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This question seems obvious, but not sure how to prove it.

Let $\mathcal{A}$ be a $\sigma$-algebra, and $X$ be a random variable. Suppose $E(X\mid A)\le1$ for any $A\in\mathcal{A}$, can we conclude that $E(X\mid \mathcal{A})\le1$ almost surely?

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    $\begingroup$ Test your inequality ${\bf E}(X|A) \leq 1$ against the events $A_\varepsilon := \{ {\bf E}(X|{\mathcal A}) \geq 1+\varepsilon \}$ for various $\varepsilon > 0$. $\endgroup$
    – Terry Tao
    Commented May 13, 2016 at 21:22
  • $\begingroup$ Thanks, Terry! To make your arguments complete: suppose $P(E(X|\mathcal{A})>1)>0$, then there exists an $\epsilon>0$ such that $A_\epsilon=\{E(X|\mathcal{A})\ge1+\epsilon\}$ has positive probability. Then $$(1+\epsilon)P(A_\epsilon)\le\int E(X|\mathcal{A})I_{A_\epsilon}dP=E(XI_{A_\epsilon})=E(X|A_\epsilon)P(A_\epsilon)\le P(A_\epsilon)$$, contradictiion! $\endgroup$
    – Jeff
    Commented May 13, 2016 at 21:41

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This is a simple question.

For any $A\in\mathcal{A}$, it holds that

$$ \int E(X\mid\mathcal{A})I_A \, dP=E(XI_A)=P(A)E(X\mid A)\le P(A), $$

hence, $E(X\mid\mathcal{A})\le1$ almost surely.. Thanks to Terry for quick guide!

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