Let \(\left(\mathscr{A}_{-n}ight)_{n \in \mathbb{N}}\) be a decreasing filtration such that \(\left.\muight|_{\mathscr{A}_{-\infty}}\) is \(\sigma\)-finite. Assume that \(\left(u_{-n}, \mathscr{A}_{-n}ight)_{n \in \mathbb{N}}\) is a backwards supermartingale which converges a.e. to a real-valued function \(u_{-\infty} \in \mathcal{L}^{1}(\mu)\) which closes the supermartingale to the left, i.e. such that \(\left(u_{-n}, \mathscr{A}_{-n}ight)_{n \in \mathbb{N} \cup\{\infty\}}\) is still a supermartingale. Then

\[\lim _{n ightarrow \infty} \int u_{-n} d \mu=\int u_{-\infty} d \mu\]