I'm studying Rudin's RCA chapter 7 and get in trouble with exercise 8.

Let $V = (a, b)$ be a bounded segment in $R^1$. Choose segments $W_n \subset V$ in such a way that their union $W$ is dense in $V$ and the set $K = V - W$ has $m(K) > 0$. Choose continuous functions $\phi_n$ so that $\phi_n(x) = 0$ outside $W_n$, $0 < \phi_n(x) < 2^{-n}$ in $W_n$. Put $\phi = \sum\phi_n$ and define $$ T(x) = \int_a^x\phi(t)dt\qquad(a<x<b) $$ Prove the following statements:

(a) $T$ satisfies the hypotheses of that $T$ is one-to-one on $X$, and $T$ is differentiable at every point of $X$, with $X = V$.

(b) $T'$ is continuous, $T'(x) = 0$ on $K$, $m(T(K)) = 0$.

(c) If $E$ is a nonmeasurable subset of $K$ (see Theorem 2.22) and $A = T(E)$, then $\chi_A$ is Lebesgue measurable but $\chi_A \circ T$ is not.

(d) The functions $\phi_n$ can be so chosen that the resulting $T$ is an infinitely differentiable homeomorphism of $V$ onto some segment in $R^1$ and (c) still holds.

Can anyone point me out about the b $m(T(K)) = 0$ and $d$? I would be very thankful for any answer.

  • $\begingroup$ Part $b$ itself has three parts. Are you stuck on all three? If so, note that the first two parts are answered wholly by methods in Rudin's Principles (i.e. "Baby Rudin")... For part $d$, are you aware of the existence of nonzero infinitely differentiable functions with compact support? $\endgroup$ – Brian Moehring Feb 23 at 7:04
  • $\begingroup$ Only the last one measure is zero $\endgroup$ – Wynne Liu Feb 23 at 7:55

For (b), to prove that $m(T(K))=0$, one need the followings:

  1. $T(K)$ is Lebesgue measurable (Why?) (Hint: prove that $T(W)$ is $F_{\sigma}$ set, and theorem 2.10 could be helpful)

  2. Construct a Lebesgue measurable function $$f(T(y))=\chi_{T(K)}(T(y))=\begin{cases}1, & y\in K\\0, & y\in K^C\end{cases}$$ (Check why this holds. Hint: for $=1$, use the fact that $T$ is one-to-one; for $=0$, use the same fact and argue by contradiction)

By theorem 7.26 in the text, since $K\subset V\Rightarrow T(K)\subset T(V)$, we have $$m(T(K))=\int_{T(V)}\chi_{T(K)}(x)dm(x)=\int_{V}\chi_{T(K)}(T(y))|J_T|dm(y)=\int_{V}\chi_{K}(y)|J_T(y)|dm(y)$$ $$=\int_{K}\chi_{K}(y)|J_T|dm(y)=\int_{K}0dm(y)=0$$ Note that this is a consequence of (iii).

For (d), choose $W_n=(\alpha_n,\beta_n)$ that are disjoint, construct $$\varphi_n(x)=\begin{cases}2^{-n}\exp\left(\frac{(\frac{\beta_n-\alpha_n}{2})^2}{(x-\frac{\alpha_n+\beta_n}{2})^2-(\frac{\beta_n-\alpha_n}{2})^2}\right), &x\in(\alpha_n,\beta_n)\\0, &\text{otherwise}\end{cases}$$

Then check that $\varphi_n(x)$ satisfies all the conditions.

Hope this helps.

  • $\begingroup$ Yes, this help me a lot. Thanks very much Mike. $\endgroup$ – Wynne Liu Feb 24 at 1:35

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