你被允许忽略的集合
若对每个 ε > 0 都能用可数多个总长度小于 ε 的区间覆盖集合 N，则称 N 为 零测集。等价地，m*(N) = 0。这类集合在最强的意义下可忽略：由单调性，零测集的任何子集仍是零测集；由 可数可加性，可数个零测集之并仍是零测集。特别地，每个 [[countable-set|可数集]] 都是零测集——包括有理数集。
Theorem. Every countable set N = {x_1, x_2, x_3, ...} has measure zero.
Fix epsilon > 0. Cover the k-th point by a tiny interval around it:
I_k = ( x_k - epsilon/2^(k+1) , x_k + epsilon/2^(k+1) ),
length(I_k) = epsilon / 2^k.
Then N is contained in the union of the I_k, and the total length is
sum_{k=1}^infinity epsilon / 2^k = epsilon * (1/2 + 1/4 + 1/8 + ...) = epsilon.
Since epsilon > 0 was arbitrary, m*(N) <= epsilon for all epsilon, so m*(N) = 0. QED
Corollary. The Dirichlet function from Guide 1 vanishes off the rationals,
a measure-zero set, so it equals 0 'almost everywhere' — and its Lebesgue
integral will be 0, exactly as intuition demanded.由此我们能定义本学科最有用的一句话。若某性质不成立的点集为零测集，则称该性质 [[almost-everywhere|几乎处处]]（a.e.）成立。“f = g 几乎处处成立”“f_n → f 几乎处处”“f 几乎处处连续”——在勒贝格理论中我们乐于丢弃零测集，因为它们从不影响积分。
一个无法被测量的集合
我们曾说勒贝格测度无法扩张到实数轴的 每一个 子集。下面给出证明，即 维塔利 构造。在 [0,1] 上定义 等价关系：x ~ y 当且仅当 x − y 为有理数。它把 [0,1] 划分成不可数多个类，每个类都是有理数的一个平移副本。借助 选择公理，从每个类中恰取一个代表元，所得集合记为 V。
现在用每个有理数 q ∈ [−1, 1] 平移 V，得到集合 V_q = V + q。两个事实让陷阱合拢。这些 V_q 两两不交（两个相差有理数的代表元会落入同一类，与每类只取其一矛盾）。且所有 V_q 之并夹在 [0,1] 与 [−1,2] 之间。勒贝格测度具有平移不变性，故所有 V_q 必有 相同 的测度 m(V)。
Suppose, for contradiction, that V is measurable with m(V) = c.
Let q_1, q_2, q_3, ... enumerate the rationals in [-1, 1] (countably many).
Set V_n = V + q_n. By translation-invariance, m(V_n) = c for every n.
The V_n are pairwise disjoint, and
[0,1] is a subset of (union of all V_n) is a subset of [-1, 2].
Apply countable additivity to the disjoint union, then monotonicity:
m([0,1]) <= sum_n m(V_n) <= m([-1,2])
1 <= sum_n c <= 3.
But sum_n c is a sum of COUNTABLY many copies of the same constant c:
* if c = 0 : sum_n c = 0, contradicting 1 <= sum_n c.
* if c > 0 : sum_n c = +infinity, contradicting sum_n c <= 3.
Either way a contradiction. Hence V is NOT measurable. QED维塔利集教给我们什么
请注意，这一构造本质上依赖 选择公理，以便一次性从不可数多个类中各取一个代表元。从未有人用明确公式给出过不可测集，事实上也不可能：与集合论其余公理相容的是“ℝ 的每个子集皆勒贝格可测”。不可测性是选择公理的产物。