Calculating the Lebesgue Integral given only the measure of a set
Solution 1:
On any measure space $(X, \mathcal{F}, \mu)$, if $f \colon X \to [0, \infty]$ is measurable, then $$\int_{X}f\,d\mu = \int_{0}^{\infty}\mu(\{x \in X : f(x) > t\})\,dt.$$ When $\mu$ is $\sigma$-finite, this is an easy consequence of Tonelli's theorem. If $\mu$ is not $\sigma$-finite, then to prove it you can first prove it when $f$ is a simple function and then use monotone convergence theorem.