In this reference, theorem 1.3.3 in page 16, it claims that if:
$$ E[|M_{n \wedge T}|^2] \leq C < \infty \implies E[M_T] = E[M_0] $$
Halfway through the proof I was stuck by this claim:
$$ E[|M_n|1\{|M_n| \geq b, T < n\}] \leq \frac{E[|M_{n \wedge T}|^2]}{b} $$
Previously, we bounded:
$$ E[|M_{n \wedge T}|^21\{T < n\}] \leq C $$
via holder inequality (put square into expectation) and monotone convergence theorem (partially to take the limit out of the expectation) on $E[|M_T|]^2$. I don't see a clear progression from this bound to the one that looks somewhat like a Markov inequality.
I tried searching up alternative sources for this proof but couldn't find this version of the theorem. Any suggestions/hints would be greatly appreciated!
Mind it's $\{T>n\}$, not $\{T<n\}$, in the reference. The function $\tilde{\mu}(A):=E[\mathbf{1}_A|M_n|],\,A \in \mathscr{F}$ is a finite measure on $(\Omega,\mathscr{F})$, because $M_n$ is $\mathscr{F}$-integrable. We write $d\tilde{\mu}=|M_n|dP$, where $|M_n|$ is then a density. By Markov's inequality, for any $B \in \mathscr{F}$ $$E[|M_n|\mathbf{1}_{\{|M_n|\geq b\}\cap B}]=\tilde{\mu}(\{|M_n|\geq b\}\cap B)\leq \frac{E^{\tilde{\mu}}[\mathbf{1}_B|M_n|]}{b}=\frac{E[\mathbf{1}_B|M_n|^2]}{b}$$ On $B=\{T>n\}$ we have $M_{n\wedge T}=M_n$. The claim follows.