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.
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, $$ \mathbb E\left[ \sup_{n\geqslant 1}\lvert X_n\rvert^p\right]\leqslant \frac{p}{p-1}\sup_{\ell\geqslant 1}\mathbb E\left[ \lvert X_{\ell}\rvert^p\right]. $$