f(X) and f(Y) are identically distributed Suppose that $X$ is a random variable defined on a probability space ( Ω , $\mathcal F$ , P ) 
and $P_X$ is the probability measure on  (  $\mathbb{R} $, $\mathcal R$)
and $Y$ is a random variable defined on  ( Ω' , $\mathcal F'$ , P' ) 
IF $ X $and $Y $ are identically distributed 
If $f:$ $\mathbb{R} $ $->$ $\mathbb{R} $  is Borel Measurable, And $X$ and $Y$ are identically distributed, then if we define  $U = f(X)$ and $V= f(Y)$.
$U$ and $V$ are also identically distributed
(Tried Solution):
After many hints and many hours all I manage to do is the below :
We know $P_x$ $=$ $P_Y$
and $P_X(B)$ , $B \in \mathcal R$ $= $$P_Y(B)$ , $B \in \mathcal R$
$P(X^{-1}(B)) = P(Y^{-1}(B))$
$ P(\omega\in Ω : X(ω) \in B)  =   P(\omega\in Ω' : Y(ω) \in B)$
Somehow we have to show that the above is equal to the below:.
$$......$$
$$......$$
$ P(\omega\in Ω : f o X(ω) \in B)  =   P(\omega\in Ω' : foY(ω) \in B)$
Hence, $ P(\omega\in Ω : U(ω) \in B)  =   P(\omega\in Ω' :V(ω) \in B)$
Thus,  $U = f(X)$ and $V= f(Y)$ are identically distributed .
Could you please explain the solution thoroughly, because I really want to understand the solution.
 A: To show that $f(X)$ and $f(Y)$ are identically distributed, we need to show that for any Borel set $B$, we have $P(f(X) \in B) = P(f(Y) \in B)$.
We have :
$$
P(f(X) \in B) = P(\{w : f(X(w)) \in B\}) = P(\{w : X(w) \in f^{-1}(B)\})
$$
where $f^{-1}(B) = \{t \in \mathbb R : f(t) \in B\}$. Note that this set is also Borel if $B$ is, since $f$ is Borel measurable. Therefore, all the above probabilities are well defined.
Finally, note that as $X$ and $Y$ are identically distributed,
$$
P(\{w : X(w) \in f^{-1}(B)\}) = P_X(f^{-1}(B))= P_Y(f^{-1}(B)) = P(\{w : Y(w) \in f^{-1}(B)\})
$$
and now go back using the logic of the first line to $P(f(Y) \in B)$ and conclude.
Now ask questions freely on this answer.
A: For every Borel set $B$:
$$P_U(B)=P_{f(X)}(B)=P(f(X)\in B)=P(X\in f^{-1}(B))=P_X(f^{-1}(B))=$$$$P_Y(f^{-1}(B))=P(Y\in f^{-1}(B))=P(f(Y)\in B)=P_{f(Y)}(B)=P_V(B)$$
The fifth equality rests on the fact that $X$ and $Y$ have identical distribution.
In the notation above e.g. $P(X\in A)$ actually abbreviates $P(\{X\in A\})$ where $\{X\in A\}$ is on its turn an abbreviation of $\{\omega\in\Omega\mid X(\omega)\in A\}$.
