Proof. Incidentally, there is some sense in which the structure of the Cantor set is fairly typical for closed sets. Theorem 3.2 Lebesgue outer measure has the following properties: (a) If A ⊆ B, then µ∗(A) ≤ µ∗(B). Countable Sets and Measure Zero Date: 05/12/2001 at 19:47:18 From: Jon Klassen Subject: Countable sets and measure zero How would you prove that if a set S is countable, then S has measure zero? Problem 14: Show that if a set Ehas positive outer measure, then there is a bounded subset of Ethat also has positive outer measure. (b) µ∗(∅) is zero. Proof: We prove by contradiction. Consider the closed intervals [ k , k+1 ] for all integers k ; there are countably many such intervals, each has measure 1, and their union is the entire real line. Example 1.1. The outer Jordan content J (E) of a set Ein Ris de ned by J (E) = inf XN j=1 jI jj where the inf is taked over every nite covering Eˆ S N j=1 I j, by intervals I j. An example of this is the rationals as a subset of the reals. Therefore it follows from Proposition 4 that the measure described by the Carath´eodory theorem is complete. So if Bwas measur-able, then A= B∪(A\B) would be measurable, too. P(C) M. But since jCj= jRj, it follows that P(C) = P(R) , and hence P(R) jMj. JPE, May 2001.

If we have a countable collection of subsets of $\mathbb{R}$, say $(A_n)_{n=1}^{\infty}$, then the Lebesgue outer measure of the union $\displaystyle{\bigcup_{n=1}^{\infty} A_n}$ will be less than or equal to the sum $\displaystyle{\sum_{n=1}^{\infty} m^*(A_n)}$.We prove this below. No. A set in a measure space has σ-finite measure if it is a countable union of sets of finite measure.

Intuition. Examples 7.3.12: Properties of Lebesgue Measure : Show that the set R and the empty set are L-measurable and find their L-measure. So I know that the Lebesgue outer measure of a set of only countably many points is 0. Find the L-measure of a singleton {x} as well as the L-measure of a countable set What is the L-measure of the set Q of all rational numbers and the set I of all irrational numbers inside [0, 1]. Very common examples of outer measures are The Lebesgue outer measure on $\mathbb R^n$, see Lebesgue measure. The set A\B⊂ Q is countable, hence measurable. (d) Lebesgue outer measure is invariant under translation, that is, for each real number x 0, µ∗(E +x 0) = µ∗(E). Ideally, every subset should be assigned a measure, but that turns out to be unrealistic in practice: see the Banach{Tarski paradox below. outer measure, and so any countable subset of R has zero outer measure, and so the subset of all rational numbers Q has outer measure zero. We say that a measure µ : M →[0,∞] is complete if every subset of a set of measure zero is measurable (and hence has measure zero). Does there exist a non-measurable subset of R whose comple-ment in R has outer measure zero?