Prove that the integral of a nonnegative lebesgue measurable function h is the integral of measures m({h>a} over [0,\infty).

Question

The problem is to prove that a nonnegative, Lebesgue measurable function $h$ satisfies the following equality: $\int h (x)dx=\int\chi_{[0,\infty)}m(${h>b}$)$ $ db$ where {h>b} inside $m$ represents the set {$x\in R^d:h(x)>b$}. Can someone please teach me how to prove this? I was thinking of first proving this when $h$ is a simple function but I am stuck even on this simplification. Also $\chi$ represents the characteristic function.

Answer 1

Let the sequence of simple functions be $$f_n(x)=\sum_{k=1}^{N(n)}\phi_{n,k}\mathbb{I}_{A_{n,k}}(x),\,A_{n,k}\cap A_{n,j}=\emptyset,\,\forall n $$ As $f_n \uparrow f$ we have $\{f_n>t\}\subseteq \{f> t\},\,\forall n\in \mathbb{N}$ and $\{f_n> t\}\subseteq \{f_{n+1}> t\}$ and we have $$\{f > t\}=\Big\{\sup_{n \in \mathbb{N}} f_n > t\Big\}=\bigcup_{n \in \mathbb{N}}\{f_n>t\}$$ So by measure continuity $$\lambda(\{f > t\})=\lim_{n \to \infty}\lambda(\{f_n>t\})$$ We have $$\lambda(\{f_n>t\})=\sum_{k:\phi_{n,k}>t}\lambda(A_{n,k})=\sum_{k=1}^{N(n)}\mathbb{I}_{\{\phi_{n,k}>t\}}(k)\lambda(A_{n,k})$$ Also $$\mathbb{I}_{\{\phi_{n,k}>t\}}(k)=\begin{cases}1& \phi_{n,k}> t\\ 0& \phi_{n,k}\leq t\end{cases}=\mathbb{I}_{[0,\phi_{n,k})}(t)$$ Therefore $$\lambda(\{f>t\})=\lim_{n \to \infty}\sum_{k=1}^{N(n)}\mathbb{I}_{[0,\phi_{n,k})}(t)\lambda(A_{n,k})$$ As $\lambda(\{f_n>t\})\leq \lambda(\{f_{n+1}>t\})$ we use MCT $$\int_{[0,\infty)}\lambda(\{f>t\})dt=\lim_{n \to \infty}\sum_{k=1}^{N(n)}\phi_{n,k}\lambda(A_{n,k})=\int_\mathbb{R} f\,d\lambda$$