Rudin Real and Complex Analysis, Theorem 8.11

Question

I can't understand the underlined statement. How does Theorem 2.20 implies that $E\times \Bbb R^s$ belongs to $\mathfrak{M}_k$?

Answer 1

Suppose $m_r(E)=0.$ We want to show $m_k(E\times \mathbb R^s)=0.$ We don't have much machinery available at this point, but here's a lemma we can prove:

Lemma: If $U$ is open in $\mathbb R^r$ and $V$ is open in $\mathbb R^s,$ then $m_k(U\times V)= m_r(U)m_s(V).$

Proof: Just before Theorem 2.2, Rudin showed that every open set is the countable pairwise disjoint union of boxes. So in this way we can write $U=\cup U_l,$ $V= \cup V_m,$ where $U_l,V_m$ are boxes in $\mathbb R^r,\mathbb R^s$ respectively. Now

$$U\times V = \bigcup_l \bigcup_m U_l\times V_m.$$

Because of the simplicity of boxes, we have $m_k(U_l\times V_m)= m_r(U_l)m_s(V_m).$ Therefore

$$m_k(U\times V) = \sum_l\sum_m m_k(U_l\times V_m) = \sum_lm_r(U_l) \sum_m m_s(V_m)=m_r(U)m_s(V).$$

Suppose now $m_r(E)=0.$ Let $\epsilon>0.$ Then there are open sets $U_j\subset \mathbb R^r$ containg $E$ such that

$$m_r(U_j)< \frac{\epsilon}{2^j}.$$

And we can choose 1-boxes $V_m$ in $\mathbb R^s$ such that $\mathbb R^s = \cup V_m.$ We then have

$$E\times \mathbb R^s \subset \bigcup_m U_m\times V_m.$$

By the lemma,

$$m_k(\bigcup_m(U_m\times V_m)) \le \sum_m m_k(U_m\times V_m) = \sum_m m_r(U_m)m_s(V_m)$$ $$ = \sum_m m_r(U_m) < \sum_m \frac{\epsilon}{2^m}=\epsilon.$$

Thus $E\times \mathbb R^s$ is contained in sets of arbitrarily small $m_k$-measure. Hence it is contained in a set of $m_k$-measure $0.$ By the completeness of $m_k,$ we have our result.

Answer 2

Hint: If we apply 2.20(b) blindly, we have the following:

1. $E \in \mathfrak{M}_r$ so there are $A,B \subseteq R^r$ with $A \subseteq E \subseteq B$, $A$ is an $F_\sigma$, $B$ is a $G_\delta,$ and $m_r(B-A) = 0$
2. Since $A \subseteq R^r$ is an $F_\sigma$, also $A \times R^s \subseteq R^k$ is an $F_\sigma$
3. Similarly, $B \times R^s \subseteq R^k$ is a $G_\delta$
4. Obviously $A \times R^s \subseteq E \times R^s \subseteq B \times R^s$
5. Also, $(B\times R^s)-(A\times R^s) = (B-A) \times R^s$

Is $m_k\big((B-A)\times R^s\big) = 0$? (remember we have $m_r(B-A)=0$)