Problems and Solutions in - Mathematics - University of Hawaii

1.7 2007 November 16 1 REAL ANALYSIS
1.7 2007 November 16
Notation: R is the set of real numbers and R n is n-dimensional Euclidean space. Denote by m Lebesgue measure on
R and mn n-dimensional Lebesgue measure. Be sure to give a complete statement of any theorems from analysis that
you use in your proofs below.
1. Let µ be a positive measure on a measure space X. Assume that E1, E2, . . . are measurable subsets of X with the
property that for n = m, µ(En ∩ Em) = 0. Let E be the union of these sets. Prove that
µ(E) =
∞
µ(En)
Solution: Define F1 = E1, F2 = E2 \ E1, F3 = E3 \ (E1 ∪ E2), . . . , and, in general,
n=1
n−1
Fn = En \ Ei (n = 2, 3, . . . ).
i=1
If M is the σ-algebra of µ-measurable subsets of X, then Fn ∈ M for each n ∈ N, since M is a σ-algebra. Also,
Fi ∩ Fj = ∅ for i = j, and F1 ∪ F2 ∪ · · · ∪ Fn = E1 ∪ E2 ∪ · · · ∪ En for all n ∈ N. Thus,
and, by σ-additivity of µ,
∞
n=1
Fn =
n=1
∞
En E,
n=1
∞
∞
µ(E) = µ( Fn) = µ(Fn).
Therefore, if we can show µ(En) = µ(Fn) holds for all n ∈ N, the proof will be complete.
Now, for each n = 2, 3, . . . ,
and
n=1
n−1
Fn = En ∩ ( Ei) c
i=1
n−1
µ(En) = µ(En ∩ ( Ei) c n−1
) + µ(En ∩ ( Ei)). (24)
Equation (24) holds because n−1
i=1 Ei is a measurable set for each n = 2, 3, . . . . Finally, note that
which implies
i=1
n−1
En ∩ ( Ei) =
i=1
n−1
µ(En ∩ ( Ei)) ≤
i=1
(En ∩ Ei),
n−1
i=1
i=1
n−1
µ(En ∩ Ei),
by σ-subadditivity. By assumption, each term in the last sum is zero, and therefore, by (23) and (24),
n−1
µ(En) = µ(En ∩ ( Ei) c ) = µ(Fn) holds for each n = 2, 3, . . . .
i=1
For n = 1, we have F1 = E1, by definition. This completes the proof. ⊓⊔
32
i=1
(23)