Proving direction of monotonicity from conditional expected value

Question

If $f$ is a monotonic function, how to prove direction of monotonicity directly from:$$E[f(X)\mid X<a]>E[f(X)\mid X\geq a]$$

Clearly, $f$ must be decreasing monotonic in X, but I'm not sure how to show this directly from the above.

Answer 1

There must exist some $c<a$ such that $f(c)\geq E[f(X)|X<a]$ (proof: suppose to the contrary that $f(c)<E[f(X)|X<a]$ for all $c<a$ - then average over the left hand side to conclude that $E[f(X)|X<a]<E[f(X)|X<a]$). Similarly, there exists $b\geq a$ such that $E[f(X)|X\geq a]\geq f(b)$ and then we can simply use the transitive property to conclude that $f(c)>f(b)$, so $f$ must be decreasing monotonic.