingales but if $X_t \to X$ almost surely then $X_t \to X$ in $L^1$ if and only if $X$ is uniformly integrable. In general, on probability spaces, convergence in $L^1$ is equivalent to uniform integrability + convergence in measure. $\endgroup$ Jul 16, 2018 at 13:07
  • $\begingroup$ @RhysSteele could you prove the "if and only if" condition you mentioned? $\endgroup$
    – FunnyBuzer
    Jul 16, 2018 at 15:59
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    $\begingroup$ This is quite a famous result - see William's "Probability with martingales" or just google it. I guarantee you'll find lots of resutls $\endgroup$
    – asdf
    Jul 16, 2018 at 16:27

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