I'm trying to solve the following exercise in Measure Theory:
Let $(X,\mathcal{F},\mu)$ be a $\sigma$ -finite measure space. Prove that for every $0\leq f\in L^{1}(\mu)$ it holds that: $\intop_{X}\, f\, d\mu=\intop_{0}^{\infty}\mu(\{f\geq t\})\, dt$
My attempt at a solution:
$\intop_{X}\, f\, d\mu=\underset{n\rightarrow\infty}{lim}\overset{\infty}{\underset{K=0}{\sum}}\frac{K}{n}\cdot\mu\{\frac{K}{n}\leq f\leq\frac{K+1}{n}\}=\underset{n\rightarrow\infty}{lim}\overset{\infty}{\underset{K=0}{\sum}}\frac{1}{n}\cdot\mu\{\frac{K}{n}\leq f\}=\underset{n\rightarrow\infty}{lim}\underset{N\rightarrow\infty}{lim}\overset{N}{\underset{K=0}{\sum}}\frac{1}{n}\cdot\mu\{\frac{K}{n}\leq f\}\overset{*}{=}\underset{N\rightarrow\infty}{lim}\intop_{0}^{N}\mu(\{f\geq t\})\, dt$
My last equation (with the asterix) is where I'd like to go but it isn't justified (why can I exchange the order of limits? And how do I justify turning the sum into an integral?). Am I in the right direction?
Would greatly appreciate some help.