Lp-bounded martingale convergence

Since the underlying measure is a probability measure, Hölder gives ${\sup }_n\mathbb{E}\mathrm{\mid }{M}_n\mathrm{\mid }\le C^{1\mathrm{/}p}$. In particular ${\sup }_n\mathbb{E}(M_n{)}^{+}\le C^{1\mathrm{/}p}$. The martingale $M$ is also a submartingale, so F1 applies directly to $M$ and gives $M_n\to M_{\mathrm{\infty }}$ almost surely for a finite integrable $M_{\mathrm{\infty }}$.

For each $N$, F2 gives ${\parallel {\max }_{k\le N}\mathrm{\mid }{M}_k\mathrm{\mid }\parallel }_p\le q{\sup }_n\mathrm{\parallel }{M}_n{\mathrm{\parallel }}_p,q=\frac{p}{p-1}\mathrm{.}$ The maxima increase to $M^{\ast }={\sup }_n\mathrm{\mid }{M}_n\mathrm{\mid }$, so F3 yields the displayed infinite-horizon bound and $M^{\ast }\in L^p$. In particular $\mathrm{\mid }{M}_{\mathrm{\infty }}\mathrm{\mid }\le M^{\ast }$ almost surely, hence $M_{\mathrm{\infty }}\in L^p$.

We have $\mathrm{\mid }{M}_n-M_{\mathrm{\infty }}{\mathrm{\mid }}^p\le (2M^{\ast }{)}^p$ and pointwise convergence to zero. Dominated convergence gives $\mathrm{\parallel }{M}_n-M_{\mathrm{\infty }}{\mathrm{\parallel }}_p\to 0$.

Fix $n$ and take $m\ge n$. F4 gives $M_n=\mathbb{E}[M_m\mid {\mathcal{F}}_n]$. Conditional $L^p$ contraction and step 2.1 imply $\mathrm{\parallel }\mathbb{E}[M_m-M_{\mathrm{\infty }}\mid {\mathcal{F}}_n]{\mathrm{\parallel }}_p\le \mathrm{\parallel }{M}_m-M_{\mathrm{\infty }}{\mathrm{\parallel }}_p⟶0.$ The left conditional expectations therefore converge to zero while $M_n$ is fixed, proving $M_n=\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_n]$ almost surely. AC has only the inherited role in F5.