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In class we learned about Lp martingale convergence theorem. I could not figure out why |Xn-X∞|<2X* implies the convergence in Lp by the dominated convergence theorem.

Hope someone could explain a bit.

Lp martingale convergence theorem: Let p ∈ (1, +∞). Let Xn be an Lp-bounded martingale. Then there exists a random variable X∞ ∈ Lp(F∞) such that Xn → X∞ almost surely and in Lp.

The first part of the proof is clear that Xn is Lp-bounded so it is L1-bounded and there exists X∞ such that Xn → X∞ almost surely and in L1.

For the final part: I checked lots of proof for this theorem and find that it used Doob's inequality to show that X∞ is indeed in Lp, and |Xn-X∞| is bounded by 2X*, for which X* = sup|Xn|. And then all the proofs just state by Dominated Convergence theorem Xn converges to X∞ in Lp, which is the part that I do not really understand and hope I could get some explanations for why it uses dominated convergence theorem to conclude the proof.

TNightS
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  • It is a bit unclear what you are asking. Can you add more context to the question? What is $X^*$? Presumably ${X_n}$ here is a martingale, with some additional assumptions such as being uniformly bounded in $L^p$? – Math1000 Nov 07 '19 at 02:29
  • Thanks for the comment, I added more details for the question. – TNightS Nov 07 '19 at 11:46
  • I nominated the question for reopening. In the future, though, be sure to include these details in your original post! Also it is highly preferred that you use MathJax to format your equations - here is a tutorial: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Math1000 Nov 07 '19 at 21:47

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Let $Y_n:= \lvert X_n-X\rvert^p$. Then $0\leqslant \sup_{n\geqslant 1} \lvert Y_n\rvert\leqslant 2^p\sup_{n\geqslant 1}\lvert X_n\rvert^p$. Since $Y_n\to 0$ almost surely, in order to apply the dominated convergence theorem, it suffices to prove that $\sup_{n\geqslant 1}\lvert X_n\rvert^p$ is integrable. To this aim, observe that by Doob's inequality, for a fixed $N$, $$ \mathbb E\left[ \max_{1\leqslant n\leqslant N}\lvert X_n\rvert^p\right]\leqslant \frac{p}{p-1}\mathbb E\left[ \lvert X_N\rvert^p\right]\leqslant \frac{p}{p-1}\sup_{\ell\geqslant 1}\mathbb E\left[ \lvert X_{\ell}\rvert^p\right] $$ hence by the monotone convergence theorem, $$ \mathbb E\left[ \sup_{n\geqslant 1}\lvert X_n\rvert^p\right]\leqslant \frac{p}{p-1}\sup_{\ell\geqslant 1}\mathbb E\left[ \lvert X_{\ell}\rvert^p\right]. $$

Davide Giraudo
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  • Thanks for the answer. Just a side question, how can we apply the dominated convergence theorem in Lp spaces? I know how it works in general cases, but I did not quite get the conditions for it in Lp spaces. – TNightS Nov 07 '19 at 13:28
  • The convergence in $L^p$ of $(X_n)$ to $X$ is nothing but the convergence in $L^1$ of $(\lvert X_n-X\rvert^p)$. – Davide Giraudo Nov 07 '19 at 14:01