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 $f$?.

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

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .