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I checked the questions pertaining to Lusin's theorem on this site, but it didn't have the answer I was looking for. But as I am carefully going through Rudin Chapter 2, it's been a challenge for me. For any of you that has the book on hand, I would like to pose this question to you. I'm a bit confused on two parts in the theorem, but everything throughout the proof seems fine to me.

On page 55 of Rudin, in the first paragraph of the proof section, he says to attach a sequence $\{s_n\}$ to $f$, as in the proof of Theorem $1.17$, and put $t_1 = s_1$ and $t_n = s_n - s_{n-1}$ for $n \geq 2$. He then concludes that $2^nt_n$ is the characteristic function of a set $T_n \subset A$.

My question is that when he says attach $\{s_n\}$ to $f$ as in 1.17, does he mean $s_n = \phi_n \circ f$ as stated in 1.17? As I was trying to work out how $2^nt_n$ is the characteristic function of a set $T_n \subset A$, I was unable to. By $T_n$ does he mean the set of points that belong to the range of $t_n$? As far as I know from 1.17, $s_n$ is just the step function that approximates $f(x) = x$, but as I tried different examples of $t_n$, I don't really know how Rudin gets $2^nt_n$ as the char. function of $T_n$. If anyone can clarify this, I would appreciate this.

One more thing. In the third paragraph of the proof section, he says "Since $2^{-n}h_n(x) = t_n(x)$ except in $V_n - K_n$", how did he come to conclude this?

Thanks again.

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Yes, he means $s_n = \varphi_n \circ f$. The value $s_n(x)$ is (by construction) obtained by rounding $f(x)$ down to the nearest multiple of $2^{-n}$ (and truncating values greater than $n$ down to $n$, but this is not of interest here, since he assumes $0 \le f < 1$). For each point $x$, the values $s_{n-1}(x)$ and $s_n(x)$ will therefore either be equal or differ by $2^{-n}$. In other words, the function $t_n$ takes only the values $0$ and $2^{-n}$, so the function $2^n t_n$ takes only the values $0$ and $1$. The points where $t_n(x)=1$ make up the set $T_n$ (by definition).

As for $h_n(x)$, it is (by construction) equal to $1$ for $x$ in $K_n$ and equal to $0$ for $x$ outside of $V_n$, and the same holds for $2^n t_n(x)$ since $K_n \subset T_n \subset V_n$. Therefore the only points where $h_n(x)$ might differ from $2^n t_n(x)$ lie in $V_n$ but not in $K_n$.

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Thank you for the explanation, Hans. Sorry about the late reply. But it helped me a lot. I'll accept this response as the answer to this question, now. Many thanks! –  MathNewbie Apr 2 '12 at 7:09
    
You're welcome! –  Hans Lundmark Apr 2 '12 at 7:25

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