Let \(\mathscr{G}\) be a sub- \(\sigma\)-algebra of \(\mathscr{A}\). Show that \(\mathbb{E}^{\mathscr{G}} g=g\) for all \(g \in L^{p}(\mathscr{G})\).

[observe that, a.e., \(g=g \mathbb{1}_{\cup_{n}}\{|g|>1 / n\}\) and \(\mu\{|g|>1 / n\}<\infty\). This emulates \(\sigma\)-finiteness.]