Let \(f : [0, 1] \to [0, \infty ]\) be measurable. Show that \(U(f) = \{(x, y) : 0 \le y \le f(x)\}\) is measurable and \[ \int f\,d\lambda = \lambda ^{2}(U(f)). \]
Question
Math 6210 • Real Analysis
Fall 2022 • Question 3
Question
Fall 2022 • Question 3
Let \(f : [0, 1] \to [0, \infty ]\) be measurable. Show that \(U(f) = \{(x, y) : 0 \le y \le f(x)\}\) is measurable and \[ \int f\,d\lambda = \lambda ^{2}(U(f)). \]