If $X_1 = X_2$ on a measurable set $B \in \mathfrak F$ then $E(X_1\mid\mathfrak F)=E(X_2\mid\mathfrak F)$ almost sure on $B$.
2026-03-25 07:38:10.1774424290
Conditional expectation almost sure
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Note what you want is the equality $$E(X_1\mid\mathscr{G})1_B=E(X_2\mid\mathscr{G})1_B.$$ Because $B\in\mathscr{G}$ and $X_11_B=X_21_B$, we know $$E(X_1\mid\mathscr{G})1_B=E(X_11_B\mid\mathscr{G})=E(X_21_B\mid\mathscr{G})=E(X_2\mid\mathscr{G})1_B.$$