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.
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.
To show that ν is a measure, we need to prove the three properties of a measure:
1. Non-negativity: ν(E) is non-negative for all measurable sets E.
2. Null empty set: ν(∅) = 0.
3. Countable additivity: For a countable collection of disjoint measurable sets {E_n}, ν(∪E_n) = Σν(E_n), where Σ denotes the sum over all n.
Let's prove each property step-by-step:
1. Non-negativity:
Since f is a non-negative measurable function, we know that f(x) ≥ 0 for all x in X.
Therefore, for any measurable set E, E can be expressed as a union of disjoint measurable sets: E = ∪E_n, where E_n = E∩{x∈X : f(x) ≥ 1/n}.
Now, let's calculate ν(E):
ν(E) = ∫E fdµ = ∫(∪E_n) fdµ = ∫(∪E_n) ∑(1/n)(f(x))dµ
Since the sets E_n are disjoint, we can apply the linearity of the integral to obtain:
ν(E) = ∑(1/n) ∫E_n (f(x))dµ
Since f(x) ≥ 0 for all x∈X, the integrand is non-negative, and therefore, ∫E_n (f(x))dµ ≥ 0.
Hence, ∑(1/n) ∫E_n (f(x))dµ ≥ 0, which implies ν(E) ≥ 0. Therefore, ν is non-negative.
2. Null empty set:
We need to show that ν(∅) = 0.
Since ∅ is an empty set, it can be expressed as the union of disjoint sets: ∅ = ∪E_n, where E_n = ∅ for all n.
Now, let's calculate ν(∅):
ν(∅) = ∫∅ fdµ = ∫(∅) ∑(1/n)(f(x))dµ = ∑(1/n) ∫(∅)(f(x))dµ = ∑(1/n)(0) = 0.
Hence, ν(∅) = 0.
3. Countable additivity:
Let {E_n} be a countable collection of disjoint measurable sets. We need to show that ν(∪E_n) = Σν(E_n).
From the property of measure, we know that the integral is countably additive over disjoint sets.
Therefore, ν(∪E_n) = ∫(∪E_n) fdµ = ∫(∪E_n) ∑(1/n)(f(x))dµ
Since the sets E_n are disjoint, we can apply the linearity of the integral to rearrange the terms:
ν(∪E_n) = ∑(1/n) ∫(∪E_n) (f(x))dµ = ∑(1/n) ∫∪E_n(f(x))dµ
By using the property of measurability, we can express the union as the union of each individual set:
ν(∪E_n) = ∑(1/n)∑ ∫E_n(f(x))dµ
Since the sets E_n are disjoint, we can swap the order of the summations:
ν(∪E_n) = Σ ∑(1/n) ∫E_n(f(x))dµ
This is equivalent to Σν(E_n), which proves countable additivity.
Hence, ν satisfies all three properties of a measure, and therefore, ν is a measure.