你被允許忽略的集合
若對每個 ε > 0 都能用可數多個總長度小於 ε 的區間覆蓋集合 N，則稱 N 為 零測集。等價地，m*(N) = 0。這類集合在最強的意義下可忽略：由單調性，零測集的任何子集仍是零測集；由 可數可加性，可數個零測集之並仍是零測集。特別地，每個 [[countable-set|可數集]] 都是零測集——包括有理數集。
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.由此我們能定義本學科最有用的一句話。若某性質不成立的點集為零測集，則稱該性質 [[almost-everywhere|幾乎處處]]（a.e.）成立。「f = g 幾乎處處成立」「f_n → f 幾乎處處」「f 幾乎處處連續」——在勒貝格理論中我們樂於丟棄零測集，因為它們從不影響積分。
一個無法被測量的集合
我們曾說勒貝格測度無法擴張到實數軸的 每一個 子集。下面給出證明，即 維塔利 構造。在 [0,1] 上定義 等價關係：x ~ y 當且僅當 x − y 為有理數。它把 [0,1] 劃分成不可數多個類，每個類都是有理數的一個平移副本。藉助 選擇公理，從每個類中恰取一個代表元，所得集合記為 V。
現在用每個有理數 q ∈ [−1, 1] 平移 V，得到集合 V_q = V + q。兩個事實讓陷阱合攏。這些 V_q 兩兩不交（兩個相差有理數的代表元會落入同一類，與每類只取其一矛盾）。且所有 V_q 之並夾在 [0,1] 與 [−1,2] 之間。勒貝格測度具有平移不變性，故所有 V_q 必有 相同 的測度 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維塔利集教給我們什麼
請注意，這一構造本質上依賴 選擇公理，以便一次性從不可數多個類中各取一個代表元。從未有人用明確公式給出過不可測集，事實上也不可能：與集合論其餘公理相容的是「ℝ 的每個子集皆勒貝格可測」。不可測性是選擇公理的產物。