Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following homework:

Let $(X, \mu, \Sigma)$ be a measure space. We define a measure preserving transformation to be a measurable map $T: X \rightarrow X$ such that for any $A \in \Sigma, \mu(T^{-1}(A)) = \mu(A)$. Such a transformation induces a transformation $U_T$ on $L^2 (X; \mu)$ given by $$ f \mapsto U_T(f) = f \circ T$$

Show that $U_T$ is a well-defined transformation from $L^2$ to itself.

My question: is it enough to show that $f \in L^2 \implies U_T(f) \in L^2$?

More generally: when I see the word "well-defined", how do I find out what it means? It means something different every time and I never really know what. Thanks for your help.

share|cite|improve this question
For your more general question, see comments in this answer. It basically asks that the formula actually defines a function from the domain set to the target set; exactly why this may be in doubt depends on the setting. Here, one possible doubt is whether the image actually lies in the target set. Another is that you are dealing with equivalence classes. – Arturo Magidin Oct 17 '11 at 13:48
up vote 6 down vote accepted

That's certainly part of it. But you might also worry that because $L^2$ consists of equivalence classes of functions — we identify square-integrable functions that agree outside of a set of measure zero — the putative image $[f \circ T]$ of a class $[f]$ might depend upon the representative $f$.

In my experience, this is what "well defined" refers to: I'm defining a map out of a set of equivalence classes using representatives of those classes, and I want to check that my choice of representative doesn't matter. For example, in basic algebra one wants to verify that if $K$ is the kernel of a group homomorphism $\varphi\colon G \to G'$ then the map $G/K \to G'$ sending $xK \mapsto \varphi(x)$ is well defined.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.