Is $E(Y|X)=0, a.s.$ equivalent to $E[Y\cdot f(X)]=0, \forall f\in \mathscr{B}$?

Question

Is the statement below true? How to prove it? $$E(Y|X)=0, a.s.$$ is equivalent to $$E[Y\cdot f(X)]=0, \forall f\in \mathscr{B}$$

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If $E(Y|X)=0, a.s.$, then $E[Y\cdot f(X)]=E\{E[Y\cdot f(X)|X]\}=E[f(X)\cdot E(Y|X)]=0$. How to prove the opposite part?

Answer 1

Take $f=I_A$. You get $EYI_A(X)=0$ which means $EYI_{X^{-1}(A)}=0$. By definition of conditional expectation this give $E(Y|X)=0$. [Note that $\sigma (X)$ is precisely the collection of sets $\{X^{-1}(A):A\in \mathcal B\}$.