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 $ X \subset [0,1]$ and $f:X \rightarrow C$ be an injective mapping into a Cantor Set $C$. How do I justify whether $f(X)$ is Lebesgue measurable or not?

The Cantor set $C$ has measure $0$ and since the mapping is injective $X = f^{-1}(C)$ has measure $0$.

share|cite|improve this question
Your statement "The cantor set has measure $0$ and since the mapping is injective, $X$ has measure $0$." is incorrect. Note that the Cantor set has the same cardinality as $\mathbb{R}$ and the interval $[0,1]$. Hence, there exists a bijective mapping from $\mathbb{R}$ (or) from $[0,1] \to F$. And clearly the measure of $\mathbb{R}$ and that of $[0,1]$ is not zero. – user17762 Nov 21 '12 at 4:38
Isn't the Lebesgue measure of the Cantor Set $0$? – Bhavish Suarez Nov 21 '12 at 12:51

Every subset of a measure zero set is Lebesgue measurable.

share|cite|improve this answer

What you mean to say is that $f(X)$ has measure $0$ because the cantor set has measure $0$ and lebesgue measure is complete.

share|cite|improve this answer
Yes, I was thinking along this line itself. However I didn't know whether it was right or not! – Bhavish Suarez Nov 21 '12 at 12:53
@BhavishSuarez Glad I could help! – Deven Ware Nov 21 '12 at 15:33

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.