Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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? –  t.b. Oct 18 '11 at 17:21
1  
@t.b sorry, my brother also uses my account , and it´s little –  August Oct 18 '11 at 17:23
3  
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. –  t.b. Oct 18 '11 at 17:31
    
@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 )= –  August Oct 18 '11 at 17:31
add comment

1 Answer

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.

share|improve this answer
    
@t.b. thanks for the ping. Looks like Zev did it already. Am cleaning up some comments. –  Willie Wong Oct 29 '11 at 16:58
    
@Willie: Okay, very good then. –  t.b. Oct 29 '11 at 17:00
add comment

Your Answer

 
discard

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.