# Doob’s Inequality II

Doob’s ${{\cal L}^p}$ Inequality for Right (Left) Continuous Martingales

In all the proofs I have seen, some of the details are elided; here is my proof with all the details.

Proof: Let ${D(m)}$ be an increasing set of finite sets whose union is dense in ${[0,t]}$ and note that by right (left) continuity, we have:

$\displaystyle \sup_{s \in [0,t]}Z_s(\omega) = \sup_m \sup_{s \in D(m)}Z_s(\omega)$

Thus:

$\displaystyle \mathop{\mathbb E} (\sup_{s \in [0,t]}Z_s(\omega)) =$

$\displaystyle \mathop{\mathbb E} (\sup_m \sup_{s \in D(m)} Z_s^p(\omega)) =$

By monotone convergence

$\displaystyle \sup_m \mathop{\mathbb E} \sup_{s \in D(m)} Z_s^p(\omega) \leq$

$\displaystyle \sup_m q^p \sup_{s \in D(m)}Z_s^p(\omega) =$

$\displaystyle q^p\sup_{s \in [0,t]}Z_s^p(\omega)$

$\Box$

Note that if ${Z_s(\omega)}$ is not continuous then all bets are off. For example, consider the finite sets on the interval ${[0,1]}$ of dyadic rationals:

$\displaystyle \begin{array}{rcl} D_0 & = & \{0,1\}\\ D_1 & = & \{0, \frac{1}{2}, 1\} \\ D_2 & = & \{0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1\} \\ & \ldots & \end{array}$

Then ${D = \cup_nD_n}$ is dense in ${[0,1]}$ and consider the function

$\displaystyle \begin{array}{rcl} Z_s(\omega) & = & 0\ {\rm if} s \in D \\ Z_s(\omega) & = & 1\ {\rm otherwise} \end{array}$

Then

$\displaystyle 1 = sup_{s \in [0,1]}Z_s(\omega) \neq \sup_m\sup_{s \in D(m)}Z_s(\omega) = 0$