Define μ(empty set)=0, μ(E)=1 if E≠empty set. Show that μ is an outer measure, and determine the measurable sets.
- EricLv 61 decade agoFavorite Answer
Claim. μ is an outer measure on X.Proof. Let's verify the five conditions.Domain.μ is defined on the domain of the collection of subsets of X.Nonnegativity.μ is clearly nonnegative.The empty set.μ(∅) = ∅ by definition.Monotonicity.Let E, F be subsets of X with E ⊆ F. There are two cases:If F = ∅, then E = ∅, μ(E) = μ(F) = μ(∅) = 0.Otherwise, F is nonempty and μ(E) ≤ 1 = μ(F).Hence μ(E) ≤ μ(F).Countable subadditivity.Let En be a countable collection of subsets of X. There are two cases:If En = ∅ for every n, then ∪n En = ∅ and μ(En) = 0 for every n, so μ(∪n En) = μ(∅) = 0 = ∑n μ(En).Otherwise, EK ≠ ∅ for some K, ∪n En ≠ ∅, μ(EK) = 1, μ(∪n En) = 1, so μ(∪n En) = 1 = μ(EK) ≤ ∑n μ(En).Hence μ(∪n En) ≤ ∑n μ(En).We conclude that μ is an outer measure on X. ∎The measurable sets: which subsets E ⊆ X satisfyμ(A) = μ(A∩E) + μ(A∖E) (*)for all A ⊆ X?Claim. The only measurable sets are the empty set and X.Proof. Clearly, the empty set and X are always measurable. If there are no other subsets of X, we are done. Otherwise, let E be a subset of X with ∅ ≠ E ≠ X. Since X, X∩E = E, and X∖E are all nonempty, we have μ(X) = μ(X∩E) = μ(X∖E) = 1. Therefore, μ(X) = 1 ≠ 2 = μ(X∩E) + μ(X∖E), and thus E is not measurable. ∎