Let(X,Θ,µ)be a measure space,f be a nonnegative measurable function onX,and ν(E)= Efdµ.Show that ν is a measure.

Answer 1

To show that ν is a measure, we need to verify the following properties:

1. Non-negativity: For any measurable set E, ν(E) = Efdµ ≥ 0.

2. Zero measure of the empty set: ν(∅) = ∅fdµ = 0.

3. Countable additivity: For any countable collection E₁, E₂, ... of pairwise disjoint measurable sets, ν(⋃ᵢ₌₁^∞ Eᵢ) = (⋃ᵢ₌₁^∞ Eᵢ)fdµ = ∑ᵢ₌₁^∞ Eᵢfdµ = ∑ᵢ₌₁^∞ ν(Eᵢ).

We can easily verify these properties as follows:

1. Non-negativity: For any measurable set E, since f is non-negative, we have 0 ≤ f(x) for all x ∈ X. Therefore, for any measurable set E, ν(E) = Efdµ = ∫ₑ fdµ ≥ 0.

2. Zero measure of the empty set: By definition, ν(∅) = ∅fdµ = ∫_∅ fdµ = 0.

3. Countable additivity: Let E₁, E₂, ... be a collection of pairwise disjoint measurable sets. Then,

ν(⋃ᵢ₌₁^∞ Eᵢ) = (⋃ᵢ₌₁^∞ Eᵢ)fdµ = ∫⋃ᵢ₌₁^∞ Eᵢ fdµ
= ∫(∑ᵢ₌₁^∞ Eᵢ) fdµ (by countable additivity of the integral)
= ∑ᵢ₌₁^∞ ∫Eᵢ fdµ (by linearity of the integral)
= ∑ᵢ₌₁^∞ ν(Eᵢ).

Therefore, ν is a measure.