I want to know about this question that is image of a measurable set under a one to one (almost everywhere) function, measurable?

Consider the following:

Let $(X, \mathcal{B}_X)$ and $(Y, \mathcal{B}_Y)$ be standard Borel spaces. Suppose $f: X \to Y$ is a measurable function that is one-to-one almost everywhere. That is, there exists a set $N \subset X$ with $\mu(N) = 0$ (where $\mu$ is some measure on $X$) such that $f$ restricted to $X \setminus N$ is injective. I want to examine if under these conditions that is the image of every measurable set under the map f measurable?

Definitions and Setup

  1. Standard Borel Spaces:

    • $(X, \mathcal{B}_X)$ and $(Y, \mathcal{B}_Y)$ are standard Borel spaces.
    • A function $f: X \to Y$ is measurable if $f^{-1}(B) \in \mathcal{B}_X$ for all $B \in \mathcal{B}_Y$.
  2. One-to-One Almost Everywhere:

    • $f$ is one-to-one almost everywhere if there exists a set $N \subset X$ with $\mu(N) = 0$ such that $f$ restricted to $X \setminus N$ is injective.

My try

We know that the forward image of a measurable set under a measurable map need not be measurable in general. However, a theorem of Lusin in classical descriptive set theory states that if $f$ is a measurable function on a standard Borel space into another such space, and if $f$ is countable-to-one in the sense that the inverse image of every singleton is at most countable, then the forward image under $f$ of any Borel set is Borel. In particular, if such an $f$ is one-to-one and onto, then it is a Borel isomorphism.


1 Answer 1


That would depend heavily on the measure. In an extreme case, if your measure $\mu$ is concentrated on a singleton then the condition "one-to-one a.e." is satisfied trivially and the counterexamples that you refer to in your try, remain valid.

  • $\begingroup$ my measure is invariant-Borel-probability and non-atomic. is this enough to say that the statement remains valid? $\endgroup$ Commented May 25 at 11:38
  • $\begingroup$ Invariant with respect to what transformation? $\endgroup$
    – Lieven
    Commented May 25 at 11:52
  • $\begingroup$ Invariant with respect to f, which means that \mu(f^{-1}(x))= \mu({x}) for every x \in X which is simply zero, since the measure is non-atomic. $\endgroup$ Commented May 25 at 11:53
  • $\begingroup$ $f$ is not a transformation, it's a mapping to a different space. The word "invariant" means that the same $\mu$ can be applied to a set and the transformed image of that set. $\mu$ is a measure on $\mathcal B_x,$ not on $\mathcal B_y.$ Unless you have an unstated hypothesis in the back of your mind that $X=Y.$ $\endgroup$
    – Lieven
    Commented May 25 at 11:57
  • 1
    $\begingroup$ Actually, if $X=Y$ and $f$ preserves a measure $\mu,$ then the image of a measurable set under $f$ is measurable by definition. You cannot have $\mu(f(B))=\mu(B)$ unless $\mu(f(B))$ is defined. $\endgroup$
    – Lieven
    Commented May 25 at 12:06

You must log in to answer this question.

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