ریاضیات

HOMEWORK 3 SOLUTIONS
MATH 5051, FALL 2018
Exercise 1 (Folland, Exercise 1.8). If (X, M, µ) is a measure space and
M, then µ (liminf Ej) ≤ liminf µ (Ej). Also, µ (limsupEj) ≥ limsupµ (Ej) provided that.
Proof. Let              , so that. Since F1 ⊂ F2 ⊂ ···,
Continuity from below implies that
.
Finally, since Fj ⊂ Ek for all k ≥ j, it follows that),
And thus), which completes the proof of the first inequality.
      Similarly, let             , so that. Since G1 ⊃
G2 ⊃ ··· and µ(G1) ∞, continuity from above implies that
.
Finally, since Gj ⊃ Ek for all k ≥ j, it follows that), and thus), which completes the proof of the second inequality.         
Exercise 2 (Folland, Exercise 1.13). Every σ-finite measure is semi finite.
Proof. If µ is an σ-finite measure on (X, M), then, where Ej ∈ M and µ (Ej) ∞. Now, suppose that µ (E) = ∞. Since
∞
E = E ∩ X = [ E ∩ Ej,
j=1
It follows that, which implies that 0 (E ∩Ej) ∞ for infinitely many j. Hence, F = E ∩ Ej satisfies F ∈ M, F ⊂ E, and
0 (F) ∞, and thus µ is semi finite.                                                                            
Exercise 3 (Folland, Exercise 1.14). If µ is a semi finite measure and µ (E) = ∞, for any C > 0 there exists F ⊂ E with C (F) ∞.
Proof. Suppose not, and let M be the supremum of µ (F) as F ranges over measurable subsets of E with 0 (F) ∞. First, we claim that there exists a set F attaining the maximum µ (F) = M. Since M is a supremum,
there exists a nested sequence of sets F1 ⊂ F2 ⊂ ··· satisfying µ(Fj) > M − 1j for all j ∈ N. Taking the countable unionM and applying continuity from below, we get µ(F) = limµ(Fj) = M. However, since µ (E \ F) = ∞, there must also exist some G ⊂ E \ F with 0 (G) ∞. But then the disjoint union F ∪ G ⊂ E has µ (F ∪ G) = µ (F) + µ (G) > M, which contradicts the assumption that M was an upper bound.