# Does the computation of the inverse function of a set that is not in the image of $f$ makes sense?

I am studying measure theory, and I was wondering if the computation of the inverse function $$f^{-1}$$ of a set that is not in the image of $$f$$ makes sense. The fact is that all the functions that I have seen in my course are defined $$(\Omega,F)\rightarrow (\mathbb{R},B(\mathbb{R}))$$. But for example, let us consider the logistic function $$f(x)=\frac{1}{1-e^{-x}}$$, would that function be defined on $$(\mathbb{R},B(\mathbb{R}))\rightarrow ([0,1],B([0,1])$$ or in $$(\Omega,F)\rightarrow (\mathbb{R},B(\mathbb{R}))$$? In the latter case, what would happened for the sets that are not in the image of $$f$$?.

• $f^{-1}(Y)=\{\,x\in X\mid f(x)\in Y\,\}$ will simply be empty if $Y$ is disjoint from the image of $f$. Commented Nov 3, 2020 at 21:44

Note that here $$f^{-1}$$ does not necessarily refer to the inverse function. Indeed, $$f$$ might not admit an inverse. Instead, $$f^{-1}$$ denotes the pre-image under $$f$$. I.e. $$f^{-1}(A)=\{x\in\Omega|\;f(x)\in A\}$$ In the case where $$A\cap f(\Omega)=\emptyset,$$ you'll simply conclude that $$f^{-1}(A)=\emptyset$$.
Note that if $$f$$ is injective and $$A\subseteq f^{-1}(\Omega)$$, then the two notions (pre-images and images under the inverse function) agree and thus, the difference is pedantic at best. However, the pre-image always makes sense for any function and thus, is a better concept in general.