$X$ uncorrelated with any function of $Y$ implies $X$ and $Y$ independent. This question is purely out of curiosity and mainly to question my intuitions about independence of random variables.
Q: Take two non trivial random variables, with non disjoint support (see edit below) $X,Y \in \mathbb{L}_2(\Omega, \mathcal{F}, \mathbb{P})$, so that projections and covariance formulas are well defined. If $X$ is uncorrelated with any function $g$ of $Y$, i.e. $\operatorname{Corr}(X,g(Y)) = 0, \: \forall \: g$ measurable, this implies $X$ and $Y$ are independent.
Is the above statement true? I could not find any post on mathstack on this.
One way I tried to prove the above is by proving the following:
Assume that $X$ and $Y$ are dependent, then there exists a function $f$ such that the correlation between $X$ and $f(Y)$ is nonzero.
Reason why $\mathbb{L}_2$ is important:
This is also the reason why we have to take the random variables in $\mathbb{L}_2$, otherwise one could find counterexamples to the second statement by taking $X$ with undefined variance or expectation and show that the covariance can never be nonzero, as it is not well defined.
Thoughts:
Any ideas or references? Maybe something additional must be assumed about the functions $g$? Maybe instead of this, one should assume that the correlation is zero with any random variable $Z$ which is $X$-measurable?
Thank you very much for your help and time.
Reason why non disjoint support is important:
EDIT. Here I post a counterexample that contradicts the second statement, if we do not assume that the random variables have non disjoint support, i.e.:
Assume that $X$ and $Y$ are dependent, then there exists a function $f$ such that the correlation between $X$ and $f(Y)$ is nonzero.
Take $([0,1], \mathcal{B}([0,1]), \lambda)$, where $\lambda$ is the Lebesgue measure. The key idea is that if they have disjoint support we can find a counterexample. Take:
$$ X(x) = \left(x - \frac{1}{2} \right) \mathbb{1}_{[0,1/2]}(x)$$
and:
$$ Y(x) = \left(x - \frac{3}{2} \right) \mathbb{1}_{[1/2,1]}(x)$$
Take any function $f$, then $f(Y(x)) = f(0)$ constant for any $x \in [0,1/2]$ thus:
$$ X(x)f(Y(x)) = f(0)X(x) \mathbb{1}_{[0,1/2]}(x)$$
which, as $\int X d\lambda = 0$, implies:
$$\int X f(Y) d \lambda = 0$$
for any $f$. This implies they are uncorrelated and it provides a counterexample.
 A: If $Y$ has only two values (for instance, if $Y$ is a Bernoulli random variable), then whenever $X,Y$ are uncorrelated, we also have $X,g(Y)$ uncorrelated for every $g$.  For when $Y$ only has two values $y_1, y_2$, then we can replace $g$ with a linear function: find constants $a,b$ with $g(y_1) = a y_1 + b$, $g(y_2) = a y_2 + b$, so that $g(Y) = aY+b$ almost surely.  Then $\operatorname{Cov}(X,g(Y)) = a \operatorname{Cov}(X,Y)=0$.
For an explicit example, let $Y$ be Bernoulli(p) for any p, $\xi$ Rademacher(1/2) independent of $Y$, and $X=\xi Y$.
If you assume that $g(Y)$ and $Z$ are uncorrelated for every $Z \in \sigma(X)$, or equivalently that $f(X)$ and $g(Y)$ are uncorrelated for all functions $f,g$, then $X,Y$ are independent.  For if not, there exist $A_1 \in \sigma(X)$, $A_2 \in \sigma(Y)$ which are not independent, which is to say that $1_{A_1}, 1_{A_2}$ are not uncorrelated.  But by definition of $\sigma(X)$, there exists a Borel set $B_1$ with $1_{A_1} = 1_{B_1}(X)$, and likewise a $B_2$ with $1_{A_2} = 1_{B_2}(Y)$.  So taking $f = 1_{B_1}$, $g = 1_{B_2}$, then $f(X), g(Y)$ are not uncorrelated.
