Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\in\mathbb{T}},\mathbb{P})$ be a filtered probability space with countable index set $\mathbb{T}$ . If $(X_t)_{t\in\mathbb{T}}$ is a submartingale, we show that \begin{equation}\label{eq:1} \mathbb{P}\left(\sup_{s\le t}X_s\ge K\right)\le K^{-1}\mathbb{E}[(X_t)_+] \end{equation}and if $X$ is a martingale or nonnegative submartingale then,

	(1)
	(2)

First, let us consider the case where $\mathbb{T}$ is finite. The first time at which $X_t\ge K$ , \begin{equation*} \tau=\inf\left\{t\in\mathbb{T}:X_t\ge K\right\} \end{equation*}is a stopping time (as hitting times are stopping times). By Doob's optional sampling theorem for submartingales $X_{\tau\wedge t}\le\mathbb{E}[X_t\mid\mathcal{F}_{\tau\wedge t}]$ and therefore, \begin{equation*} K\mathbb{P}(\tau\le t) \le\mathbb{E}[1_{\{\tau\le t\}}X_{\tau\wedge t}] \le\mathbb{E}[1_{\{\tau\le t\}}X_{t}] \end{equation*}However, $\tau\le t$ if and only if $\sup_{s\le t}X_s\ge K$ giving, \begin{equation}\label{eq:4} \mathbb{P}\left(\sup_{s\le t}X_s\ge K\right)\le K^{-1}\mathbb{E}[1_{\{\sup_{s\le t}X_s\ge K\}}X_t], \end{equation}where the supremum is understood to be over $s\in\mathbb{T}$ . Now suppose that $\mathbb{T}$ is countable. Then choose finite subsets $\mathbb{T}_n\subseteq\mathbb{T}$ which increase to $\mathbb{T}$ as $n$ goes to infinity. Replacing $\mathbb{T}$ by $\mathbb{T}_n$ in inequality () and using the monotone convergence theorem to take the limit $n\rightarrow\infty$ extends () to arbitrary uncountable index sets. Then, inequality () follows immediately from ().

Now, suppose that $X$ is a martingale. Jensen's inequality gives \begin{equation*} \mathbb{E}[|X_t|\mid\mathcal{F}_s]\ge \left|\mathbb{E}[X_t\mid\mathcal{F}_s]\right|=|X_s| \end{equation*}for any $s<t$ , so $|X|$ is a nonnegative submartingale. Therefore, it is enough to prove inequalities (1) and (2) for $X$ a nonnegative submartingale, and the martingale case follows by replacing $X$ by $|X|$ .

So, we take $X$ to be a nonnegative submartingale in the following. In this case, (1) just reduces to () and it only remains to prove inequality (2).

For $p>1$ , multiply () by $K^{p-1}$ and integrate up to some limit $L>0$ , \begin{equation}\label{eq:5} \int_0^LK^{p-1}\mathbb{P}(X^*_t\ge K)\,dK\le \int_0^LK^{p-2}\mathbb{E}[1_{\{X^*_t\ge K\}}X_t]\,dK. \end{equation}The left hand side of this inequality can be computed by commuting the order of integration with respect to $\mathbb{P}$ and $dK$ (Fubini's theorem), \begin{equation*}\begin{split} \int_0^L K^{p-1}\mathbb{P}(X^*_t\ge K)\,dK &=\mathbb{E}\left[\int_0^L K^{p-1}1_{\{X^*_t\ge K\}}\,dK\right]\\ &=\frac{1}{p}\mathbb{E}[(L\wedge X^*)^p]. \end{split}\end{equation*}The right hand side of () can be computed similarly, \begin{equation*}\begin{split} \int_0^LK^{p-2}\mathbb{E}[1_{\{X^*_t\ge K\}}X_t]\,dK &=\mathbb{E}\left[X_t\int_0^LK^{p-2}1_{\{X^*_t\ge K\}}\,dK\right]\\ &=\frac{1}{p-1}\mathbb{E}[X_t(L\wedge X^*_t)^{p-1}]. \end{split}\end{equation*}Putting these back into (), \begin{equation}\label{eq:6} \Vert L\wedge X^*_t\Vert_p^p\le\frac{p}{p-1}\mathbb{E}[X_t(L\wedge X^*_t)^{p-1}]. \end{equation}Now let $q=p/(p-1)$ , so that $p,q$ are conjugate and the Hölder inequality gives \begin{equation*} \mathbb{E}[X_t(L\wedge X^*_t)^{p-1}] \le\Vert X_t\Vert_p \Vert (L\wedge X^*_t)^{p-1}\Vert_q =\Vert X_t\Vert_p \Vert L\wedge X^*_t\Vert_p^{p-1}. \end{equation*}Substituting into (), the finite term $\Vert L\wedge X^*_t\Vert_p^{p-1}$ cancels to get \begin{equation*} \Vert L\wedge X^*_t \Vert_p\le \frac{p}{p-1}\Vert X_t\Vert_p, \end{equation*}and the result follows by letting $L$ increase to infinity.