Say I have a probability distribution $P_X(x)$ and I am computing the conditional expectation of some function $f(x)$ as :
$$E[f| P>a] = \int_{x=P^{-1}(a)}^{x=\infty} f(x)p_X(x)dx$$
(where $a$ is some number between $0$ and $1$). Note I am abusing notation a bit as follows: $x$ is a vectors which makes $P_X(x)$ a multidimensional distribution, and $p_X(x)$ is the density function corresponding to $P_X(x)$.
Now say there is a function $g$ which maps $y$ to $x$ i.e. $x =g(y)$. Under this transformation the probability distribution is also transformed as $$p_X(x)dx = q_Y(y)dy$$
Then my question is can I write the following:
$$E[f | P>a] = \int_{y=Q^{-1}(a)}^{y=\infty} f(g(y))q_Y(y)dy$$
where as before $q_Y(y)$ is the density function corresponding to $Q_Y(y)$.
The main conceptual issue that I am trying to clarify here is: Someone asks me to compute $E[f | P>a]$ in a region where the probability distribution is above $a$. I somehow come across a function $g(y)$ and the corresponding $q_Y(y)$ whose density I know. I do not know what $p_X(x)$ is but I do know what $q_Y(y)$ is (i.e in principal $p_X(x)$ is computable but maybe it is hard to compute). Then the above statement is saying that I do not need to know $p_X(x)$, I can perform the computation in the $y$-space and it will give me $E[f | P>a]$.
Your question is whether it is true that $$ E[f(X)\mid P(X)>a] = \int_{P^{-1}(a)}^\infty f(x)p_X(x)\, \mathrm dx \stackrel?= \int_{Q^{-1}(a)}^\infty f(y)q_Y(y)\, \mathrm dy $$ for a substitution $g(x) = y$.
Short answer: No. For a counterexample, consider $f(x) = x$ and $a = \infty$. Then
$$ \lim_{a\to\infty} E\bigl[f(X)\mid P(X)>a\bigr]= E[f(X)] = E[X], $$ which is just the expectation. Now, if $p_X(x) = (1+x)^2$, a $\operatorname{Pareto(1,1)}$ r.v. and $q_Y(y) = \exp(-y)$ an $\exp(1)$ r.v. then the former has no finite expectation while the latter has an expectation of $1$; a contradiction to (1).
You can, however, use substitution of variable to show that
$$ \int_{P^{-1}(a)}^\infty f(x)p(x)\, \mathrm dx = \lim_{b\to\infty}\int_{Q^{-1}(a)}^{g^{-1}(b)} f(g(y)) q(y)\, \mathrm dy, $$ where $Q^{-1}(a) = g^{-1}(P^{-1}(a))$. This puts restrictions on $g$.