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Let $(X, \mathcal F, \mu)$ be a measure space and $f\in M^+(X,\mu)$ (the measurable non-negative functions), and $t>0$. Now let $$S_f(t)=\{x\in X:f(x)>t\} \quad \Psi_f(t)=\mu(S_f(t))$$ Prove that $$\int_Xfd\mu=\int_0^\infty \Psi_f(t)dt$$

I guess the non-trivial part is to prove it for simple functions. The rest follows from B. Levy's theorem. Namely if $\{f_n\}_{n=0}^\infty$ a monotone increasing sequence of measurable functions, then it has a limit $$f:f\in M^+ \land \lim_{n\to \infty}\int_Xf_nd\mu=\int_Xfd\mu$$

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See Two definitions of Lebesgue integration –  Martin Feb 6 '13 at 10:13

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Using Tonelli's/Fubini's theorem, we have $$ \begin{align} \int_0^\infty \Psi_f(t)\,\mathrm dt&=\int_0^\infty\int_X 1_{S_f}(t)\,\mathrm \mu(\mathrm dx)\,\mathrm dt=\int_X\,\int_0^\infty 1_{S_f}(t)\,\mathrm dt\,\mu(\mathrm dx)\\ &=\int_X\int_0^\infty 1_{[0,f(x)]}\,\mathrm dt\,\mathrm \mu(\mathrm dx)=\int_X f(x)\,\mu(\mathrm dx), \end{align} $$ which is justified by the assumption that $f$ is non-negative. However, you need your measure space to be $\sigma$-finite in order for Tonelli/Fubini to work.

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Yes, thank you. $1_Y$ must be the characteristic function of Y, but what do you mean by $\mu(dx)$? This is the first time I see this notation. It must be what I mean by $d\mu$, must it not? –  Student Feb 6 '13 at 13:36
    
Yes, $\mathrm d\mu$ and $\mu(\mathrm d x)$ is the same. I just prefer writing $\mu(\mathrm dx)$ whenever I write $f(x)$. For example $$ \int f(x)\,\mu(\mathrm dx)\quad \text{vs.}\quad\int f(x)\,\mathrm d\mu$$ –  Stefan Hansen Feb 6 '13 at 16:03

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