Given $X \subset R$ we say that $X$ has no content (or null content) and we denote it by $c(X) = 0 $ if given any $\varepsilon > 0$, there exist a finite collection of intervals $\{ I_k \}$ such that $X \subset \bigcup\limits_{k = 1}^n I_k$ and $\sum |I_k| < \varepsilon$, where clearly $|I_k|$ denotes the length of the interval. Now we say that $X$ has measure $0$, we accept any countable collection of intervals, under the same hypothesis.

Prove that the null content is preserved by continuous functions $f: \mathbb R \to \mathbb R$, and give an example of a continuous function $f$ such that $f(K)$ has nonzero measure, where $K$ denotes the cantor third set.

I don't know how to do it.

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    $\begingroup$ August, just my impression: you're jumping around quite a bit between many topics in real analysis, recently; some of them quite advanced, some quite elementary. Maybe you should try and focus a little more? $\endgroup$ – t.b. Oct 18 '11 at 17:21
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    $\begingroup$ @t.b sorry, my brother also uses my account , and it´s little $\endgroup$ – August Oct 18 '11 at 17:23
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    $\begingroup$ No problem, it's just a little puzzling. Maybe the two of you should think about having two separate accounts, I for one try to tune my answers towards what I expect a user to know from previous questions and I'm pretty sure others do that as well. $\endgroup$ – t.b. Oct 18 '11 at 17:31
  • $\begingroup$ @t.b Ok t.b we´ll do it . An example , for example a continuos surjection onto [0,1]. And now the proof , i was afraid with the problem , sorry )= $\endgroup$ – August Oct 18 '11 at 17:31

I'm sorry, I cannot post this as a comment since I don't have enough points, but, the function that takes the Cantor set to the unit interval is continuous, and, by Tietze extension, could be extended to a continuous function f: $\mathbb R\rightarrow \mathbb R$. The Cantor set itself has zero content, but its image is the unit interval, which does not.

BTW: I used to have an account here, but my computer crashed, and I lost the login info.

  • $\begingroup$ @t.b. thanks for the ping. Looks like Zev did it already. Am cleaning up some comments. $\endgroup$ – Willie Wong Oct 29 '11 at 16:58
  • $\begingroup$ @Willie: Okay, very good then. $\endgroup$ – t.b. Oct 29 '11 at 17:00

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