Proof of Doob's L2 Maximal Inequality in Discrete Time

Part 1: nonnegative submartingales. By Doob's Maximal Inequality for Square-Integrable Submartingales, $M^{*}$ is a square-integrable random variable, and for every $u>0$,

Since every $M_{t_k}\ge0$ pointwise, also $M^{*}\ge0$ pointwise. By Layer-Cake Formula for the Second Moment applied to $Y=M^{*}$,

with $m$ Lebesgue measure and $\varphi$ measurable.

As in the proof of the layer-cake lemma, form the product measure $P\otimes m$ (Product Sigma-Algebra, Existence and Uniqueness of the Product Measure; $P$ is finite and $m$ is $\sigma$-finite since $m([-n,n])=2n$), and let

the union over positive rationals $q$, which lies in $\mathcal{F}\otimes\mathcal{B}(\mathbb{R})$ by the density of the rationals, the countability of the rationals, and Product Sigma-Algebra. Define $h(\omega,u)=2M_{t_n}(\omega)$ for $(\omega,u)\in E'$ and $h=0$ otherwise; $h\ge0$ since $M_{t_n}\ge0$. For $a\ge0$,

and $\{h>a\}=\Omega\times\mathbb{R}$ for $a<0$; by the half-line criterion of Measurable Function and Real-Valued Measurable Function, $h$ is measurable. By the Tonelli part of Tonelli and Fubini Theorems:

$u$-slices. For $u\le0$ the slice is $0$; for $u>0$ it is the function $2M_{t_n}\mathbf{1}_{\{M^{*}>u\}}$, with integral $\psi(u)=2\,\mathbb{E}[M_{t_n}\mathbf{1}_{\{M^{*}>u\}}]\ge0$. Tonelli asserts that $\psi$ (extended by $0$ for $u\le0$) is measurable and $\int_{\mathbb{R}}\psi\,dm=\int h\,d(P\otimes m)$.

$\omega$-slices. Fix $\omega$ and put $c=2M_{t_n}(\omega)\ge0$, $d=M^{*}(\omega)\ge0$. The slice $h(\omega,\cdot)$ is the simple function $c\,\mathbf{1}_{(0,d)}$, with integral $c\,m((0,d))=c\,d$ by the interval property of Existence of Lebesgue Measure on the Real Line (for $d=0$ the interval is empty and the integral is $0=c\,d$). Hence the other iterated integral is $\int_{\Omega}2M_{t_n}M^{*}\,dP$, and Tonelli gives

the product $M_{t_n}M^{*}$ being integrable as a product of square-integrable random variables (Square-Integrable Random Variables and the Mean-Square Inner Product).

By the displayed weak-type inequality, $\varphi\le\psi$ pointwise on $\mathbb{R}$ (both vanish for $u\le0$), so by monotonicity of the integral (Linearity and Monotonicity of the Lebesgue Integral) and the Cauchy-Schwarz inequality for the mean-square inner product,

with the mean-square norm $\lVert X\rVert_{2}=\mathbb{E}[X^{2}]^{1/2}$. If $\lVert M^{*}\rVert_{2}=0$, then $\mathbb{E}[(M^{*})^{2}]=0\le4\,\mathbb{E}[M_{t_n}^{2}]$. Otherwise, dividing the inequality $\lVert M^{*}\rVert_{2}^{2}\le2\lVert M_{t_n}\rVert_{2}\lVert M^{*}\rVert_{2}$ by $\lVert M^{*}\rVert_{2}>0$ gives $\lVert M^{*}\rVert_{2}\le2\lVert M_{t_n}\rVert_{2}$, and squaring both (nonnegative) sides gives $\mathbb{E}[(M^{*})^{2}]\le4\,\mathbb{E}[M_{t_n}^{2}]$.

Part 2: martingales. Define $N_t=|M_t|$ pointwise for every $t\ge0$. We check that $N=(N_t)_{t\ge0}$ is a square-integrable submartingale with $N_t\ge0$ pointwise.

Adapted and square-integrable. For $a\ge0$, $\{N_t>a\}=\{M_t>a\}\cup\{M_t<-a\}\in\mathcal{F}_t$ by clause (i) of Square-Integrable Martingale, Submartingale, and Supermartingale and Filtration, Adapted Process, and Natural Filtration, and $\{N_t>a\}=\Omega$ for $a<0$; by the half-line criterion of Measurable Function and Real-Valued Measurable Function, $N_t$ is $\mathcal{F}_t$-measurable. Since $N_t^{2}=M_t^{2}$ pointwise, $N_t$ is square-integrable.

Submartingale property. Fix real $0\le s\le t$ and $A\in\mathcal{F}_s$, and split $A^{+}=A\cap\{M_s>0\}$ and $A^{-}=A\cap\{M_s\le0\}$, both in $\mathcal{F}_s$ since $M_s$ is $\mathcal{F}_s$-measurable and a $\sigma$-algebra contains complements and intersections. Pointwise $N_t\ge M_t$ and $N_t\ge-M_t$; all products below are integrable since multiplying by indicators only decreases absolute values. On $A^{+}$ one has $M_s=N_s$ pointwise, so by monotonicity of the integral (Linearity and Monotonicity of the Lebesgue Integral) and the averaged martingale identity of Square-Integrable Martingale, Submartingale, and Supermartingale applied to $A^{+}\in\mathcal{F}_s$,

On $A^{-}$ one has $-M_s=N_s$ pointwise, so by monotonicity, linearity, and the averaged identity applied to $A^{-}\in\mathcal{F}_s$,

Adding the two inequalities and using $\mathbf{1}_{A}=\mathbf{1}_{A^{+}}+\mathbf{1}_{A^{-}}$ pointwise with linearity of the integral,

which is the averaged submartingale inequality of Square-Integrable Martingale, Submartingale, and Supermartingale.

Conclusion. The process $N$ is a square-integrable submartingale with $N_t(\omega)\ge0$ everywhere, and its running maximum at the times $t_0<\dots<t_n$ is $\overline{M}$ pointwise. By Doob's Maximal Inequality for Square-Integrable Submartingales, $\overline{M}$ is a square-integrable random variable, and by Part 1 applied to $N$,

since $N_{t_n}^{2}=M_{t_n}^{2}$ pointwise. $\blacksquare$