Let (mathscr{H} subset mathscr{G}) be two sub- (sigma)-algebras of (mathscr{A}). Show that [mathbb{E}^{mathscr{G}} mathbb{E}^{mathscr{H}} u=mathbb{E}^{mathscr{H}} mathbb{E}^{mathscr{G}} u=mathbb{E}^{mathscr{H}}

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Let \(\mathscr{H} \subset \mathscr{G}\) be two sub- \(\sigma\)-algebras of \(\mathscr{A}\). Show that

\[\mathbb{E}^{\mathscr{G}} \mathbb{E}^{\mathscr{H}} u=\mathbb{E}^{\mathscr{H}} \mathbb{E}^{\mathscr{G}} u=\mathbb{E}^{\mathscr{H}} u\]

for all \(u \in L^{p}(\mathscr{A})\), resp. for all \(u \in M^{+}(\mathscr{A})\), provided that \(\left.\muight|_{\mathscr{H}}\) is \(\sigma\)-finite.

[ if \(\left.\muight|_{\mathscr{H}}\) is not \(\sigma\)-finite, the set \(L^{p}(\mathscr{H})\) can be very small ...]

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