The idea of a measure is that we can associate boxes with a volume. Here, we define a box $B \in \R^n$ of side length $\delta > 0$ as
$$ \{x\in \R^n\mid a_i<x_i<a_i+\sigma , i=1,\dots, n\}. $$
Definition 4.1.1 (Measure 0). A set $X\subset \R^n$ has measure 0 if for every $\epsilon > 0$, there exists an countable collection of open boxes $\{B_i\}_{i\in I}$ such that
$$ X \subset \bigcup_{i \in I} B_i \text{ and } \sum_{i\in I} \text{vol}_n(B_i) \leq \epsilon. $$
Definition 4.2.2 (Almost everywhere, almost all). the term almost everywhere, abbreviated a.e., means except on a set of measure 0. ****
The difference between measure and volume is that is is possible to have a set of measure 0 without defined volume. However a set with volume 0 necessarily has measure 0. Volume 0 $\implies$ measure 0.
Example. Any countable set in $\R^n$ has measure 0 but does not have defined volume since there are no dyadic cubes. This is because if you put a box around each number, each box will converge to measure 0, so the whole set has measure 0.
Example 4.4.3 (Rationals in [0,1]). The set of rationals in [0, 1] have measure 0. You can list them out in order $1, 1/2, 1/3, 2/3, 1/4, 2/,4, 3/4, \dots$.
Suppose we centered an interval around each rational with decreasing radius. $\epsilon / 2, \epsilon / 4, \epsilon / 8 , \dots$. The sum of the lengths of these intervals is $\epsilon (1/2 + 1/4 + \dots) = \epsilon.$ The union of these would intervals would be $< \epsilon$ because of overlapping. Because $\epsilon$ can be arbitrarily small, $U_\epsilon$ has measure 0.
Theorem 4.4.4 (A countable union of sets of measure 0 has measure 0). Let $i\mapsto X_i$ be a sequence of sets of measure 0. Then $X_1 \cup X_2 \cup \dots$ is a set of measure 0.
Proof. By the definition of measure 0, each set $X_j$ has an infinite sequence $i\mapsto B_{j,i }$ of boxes such that
$$ X_j \subset B_{j,1}\cup B_{j,2} \cup \dots , $$
and
$$ \sum \text{vol} B_{j,i} \leq \frac{\epsilon}{2^j}. $$