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I am trying to prove that a certain martingale $(R_t)_{t\geq 0}$ is uniformly integrable over a finite time interval $[0,T]$.

Now I know that the definition of uniform integrability is that $\lim_{a\to\infty}\sup_{t\geq 0}\mathbb{E}[|R_t|\mathbb{I}_{|R_t|>a}]=0$ where $\mathbb{I}$ is the indicator function.

I know also that in general showing $\sup_{t}\mathbb{E}{|R_t|}<\infty$ is not sufficient to show $R$ is uniformly integrable (but $\sup_t \mathbb{E}|R_t|^2 < \infty$ is sufficient) but what about in the case of a bounded time interval $[0,T]$. Is $\mathbb{E}[\sup_{t\leq T} |R_t|]<\infty$ sufficient to show uniform integrability on $[0,T]$; i.e. that $\lim_{a\to\infty}\sup_{t\leq T}\mathbb{E}[|R_t|\mathbb{I}_{|R_t|>a}]=0$?

It feels intuively like the reason $\sup_t\mathbb{E}|R_t|<\infty$ fails as a condition is because in some vague sense: "when $t$ is very large $|R_t|$ has a much larger chance of being >a" but I don't know how I would go about proving something like this?

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  • $\begingroup$ If $R_t \to R_T$ in $L^1$-norm then it's equivalent to the fact that the family $R_t$ converges in probability to $R_T$ and is Uniformly Integrable. This is sometimes referred to as Vitali's theorem unless mistaken en.wikipedia.org/wiki/Vitali_convergence_theorem (I think you can find proofs of this fact on MSE). Your question seems involved to me, so if you can show $L^1$ cv in your case then you don't need to go through the complex argument you are looking for. $\endgroup$
    – TheBridge
    May 23 at 9:35

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If you have a (deterministic) time interval $[0,T]$, and a positive martingale $(R_t)_{0 \le t \le T}$ in some filtration $(\mathcal{F}_t)_{0 \le t \le T}$, then for every $t \in [0,T]$, $R_t = E[R_T|\mathcal{F}_t]$. This forces uniform integrability, since given any integrable random variable $Z$, and any family of sub-$\sigma$ fields $(\mathcal G_i)_{i \in I}$, the family of conditional expectations $(E[Z|\mathcal G_i])_{i \in I}$ is uniformly integrable.

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