5

I'm reading Rudin's Real and Complex Analysis and I'm puzzled about theorem 2.20, in which the Lebesgue measure is constructed via the Riesz representation theorem. Specifically, parts (c) and (e) of theorem 2.20 are on the translation-invariance of the Lebesgue measure, and how composition of the Lebesgue measure with a linear transformation is a scalar multiple of the Lebesgue measure, respectively.

Proving these properties requires that translations of Lebesgue measurable sets are Lebesgue measurable, as are linear transformations of Lebesgue measurable sets. One might be able to prove these facts by showing such operations are continuous, since continuous functions are Borel measurable, and the Lebesgue measure is regular. I have some doubts that this is the best approach, since Rudin proves that bounded linear transformations are continuous much later in Chapter 5. Are there other ways to prove that linear transformations of measurable sets is measurable that might reasonably be adopted by a student of Rudin who hasn't gotten to Chapter 5?

user207295
  • 51

1 Answers1

4

I will attempt to fill in some details for RCA 2.20.

For part c:

We wish to verify the definition $\lambda(E) = m(x + E)$, for fixed $x \in \mathbb{R}^k $ using only material in Rudin prior to this point.

First note that if $E$ is open, $x + E$ is open and thus measurable (since every Borel set is m-measurable by the Riesz Representation Theorem). Next, we know $m(E) = m(x + E)$ for open $E$ using the fact that they agree on boxes and 2.19d.

Also, we know that if $E$ is closed, then certainly $x + E$ is closed. This leads to an important observation about $F_\sigma$'s and $G_\delta$'s. Let $\{F_i\}_{i=1}^\infty$ be a collection of closed sets and let $A = \bigcup_{i=1}^\infty F_i$. We have $$x + A = x + \bigcup_{i=1}^\infty F_i = \bigcup_{i=1}^\infty (x + F_i)$$ Since the translation of closed sets are closed, we have shown that the translation of an $F_\sigma$ is an $F_\sigma$ and thus is measurable. Similarly, the translation of a $G_\delta$ is a $G_\delta$ and thus is measurable.

That m is translation invariant for $G_\delta$'s and $F_\sigma$'s follows from the fact that m is regular and the fact that m is translation invariant on open sets, as previously discussed.

Finally, take an arbitrary measurable set $E$. By part b of the theorem there exists an $F_\sigma$, $A$, and a $G_\delta$, $B$, such that $A \subset E \subset B$ (Rudin's subset notation) and $m(B - A) = 0$. This implies that $x + A \subset x + E \subset x + B$. Since $x + A$ is an $F_\sigma$ and $x + B$ is a $G_\delta$ and $m(B-A) = m((x+B) - (x+A)) = 0$, we see immediately that part b of the theorem implies that $x + E$ is measurable and $m(E) = m(A) = m(x + A) = m(x + E)$, since m is complete.

For part e:

This proceeds more or less analogously to the previous part.

We begin with the preliminary observation that since the Borels are translation invariant (since translation is continuous) and since $T(E)$ is Borel if $E$ is Borel as noted by Rudin, the proof goes through for the restriction of $\mu$ to the Borel sigma algebra.

Next, notice that since both T and its inverse are continuous, we know that $T(E)$ is open if $E$ is open and closed if $E$ is closed.

By elementary set theory, then, $T$ maps $F_\sigma$'s to $F_\sigma$'s. Using the fact that $T$ is injective, we also see that $T$ maps $G_\delta$'s to $G_\delta$'s.

Let $E$ be an arbitrary measurable set. By part b of the theorem, there exists an $F_\sigma$, $A$, and a $G_\delta$, $B$, such that $A \subset E \subset B$ (Rudin's subset notation) and $m(B - A) = 0$. This implies $T(A) \subset T(E) \subset T(B)$. Since $T(A)$ is an $F_\sigma$ and $T(B)$ is a $G_\delta$ and $m(T(B) - T(A)) = \Delta(T)m(B-A) = 0$ (we used the fact that the proof went through for the restriction of $\mu$ to the Borel sigma algebra here), we see that part b of the theorem implies that $T(E)$ is measurable.

Coriolanus
  • 1,725
  • 2
  • 8
  • 16