3.4 A point from two averages
Let \((\Omega , \mu )\) be a measure space and let \(f, g : \Omega \to \mathbb {R}\) be integrable.
(a) If \(\int _\Omega g \, d\mu {\lt} \int _\Omega f \, d\mu \), then \(g(x) {\lt} f(x)\) for at least one \(x \in \Omega \).
(b) If \(f\) and \(g\) are nonnegative with \(\int _\Omega g \, d\mu \leq \int _\Omega f \, d\mu \) and \(\int _\Omega f \, d\mu {\gt} 0\), then there is an \(x \in \Omega \) with \(g(x) \leq f(x)\) and \(f(x) {\gt} 0\).
(a) If \(f \leq g\) everywhere then \(\int f \leq \int g\).
(b) If \(f \leq g\) everywhere then the two averages agree, so the nonnegative function \(g - f\) has average zero and vanishes almost everywhere. At every point where \(f\) and \(g\) agree, the assumption forces \(f\) to vanish, so \(f\) vanishes almost everywhere, contradicting \(\int f {\gt} 0\).
The paper states step (a) as: the set of admissible points has positive measure. One point is all the construction uses. When the two averages are equal, which happens in the doubly degenerate case in the proof of Theorem 1.1, a strict inequality is no longer available and (b) takes over. Both the nonnegativity of \(g\) and the conclusion \(f(x) {\gt} 0\) in (b) are needed: for \(f = (1, 0)\) and \(g = (2, -1)\) on two points with the counting measure both averages are \(1\), and the only point with \(g \leq f\) has \(f = 0\).