# Image of measurable sets under one to one (a.e) functions

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.

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.
• $f$ 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 $X=Y.$ Commented May 25 at 11:57
• 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. Commented May 25 at 12:06