How the author uses MCT to complete his proof of Doob’s martingale maximal inequalities?

Question

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space and $M=(M_t, t\ge 0)$ a continuous martingale with respect to a filtration $(\mathcal F_t, t\ge 0)$. Let $T>0$ and $p > 1$. Let $X := \sup_{t \in [0, T]} |M_t| \ge 0$. Because $M$ has continuous paths, $X$ is Borel measurable.

I'm reading a proof of below theorem from this link.

Theorem If $\mathbb E [ |M_T|^p] < \infty$ then $$ \mathbb E [ X^p ] \le \bigg (\frac{p}{p-1} \bigg)^p \mathbb E [ |M_T|^p ]. $$

Proof Fix $\lambda>0$. Then $$ \mathbb P [ X \ge \lambda ] \le \frac{1}{\lambda} \mathbb E [ |M_T| 1_{\{X \ge \lambda\}}] . $$

So $$ \int_0^{+\infty} \lambda^{p-1} \mathbb P [X \ge \lambda] \mathrm d \lambda \le \int_0^{+\infty} \lambda^{p-2} \mathbb E [ |M_T| 1_{\{X \ge \lambda\}} ] \mathrm d \lambda. $$

First, $$ \begin{align} \int_0^{+\infty} \lambda^{p-1} \mathbb P [X \ge \lambda] \mathrm d \lambda &= \int_0^{+\infty} \lambda^{p-1} \int_\Omega 1_{\{X \ge \lambda\}} (\omega) \mathrm d \mathbb P (\omega) \mathrm d \lambda \\ &= \int_\Omega \bigg [ \int_0^{X (\omega)} \lambda^{p-1} \mathrm d \lambda \bigg ]\mathrm d \mathbb P (\omega) \quad \text{by Tonelli's theorem}\\ &= \frac{\mathbb E [X^p]}{p}. \end{align} $$

Second, $$ \begin{align} \int_0^{+\infty} \lambda^{p-2} \mathbb E [ |M_T| 1_{\{X \ge \lambda\}} ] \mathrm d \lambda &= \int_0^{+\infty} \lambda^{p-2} \int_\Omega |M_T (\omega)| 1_{\{X \ge \lambda\}} (\omega) \mathrm d \mathbb P (\omega) \mathrm d \lambda \\ &= \int_\Omega |M_T (\omega)| \bigg [ \int_0^{X (\omega)} \lambda^{p-2} \mathrm d \lambda \bigg ] \mathrm d \mathbb P (\omega) \quad \text{by Tonelli's theorem}\\ &= \frac{\mathbb E [ X^{p-1} |M_T|]}{p-1}. \end{align} $$

Hence $$ \mathbb E [X^p] \le \frac{p}{p-1} \mathbb E [ X^{p-1} |M_T| ]. $$

By Hölder's inequality, $$ \mathbb E [ X^{p-1} |M_T| ] \le (\mathbb E [ |M_T|^p ])^{\frac{1}{p}} (\mathbb E [ X^p ])^{\frac{p-1}{p}}. $$

So $$ \mathbb E [X^p] \le \frac{p}{p-1} (\mathbb E [ |M_T|^p ])^{\frac{1}{p}} (\mathbb E [ X^p ])^{\frac{p-1}{p}}. $$

1. We consider $\mathbb E [X^p] < \infty$. Then $$ \mathbb E [X^p] \le \bigg (\frac{p}{p-1} \bigg)^p \mathbb E [ |M_T|^p ]. $$

2. We consider $\mathbb E [X^p] = \infty$. For $n \in \mathbb N$, let $$ \tau_n := T \wedge \inf \{t \ge 0 : |M_t| \ge n\}. $$

Then $\tau_n$ is a stopping time w.r.t. $(\mathcal F_t, t\ge 0)$ such that $\tau_n \le T$. By Doob’s optional stopping theorem, $(M_{t \wedge \tau_n}, t \ge 0)$ is a martingale w.r.t. $(\mathcal F_t, t\ge 0)$. Notice that $$ |M_{t \wedge \tau_n}| \le \max \{n, |M_T|\} \quad \forall t \ge 0, $$

Let $X_n := \sup_{t \in [0, T]}|M_{t \wedge \tau_n}|$. Then $\mathbb E [X_n^p] < \infty$. By (1.), we get $$ \mathbb E [X_n^p] \le \bigg (\frac{p}{p-1} \bigg)^p \mathbb E [ |M_{T \wedge \tau_n}|^p ]. $$

We have $\tau_n \overset{n \to \infty}{\longrightarrow} T$, so $t \wedge \tau_n \overset{n \to \infty}{\longrightarrow} t$ for all $t \in [0, T]$. Because $M$ has continuous paths, $M_{t \wedge \tau_n} \overset{n \to \infty}{\longrightarrow} M_t$ a.s. for all $t \in [0, T]$.

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Could you explain how the author uses the monotone convergence theorem to complete the proof of part (2.)?

Answer 1

We got $$ \mathbb E [X_n^p] \le \bigg (\frac{p}{p-1} \bigg)^p \mathbb E [ |M_{T \wedge \tau_n}|^p ]. $$ As $T \wedge \tau_n$ is a stopping time we have: $M_{T \wedge \tau_n}$ is martingale and $|M_{T \wedge \tau_n}|^p $ is submartingale (because $f(x) = |x|^p$ is a convex function). Thus $E [ |M_{T \wedge \tau_n}|^p ] \le E [ |M_{T}|^p ]$, because $T$ and $T \wedge \tau_n$ are stopping times such that $T \wedge \tau_n \le T$. So, We got $$ \mathbb E [X_n^p] \le \bigg (\frac{p}{p-1} \bigg)^p \mathbb E [ |M_{T \wedge \tau_n}|^p ] \le \bigg (\frac{p}{p-1} \bigg)^p \mathbb E [ |M_{T}|^p ]. $$ As $X_n^p = |...|^p \ge 0$ it's sufficient to show that $X_n^p$ is nondecreasing, because in this case we may apply monotone convergence theorem and get $$\mathbb E [X_n^p] \to \mathbb E [(\lim X_n)^p] = \mathbb E \sup_{t \in [0, T]}|M_{t}|^p $$

So it's sufficient to show that $X_n$ is nondecreasing - it can be done by definition.