Proof of Doob's L2 Maximal Inequality for Bounded Right-Continuous Martingales on a Compact Time Interval

Step 1: extension of the martingale to all nonnegative times. Define, for $t\ge0$,

$$\mathcal{F}^{\mathrm{ext}}_t=\mathcal{F}_{\min(t,T)},\qquad M^{\mathrm{ext}}_t=M_{\min(t,T)} .$$

The family $(\mathcal{F}^{\mathrm{ext}}_t)_{t\ge0}$ is a filtration, since $r\le t$ implies $\min(r,T)\le\min(t,T)$ and hence $\mathcal{F}^{\mathrm{ext}}_r\subseteq\mathcal{F}^{\mathrm{ext}}_t$. Each $M^{\mathrm{ext}}_t$ is measurable with respect to $\mathcal{F}^{\mathrm{ext}}_t$ and square-integrable, being one of the $M_s$ with $s\in[0,T]$. For $0\le r\le t$ the conditional expectation of $M^{\mathrm{ext}}_t$ given $\mathcal{F}^{\mathrm{ext}}_r$ equals $M^{\mathrm{ext}}_r$: if $t\le T$ this is the martingale property of $M$ on $[0,T]$; if $r\le T\le t$ it is the martingale property of $M$ for the pair $r\le T$, because $M^{\mathrm{ext}}_t=M_T$ and $\mathcal{F}^{\mathrm{ext}}_r=\mathcal{F}_r$; and if $T\le r$ both sides equal $M_T$, which is measurable with respect to $\mathcal{F}_T=\mathcal{F}^{\mathrm{ext}}_r$. Hence $M^{\mathrm{ext}}$ is a square-integrable martingale indexed by $t\ge0$.

Step 2: the inequality on finite dyadic grids. For a natural number $n$ let $D_n=\{kT2^{-n}:k\in\{0,1,\dots,2^n\}\}$, whose elements, listed in increasing order, form a finite family of sample times $0=t_0<t_1<\dots<t_{2^n}=T$. Applying Doob's L2 maximal inequality to the square-integrable martingale $M^{\mathrm{ext}}$ and these sample times gives that $\max_{t\in D_n}|M_t|$ is square-integrable with

$$\mathbb{E}\Big[\max_{t\in D_n}|M_t|^2\Big]\le4\,\mathbb{E}\big[M_T^2\big] .$$

The maximum of finitely many random variables is a random variable, since $\{\max(U,V)>c\}=\{U>c\}\cup\{V>c\}$ and one may iterate.

Step 3: passage to the supremum. The sets $D_n$ increase with $n$ and their union is $D$, so the random variables

$$W_n=\max_{t\in D_n}\big(|M_t|\,\mathbf{1}_{\Omega_0}\big)^2$$

are nondecreasing in $n$, and $W_n\le\max_{t\in D_n}|M_t|^2$, so $\mathbb{E}[W_n]\le4\,\mathbb{E}[M_T^2]$ by Step 2. They converge pointwise to $\overline{M}^{\,2}$: indeed $W_n\le\overline{M}^{\,2}$ for every $n$, while for $\varepsilon>0$ there is, by the definition of the supremum, some $t\in D$ with $|M_t|\,\mathbf{1}_{\Omega_0}>\overline{M}-\varepsilon$ when $\overline{M}>0$, and this $t$ lies in $D_n$ for all large $n$; the case $\overline{M}=0$ is immediate. By the monotone convergence theorem applied to the nondecreasing nonnegative sequence $(W_n)$, together with the bound $\mathbb{E}[W_n]\le4\,\mathbb{E}[M_T^2]$ just obtained,

$$\mathbb{E}\big[\overline{M}^{\,2}\big]=\lim_{n\to\infty}\mathbb{E}[W_n]\le4\,\mathbb{E}\big[M_T^2\big] .$$

Here $\overline{M}$ is a random variable by the supremum lemma, whose hypotheses hold by assumption. That lemma also gives $\overline{M}(\omega)=\sup_{t\in[0,T]}|M_t(\omega)|$ for every $\omega\in\Omega_0$, and since $P(\Omega_0)=1$ the expectation of the pathwise supremum, computed on $\Omega_0$, coincides with $\mathbb{E}[\overline{M}^{\,2}]$. $\blacksquare$