confusion : law of f(x) Let $x$ a random variable and $y=f(x)$. 
$x$ has distribution $\phi_X$ and $y$ has distribution $\phi_Y$.
What is the right way of writing the law of $y$? :
(1)
$$P(Y \in A)=\int_{A} \phi_Y(y) dy = \int 1_{\{x\in f^{-1}(A)\}} \phi_Y[f(x)] f'(x) dx$$
or :  
(2)
$$P(Y \in A)=P(f(X) \in A)=\int 1_{\{x\in f^{-1}(A)\}}  \phi_X(x) dx$$
I want to be sure that both are correct. if not could you me tell me why?
 A: The following assumes $f$ monotone (as it must be the case if it's bijective)
You have $$P(Y \in A) = \int_A \phi_Y(y)dy$$ Setting $y = f(x)$ we get $dy = f'(x) dx$, hence 
$$P(Y \in A) = \int 1_{\{x \in f^{-1}(A)\}} \phi_Y(f(x)) |f'(x)| dx$$
(Note the absolute value;it is necessary because if $f$ is increasing, then the extreme of integration are in the correct order and $f'(x) > 0$, so we're good. But if it is decreasing, $f'(x) < 0$ but we have to exchange the extreme of integration; a minus pops up making it $|f'(x)|$)  
But the two formula must both be right; that is it, for every event $A$, it must hold that 
$$\int 1_{\{x \in f^{-1}(A)\}} \phi_Y(f(x)) |f'(x)| dx = \int 1_{\{x\in f^{-1}(A)\}}  \phi_X(x) dx$$
This is enough to imply that the integrands must be equal, that is
$$\phi_Y(f(x)) |f'(x)| = \phi_X(x)$$
Now setting $y = f(x)$ we get 
$$\phi_Y(y) |f'(f^{-1}(y))| = \phi_X(f^{-1}(y))$$
Calling for simplicity $h(y) = f^{-1}(y)$, we notice that $\displaystyle |f'(f^{-1}(y))| = \frac 1{|h'(y)|}$ (usual formula for derivative of inverse) so we get 
$$\phi_Y(y) = \phi_X(h(y)) \cdot |h'(y)|$$
As it is indeed the case. 
