What do I need as assumptions on $f$ to get $P(X \in A) = P (f(X) \in f(A))$ for a (probability) measure $P$? Good day,
In class we had a problem where we used the property 
$$P(|X_n| > \epsilon ) = P\left(\frac{|X_n|}{1+|X_n|} > \frac{\epsilon}{1+\epsilon} \right)$$ 
and said that this holds since $f(x)=\frac{x}{1+x}$ is strictly increasing. 
I don't really know why so my general question is:

What do I need as assumptions on $f$ to get $P(X \in A) = P (f(X) \in f(A))$ for a (probability) measure $P$? And why exactly?

Maybe this holds for a general measure so I put "probability" in brackets.
I should need sth like this: $$P(f(X) \in f(A))=P(X \in f^{-1}(f(A))=P(X \in A)$$
i.e. $f$ should be left-invertible (please say whether this is enough). But I am still not convinced that I can do this inside of $P$, seems quite sketchy/rigorous to me.
So let's prove

$P(f(X)\in A)=P(X \in f^{-1}(A))$

Let $\mu_Y$ denote the distribution of a random variable $Y$ via $\mu_Y(A)=P(Y \in A)$. Then
$$P(f(X) \in A)= \int_A \mu_{f(X)}(dx) \color{red}{=} \int_{f^{-1}(A)} \mu_{X}(dx)=P(X \in f^{-1}(A))$$
where I am not sure about "$\color{red}{=}$". Can someone please fill this step for me?
Thanks a lot,
Marvin

In general: $$A\subseteq f^{-1}f(A)$$ 
This because $$x\in A\implies f(x)\in f(A)\implies x\in f^{-1}f(A)$$
Consequently:$$P(X\in A)\leq P(X\in f^{-1}(f(A)))=P(f(X)\in f(A))$$
If $f$ is injective (left-invertible is the same) then: $$A= f^{-1}f(A)$$ for every set $A$.

Extreme case: $f$ is a constant function. 
Then for every non-empty $A$ we have $\{f(X)\in f(A)\}=\Omega$ and consequently $P(f(X)\in f(A))=1$. 
But if $X$ is not degenerated then we can find non-empty sets $A$ with $P(X\in A)<1$.
