Let $(\mathcal{F})_{t\in\mathbb{T}}$ be a right-continuous filtration on the measurable space $(\Omega,\mathcal{F})$ It is assumed that $\mathbb{T}$ is a closed subset of $\mathbb{R}$ and that $\mathcal{F}_t$ is universally complete for each $t\in\mathbb{T}$

If $A\subseteq\mathbb{T}\times\Omega$ is a progressively measurable set, then we show that its début \begin{equation*} D(A)=\inf\left\{t\in\mathbb{T}:(t,\omega)\in A\right\} \end{equation*}is a stopping time.

As $A$ is progressively measurable, the set $A\cap((-\infty,t)\times\Omega)$ is $\mathcal{B}(\mathbb{T})\times\mathcal{F}_t$ measurable. By the measurable projection theorem it follows that \begin{equation*} \left\{D(A)<t\right\}=\left\{\omega\in\Omega:(s,\omega)\in A\cap((-\infty,t)\times\Omega)\textrm{ for some }s\in\mathbb{T}\right\} \end{equation*}is in $\mathcal{F}_t$ If there exists a sequence $t_n\in\mathbb{T}$ with $t_n>t$ and $t_n\rightarrow t$ then \begin{equation*} \left\{D(A)\le t\right\}=\bigcap_n\left\{D(A)<t_n\right\}\in\bigcap_n\mathcal{F}_{t_n}=\mathcal{F}_{t+}=\mathcal{F}_t. \end{equation*}On the other hand, if $t$ is not a right limit point of $\mathbb{T}$ then \begin{equation*} \{D(A)\le t\}=\{D(A)<t\}\cup\{\omega\in\Omega:(t,\omega)\in A\}\in\mathcal{F}_t. \end{equation*}In either case, $\{D(A)\le t\}$ is in $\mathcal{F}_t$ so $D(A)$ is a stopping time.