= Solution
For the continuous <adapted process> $X$, the supremum on each compact time interval is attained. Let $D_t=(\mathbb Q\cap[0,t])\cup\{t\}$. Continuity gives $\sup_{s\leq t}X_s=\sup_{r\in D_t}X_r$, and hence
$$
\{T\leq t\}=\{X_t^*\geq\lambda\}
=\bigcap_{m=1}^\infty\bigcup_{r\in D_t}\{X_r>\lambda-1/m\}\in\mathcal F_t.
$$
Each event inside the union is in $\mathcal F_r\subseteq\mathcal F_t$, so this proves carefully that $T$ is a <stopping time>. Merely checking rational times for $X_r\geq\lambda$ would miss a path that touches the level at an isolated irrational time; the approximation from below avoids this issue. This is the <closed-level hitting times of continuous adapted processes> criterion. For $\lambda=0$, nonnegativity makes $T=0$ and the later inequality is immediate.
For $\lambda>0$, put $A=\{T\leq t\}$ and $\tau=T\wedge t$. On $A$, continuity implies $X_\tau\geq\lambda$. To justify the <optional sampling> step without extra filtration assumptions, round $\tau$ upwards to a finite dyadic grid in $[0,t]$, obtaining stopping times $\tau_n\downarrow\tau$. The event $A$ belongs to $\mathcal F_\tau$, and hence to $\mathcal F_{\tau_n}$: directly, $A\cap\{\tau\leq u\}=\{T\leq u\}$ for $u<t$, while for $u\geq t$ it is $A$. Discrete <optional sampling> for the sampled <submartingale> gives
$$
\mathbb E(X_t\mathbf1_A)\geq\mathbb E(X_{\tau_n}\mathbf1_A).
$$
As $n\to\infty$, continuity gives $X_{\tau_n}\to X_\tau$. Nonnegativity and the <Fatou lemma> therefore yield
$$
\mathbb E(X_t\mathbf1_A)\geq\mathbb E(X_\tau\mathbf1_A)\geq\lambda\mathbb P(A).
$$
Consequently the <Doob maximal inequality for a nonnegative submartingale> takes the precise form
$$
\boxed{\lambda\mathbb P(X_t^*\geq\lambda)\leq\mathbb E\bigl(X_t\mathbf1_{\{X_t^*\geq\lambda\}}\bigr).}
$$
For the second-moment bound, suppose first that $\mathbb E X_t^2<\infty$; otherwise the inequality with an infinite right side is automatic. Set $M_R=X_t^*\wedge R$. The <layer cake representation> and the preceding tail estimate give
$$
\begin{aligned}
\mathbb E M_R^2
&=2\int_0^R\lambda\mathbb P(X_t^*\geq\lambda)\,d\lambda\\
&\leq2\int_0^R\mathbb E\bigl(X_t\mathbf1_{\{X_t^*\geq\lambda\}}\bigr)d\lambda
=2\mathbb E(X_tM_R)\\
&\leq2\|X_t\|_2\|M_R\|_2.
\end{aligned}
$$
The integrands are nonnegative, so the interchange is justified by the <Tonelli theorem>. Since $M_R$ is bounded, division is legitimate when its norm is nonzero, giving $\|M_R\|_2\leq2\|X_t\|_2$; the zero-norm case is immediate. Finally, <monotone convergence> as $R\to\infty$ gives
$$
\boxed{\|X_t^*\|_2\leq2\|X_t\|_2.}
$$
This <truncated layer-cake proof of the L2 maximal inequality> proves square integrability of the maximum rather than assuming it during the <Cauchy-Schwarz inequality> step.
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