# Injective mapping into the Cantor set

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 $0$.

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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
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## 2 Answers

Every subset of a measure zero set is Lebesgue measurable.

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

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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
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