Let(X,Θ,µ) be a measure space,f a nonnegative measurable function on X.Show that Xfdµ=0 if and only if f=0 µ a.e. on X.
First, assume that Xfdµ=0. We need to show that f=0 µ a.e. on X.
Let A ={x∈X : f(x) > 0 }. Suppose µ(A) > 0. Then we can find a measurable subset B of A such that 0 < µ(B) < ∞, because µ is a non-negative measure. Define the function g as g(x) = µ(B)^-1 for x∈B and g(x) = 0 otherwise. Then g is a non-negative measurable function and g(x)f(x) > 0 for x∈B. Therefore, we have
∫X g(x)f(x) dµ ≥ ∫B g(x)f(x) dµ > 0
which contradicts the assumption that Xfdµ=0. Hence, µ(A) = 0 and we conclude that f=0 µ a.e. on X.
Now we need to prove the converse. Assume that f=0 µ a.e. on X. We need to show that Xfdµ=0.
Since f=0 µ a.e. on X, we can find a measurable set A such that µ(A) = 0 and f(x) = 0 for all x∈X\A. Then
∫X fdµ = ∫A fdµ + ∫X\A fdµ
= ∫A 0 dµ + ∫X\A 0 dµ (by definition of A)
= 0
Hence, we have Xfdµ=0.
Therefore, we have proved both directions of the statement, i.e., Xfdµ=0 if and only if f=0 µ a.e. on X.
To prove the statement "Xfdµ=0 if and only if f=0 µ a.e. on X," we need to show both implications.
1. Xfdµ = 0 implies f = 0 µ a.e. on X:
Assume Xfdµ = 0. We want to show that f = 0 µ almost everywhere on X.
Since f is a nonnegative measurable function, we can consider the set A = {x ∈ X : f(x) > 0}. We aim to prove that µ(A) = 0.
Suppose, for contradiction, that µ(A) > 0. Then, there exists a measurable subset B of A such that 0 < µ(B) < ∞, since µ is a measure. Now, consider the function g defined as follows:
g(x) = {
1 if x ∈ B,
0 otherwise.
}
It is clear that g is a measurable function and g(x) ≤ f(x) for all x ∈ X. Therefore, we have:
Xgdµ ≤ Xfdµ = 0.
However, we also have:
Xgdµ = X(B)dµ ≥ µ(B) > 0,
which leads to a contradiction.
Hence, we must have µ(A) = 0, which implies f = 0 µ almost everywhere on X.
2. f = 0 µ a.e. on X implies Xfdµ = 0:
Assume f = 0 µ almost everywhere on X. We want to show that Xfdµ = 0.
Consider the set A = {x ∈ X : Xfdµ > 0}. We aim to prove that µ(A) = 0.
Suppose, for contradiction, that µ(A) > 0. Then, there exists a measurable subset B of A such that 0 < µ(B) < ∞, since µ is a measure. Since f = 0 µ almost everywhere on X, we have:
µ({x ∈ B : f(x) > 0}) = 0.
However, this contradicts the fact that Xfdµ > 0 on B, as this would imply µ({x ∈ B : f(x) > 0}) > 0.
Hence, we must have µ(A) = 0, which implies Xfdµ = 0.
Therefore, we have successfully proven both implications, and the statement "Xfdµ=0 if and only if f=0 µ a.e. on X" is verified.