If f (x) > 0 for all x S and is -finite, then S

If f (x) > 0 for all x ∈ S and μ is σ-finite, then 

S f dμ = 0 implies

μ(S) = 0.

[Let Sn be the subset of S on which f (x) ≥ 1/n. Then μ(S) ≤ μ(Sn) and

μ(Sn) ≤ n



Sn f dμ ≤ n



S f dμ = 0.]