Let \(f,g\in L^1([a,b])\) be real-valued with \[ \int_{a}^{b} f(x) ~d x=\int_{a}^{b} g(x) ~d x. \] Show that either
- \(f(x) = g(x)\) almost everywhere, or
- There exists a measurable set \(E \subset [a, b]\) such that \[ \int _{E} f(x) \, dx > \int _{E} g(x) \, dx \]
- Monotonicity of the Lebesgue integral: \(f\leq g\) on \(A\) \(\implies \int_A f \leq \int_A g\)
Take the assumption and the negation of (1) and show (2). The obvious move: define the set \(A\) where they differ. The non-obvious move: split \(A\) itself up to get a strict inequality.
Solution
Let \[ h=f-g. \] Then \(h\in L^1([a,b])\) and \[ \int_a^b h\,dx=0. \] If \(h=0\) almost everywhere, then \(f=g\) almost everywhere and alternative (1) holds.
Assume instead that \(h\ne0\) on a set of positive measure. Write \[ E_+=\{h>0\}, \qquad E_-=\{h<0\}. \] If \(m(E_+)=0\), then \(h\le0\) almost everywhere and \(h<0\) on a set of positive measure. Consequently \[ \int_a^b h\,dx<0, \] contradicting \(\int h=0\). Thus \(m(E_+)>0\).
Because \(h>0\) on \(E_+\), \[ E_+=\bigcup_{k=1}^\infty\{h\ge 1/k\}. \] Hence some \(k\) satisfies \[ m(\{h\ge1/k\})>0. \] It follows that \[ \int_{E_+}h\,dx \ge \frac1k\,m(\{h\ge1/k\})>0. \] Taking \(E=E_+\) gives \[ \int_E f\,dx-\int_E g\,dx =\int_E h\,dx>0. \] Therefore alternative (2) holds.