Showing that the independence of X and Y implies that the $Cov(g(X), h(Y)) = 0)$

Question

I have been thinking about this. I already know if $X$ and $Y$ are independent $$E(XY) = E(X)E(Y)$$

therefore, if $$Cov(X,Y) = E(XY) - E(X)E(Y)$$ then $$Cov(X,Y) = 0$$

The question I have is as follows: is there a way to use a proof that includes $h(X)$ and $g(Y)$ that can be deduced to the same answer? I feel like the answer is in my question. It may be be right in front of me, but I am failing to make any some sort of connection.

Help would be extremely appreciated.

Answer 1

Independence of X and Y means that the $\sigma$-fields $\sigma(X)$ and $\sigma(Y)$ are independent. As $g(X)$ is $\sigma(X)$-measurable and $h(Y)$ is $\sigma(Y)$-measurable (for measurable maps $g,h$), it follows that $g(X)$ and $h(Y)$ are independent. In particular $Cov(g(X),h(Y))=0$ (the covariance is well defined because $g,h$ are bounded).