Law of iterated expectation problem

Question

When $E(u \mid x)=0$, $E(u)=E(E(u \mid x))=0$.
Then why $E(ux)=0$ by law of iterated expectation?
The book says because it is a form of $E(h(x)E(u\mid x)$) but still I can't understand.

Answer 1

You can make the calculations $$ E(UX) = E( E(UX\mid X)) = E( X E(U\mid X)) = E( X (0)) = E(0) = 0, $$ since $E(U\mid X)=0$, yielding the desired result. The reason you can move $X$ outside the conditional expectation is a fact most naturally stated for a more abstract notion of conditional expectations with respect to $\sigma$-algebras, but you can also just take it as a case of the general rule $$ E(h(X)U\mid X) = h(X) E(U\mid X), $$ where $h$ is some mapping.