Suppose $f: X \subset R^n \to R^n$ is a smooth function (for example $C^2$ function), and for each $y \in R^n$, the set $f^{-1}(y)$ is finite. Do we have $f(A)$ is a positive measure set if $A$ has positive measure?
If $C^2$ function is not enough, how about analytic function?
Counterexample for smooth $f:$ Take $n=1.$ Let $U$ be an open subset of $\mathbb R,$ with $m(U)<1,$ that contains the rationals. Let $A= \mathbb R\setminus U.$ Then there exists $g\in C^\infty(\mathbb R)$ with $g= 0$ on $A,$ $g>0$ off $A.$ Define
$$f(x) = \int_0^x g(t)\,dt.$$
Then $f\in C^\infty(\mathbb R),$ $f$ is strictly increasing, and $f'=g=0$ on $A.$ It is well known that a smooth function whose derivative is $0$ on a set $E$ maps that set to a set of measure $0.$ Thus $m(f(A)) = 0,$ even though $m(A)=\infty.$
You can use this example to create higher dimensional examples. For example, $F(x,y) = (f(x),y)$ in $\mathbb R^2.$
For real analytic maps, I think the answer is yes, but I don't have a proof except for $n=1.$ Proof for $n=1:$ Suppose $f:\mathbb R\to \mathbb R$ is real analytic and satisfies your hypothesis. Then $f$ is nonconstant, hence the zero set $Z$ of $f'$ is discrete by the identity principle. From this the result follows quickly.
