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Measure Zero, Almost Everywhere, and a Non-Measurable Set

Two ideas that quietly run the whole subject: sets of measure zero (so small they can be ignored) and the phrase “almost everywhere.” Then the shock: using the axiom of choice we build a Vitali set that cannot be measured at all, proving no perfect measure on every subset exists.

Sets you are allowed to ignore

A set N has measure zero if for every ε > 0 it can be covered by countably many intervals of total length below ε. Equivalently, m*(N) = 0. Such sets are negligible in the strongest sense: by monotonicity any subset of a null set is null, and by countable additivity a countable union of null sets is null. In particular every [[countable-set|countable set]] has measure zero — including the rationals.

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.
The geometric-series trick: cover the k-th point with length ε/2^k so the total is ε.

This lets us define the most useful phrase in the subject. A property holds [[almost-everywhere|almost everywhere]] (a.e.) if the set of points where it fails has measure zero. “f = g almost everywhere,” “f_n → f almost everywhere,” “f is continuous almost everywhere” — in Lebesgue theory we cheerfully discard measure-zero sets, because they never affect an integral.

A set that cannot be measured

We promised Lebesgue measure cannot extend to every subset of the line. Here is the proof, the Vitali construction. Define an equivalence relation on [0,1] by x ~ y iff x − y is rational. This splits [0,1] into uncountably many classes, each a translated copy of the rationals. Using the axiom of choice, pick exactly one representative from each class and call the resulting set V.

Now translate V by each rational q in [−1, 1], giving sets V_q = V + q. Two facts make the trap close. The V_q are pairwise disjoint (two representatives differing by a rational would lie in the same class, contradicting that we chose only one per class). And the union of all the V_q sits between [0,1] and [−1,2]. Lebesgue measure is translation-invariant, so all the V_q must have the same measure 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
The Vitali set: translation-invariance plus countable additivity force an impossible value for m(V).

What the Vitali set teaches

Notice that the construction leaned essentially on the axiom of choice to select one representative from uncountably many classes at once. No one has ever exhibited a non-measurable set by an explicit formula, and in fact one cannot: it is consistent with the other axioms of set theory that every subset of ℝ is Lebesgue measurable. Non-measurability is a creature of choice.