Proof of Doob's Maximal Inequality for Square-Integrable Submartingales

Throughout, clauses (i)–(iii) refer to Square-Integrable Martingale, Submartingale, and Supermartingale, and integrals, linearity, and monotonicity of integrals are those of Linearity and Monotonicity of the Lebesgue Integral; the integral of a simple function $c\,\mathbf{1}_{A}$ is $c\,P(A)$.

Claim 1. For every real $a$,

$$\{M^{*}>a\}=\bigcup_{k=0}^{n}\{M_{t_k}>a\},$$

since the largest of the values $M_{t_0}(\omega),\dots,M_{t_n}(\omega)$ exceeds $a$ exactly when at least one of them does. Each $\{M_{t_k}>a\}$ is an event, and a $\sigma$-algebra contains finite unions, so $M^{*}$ is measurable by the half-line criterion of Measurable Function and Real-Valued Measurable Function. Pointwise, $M^{*}(\omega)$ equals $M_{t_{k}}(\omega)$ for some $k$ (depending on $\omega$), so

$$(M^{*})^{2}\le M_{t_0}^{2}+M_{t_1}^{2}+\dots+M_{t_n}^{2}$$

pointwise, the omitted terms being nonnegative. Each $M_{t_k}^{2}$ is integrable by clause (ii) and Square-Integrable Random Variables and the Mean-Square Inner Product, the finite sum is integrable by linearity, and monotonicity gives $\int(M^{*})^{2}\,dP<\infty$; hence $M^{*}$ is square-integrable.

Claim 2, threshold $\ge\lambda$. Fix $\lambda>0$ and set $A=\{M^{*}\ge\lambda\}$; this is an event, being the complement of the event $\{M^{*}<\lambda\}$, which lies in $\mathcal{F}$ because $M^{*}$ is measurable. Define, for $0\le k\le n$,

$$A_{k}=\{M_{t_k}\ge\lambda\}\cap\bigcap_{j=0}^{k-1}\{M_{t_j}<\lambda\},$$

the intersection over $j$ being $\Omega$ when $k=0$. The sets $A_{0},\dots,A_{n}$ are pairwise disjoint (on $A_{k}$ the index $k$ is the least index whose value reaches $\lambda$) and their union is $A$: if $M^{*}(\omega)\ge\lambda$, then $M_{t_k}(\omega)\ge\lambda$ for at least one $k$, and $\omega$ lies in $A_{k}$ for the least such $k$; conversely every $A_{k}$ is contained in $A$.

By clause (i) and Filtration, Adapted Process, and Natural Filtration, each $M_{t_j}$ is $\mathcal{F}_{t_j}$-measurable, and the filtration is increasing, so $\mathcal{F}_{t_j}\subseteq\mathcal{F}_{t_k}$ for $j\le k$; since a $\sigma$-algebra contains complements and finite intersections, $A_{k}\in\mathcal{F}_{t_k}$.

Pointwise, $\lambda\,\mathbf{1}_{A_k}\le M_{t_k}\mathbf{1}_{A_k}$: on $A_{k}$ one has $M_{t_k}\ge\lambda$, and off $A_{k}$ both sides vanish. The left side is a simple function with integral $\lambda\,P(A_k)$; the right side is integrable since $|M_{t_k}\mathbf{1}_{A_k}|\le|M_{t_k}|$ and square-integrable random variables are integrable (Square-Integrable Random Variables and the Mean-Square Inner Product). By monotonicity and then the averaged submartingale inequality of Square-Integrable Martingale, Submartingale, and Supermartingale applied with $s=t_{k}$, $t=t_{n}$, and the event $A_{k}\in\mathcal{F}_{t_k}$,

$$\lambda\,P(A_{k})\le\mathbb{E}\bigl[M_{t_k}\mathbf{1}_{A_k}\bigr]\le\mathbb{E}\bigl[M_{t_n}\mathbf{1}_{A_k}\bigr]\qquad(0\le k\le n).$$

Summing over $k$: the left sides sum to $\lambda\,P(A)$ by the additivity of the measure $P$ on the disjoint union $A=\bigcup_k A_k$; the right sides sum to $\mathbb{E}[M_{t_n}\mathbf{1}_{A}]$ by linearity of the integral, since $\mathbf{1}_{A}=\sum_{k=0}^{n}\mathbf{1}_{A_k}$ pointwise. This proves the first inequality.

Claim 2, threshold $>\lambda$. The same argument applies verbatim with

$$A'=\{M^{*}>\lambda\},\qquad A'_{k}=\{M_{t_k}>\lambda\}\cap\bigcap_{j=0}^{k-1}\{M_{t_j}\le\lambda\}:$$

the sets $A'_{k}$ are pairwise disjoint events with union $A'$ and $A'_{k}\in\mathcal{F}_{t_k}$ as before, and on $A'_{k}$ one has $M_{t_k}>\lambda$, so the pointwise bound $\lambda\,\mathbf{1}_{A'_k}\le M_{t_k}\mathbf{1}_{A'_k}$ again holds, and the chain of inequalities and the summation are unchanged. $\blacksquare$