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Need this as a part of a bigger thing I'm trying to prove, but couldn't have it still, after a long time trying.

$A,B,C$ are three independent random variables. We define additional two random variables: $f(a) = P(a > B), g(a) = P(a > C)$. Prove that $COV[f,g]=0$.

Note that $f,g$ are not necessarily independent...

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    I do not understand. $\mathbb{P}(a>B)$ is the probability that r.v. B falls below constant $a$. That is, it's a number $F_{B}(a)$? You must mean something else. – 2017-01-25
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    for every given a, $P(a > B)$ is a number, indeed. so $f(a) = P(a, B)$ is a function of $a$ (and a random variable). – 2017-01-25
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    OK, let me see if I understand. You have r.v. $A$. You construct two further r.v.s $A'=F_{B}(A)$ and $A''=F_{C}(A)$, where $F_{B}$ and $F_{C}$ happen to be cdfs of two r.v.s $B$ and $C$. Then you want to prove that $A'$ and $A''$ have zero covariance. – 2017-01-25
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    yes, this is exactly what I mean. – 2017-01-25

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Suppose you are trying to prove: given r.v. $A$, construct two r.v.s $A'=F_{B}(A)$ and $A''=F_{C}(A)$, where $F_{B}$ and $F_{C}$ happen to be cdfs of two r.v.s $B$ and $C$, then $A'$ and $A''$ have zero covariance.

I do not think this is correct. Take $A$, $B$, $C$ uniform on $[0,1]$. Then cdf is $F_{B}(x)=F_{C}(x)=x$ on $[0,1]$ and hence $A=A'=A''$.