> 
> 
> > 
> > Let $f\colon X\to\overline{\mathbb{R}}\_{\geq 0}$ be a measurable function. Show that
> >  $$
> > \int f\, d\mu=\int\limits\_0^{\infty}\mu(\left\{x\in X: f(x)>t\right\})\, dt.
> > $$
> >  (The right integral is to be read as an Lebesgue-integral.)
> > 
> > 
> > 
> 
> 
> 

Hello, my idea is to approximate $f$ by the functions
$$
f\_n:=2^{-n}\sum\limits\_{k=1}^{\infty}\chi\_{\left\{x\in X: f(x)> k/2^n\right\}}.
$$

With this approximation by simple functions I got
$$
\int\limits\_X f\, d\mu=\lim\limits\_{n\to\infty}\int\limits\_X f\_n\, d\mu=\lim\limits\_{n\to\infty}\sum\limits\_{k=1}^{\infty}\frac{k}{2^n}\cdot\mu\left(\left\{x\in X: f\_n(x)=k/2^n\right\}\right)\\=\lim\limits\_{n\to\infty}\sum\limits\_{k=1}^{\infty}\frac{k}{2^n}\cdot\mu\left(\left\{x\in X:f(x)>k/2^n\right\}\right)
$$

This is the point where I do not come along anymore... is that right to this point and if yes: How can I continue in order to get the desired right side?

With regards