Proposition

Use induction on $n$ to show that $[a_1, b_1] \times \cdots \times [a_n, b_n]$ is not a set of measure 0 (or content 0) if $a_1 < b_i$ for each $i$.

Solution

By Theorem 3-6[Spivak], it suffices to prove this for measure 0. By Theorem 3-5[Spivak], the statement is true when $n = 1$.

Suppose that we have proved this up to some $n \in \mathbb{N}$. Let $A = [a_1, b_1] \times \cdots \times [a_n, b_n]$ and $B = [a_{n + 1}, b_{n + 1}]$. Let $f: A \times B \rightarrow \mathbb{R}$ be defined such that $f(a, b) = 1$ for all $(a, b) \in A \times B$. By Theorem 3.8, $f$ is integrable over $A \times B$ and $b \mapsto f(a, b)$ is integrable over $B$ for any fixed $a \in A$.

By the inductive hypothesis, $A$ and $B$ do not have measure 0. By taking the contrapositive of a proposition we showed before, $\int_A dx \ne 0$ and $\int_B dx \ne 0$. Therefore, $\int_{A \times B} f \ne 0$. By taking the contrapositive of a proposition we showed before, $A \times B$ does not have measure 0.