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

Let $f,g:\mathbb{R}\rightarrow\mathbb{R}$ be Lebesgue measurable functions, and let $D$ be a subset of $\mathbb{R}$. If $f(D)$ and $g(D)$ are measurable subsets of $\mathbb{R}$, and $(f-g)(D):=\{f(x)-g(x):x\in D\}$ is a measurable subset of $\mathbb{R}$ with measure $0$, is it necessarily true that $\mu(f(D))=\mu(g(D))$ (where $\mu$ denotes Lebesgue measure)?

This question was originally posed to me without the measure being specified as Lebesgue, and I was able to find a counter-example (using "unit mass" measure). However, for Lebesgue measure it appears to be true... it is clearly true if $f$ and $g$ are simple functions... although I don't see how a more general proof would go.

Thanks in advance for any help.

share|cite|improve this question
up vote 4 down vote accepted

No. Consider $f(x) = x$ and $g(x) = x - \lfloor x \rfloor$, the fractional part.

share|cite|improve this answer
Of course! Thank you! – John Adamski Aug 21 '12 at 22:41

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.