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.

1 answer

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.