I'm working on a problem from Stein and Shakarchi's Real Analysis book (Exercise 6.7.13). I'm having some trouble understanding the problem and am looking for some insight.
Exercise 13: Let $m_j$ be the Lebesgue measure for the space $\mathbb{R^{d_j}}$, $j = 1,2$. Consider the product $\mathbb{R^d} = \mathbb{R^{d_1}} \times \mathbb{R^{d_2}}$ with $m$ the Lebesgue measure on $\mathbb{R^d}$. Show that $m$ is the completion (in the sense of exercise 2) of the product measure $m_1 \times m_2$
Exercise 2, which is mentioned in the problem above: One can define the completion of a measure space $(\mathcal{X}, \mathcal{M}, \mu)$ as follows. Let $\mathcal{\overline{M}}$ be the collection of sets of the form $E \cup Z$, where $E \in \mathcal{M}$, and $Z \subset F$ with $F \in \mathcal{M}$ and $\mu(F) = 0$. Define $\overline{\mu}(E \cup Z) = \mu(E)$.
$\mathcal{\overline{M}}$ is the smallest $\sigma$-algebra containing $\mathcal{M}$ and all subsets of elements of $\mathcal{M}$ of measure $0$, and the function $\overline{\mu}$ is a measure on $\mathcal{\overline{M}}$ and is complete.
My thinking: My understanding of a measure being complete is that the measure of subsets of sets of measure $0$ also have measure $0$. This coincides with the definition in exercise 2 in the sense that "adding" a subset of a set of measure $0$ onto any measurable set doesn't change the measure, so it makes sense to call those measure $0$ sets. To solve exercise 13, I need to show that if I take the union of a measurable set $E \in M$ and an element of $\mathbb{R^{d_1}} \times \mathbb{R^{d_2}}$ that is a subset of a set of measure $0$, then the measure of the union is the same as the measure of $E$. In this sense, subsets of measure $0$ have measure $0$ under $\mu$. Am I thinking about this properly, or am I confused?
Thanks!