Hello I would like to show that a nonnegative, Lebesgue measurable function $g$ satisfies the following equality: $\int g (x)dx=\int\chi_{[0,\infty)}m(${g>a}$)$ $ da$ where {g>a} inside $m$ represents the set {$x\in R^d:g(x)>a$}.
My thoughts are that if I can show this is true for simple functions $s$, then since $g$ is a nonnegative function, we know there exists a sequence of increasing simple functions, say {$\phi_n$}, such that lim $\phi_n$ = $g$. Then use some limit theorem to say this is true for $g$ like the monotone convergence theorem?
Please help me show that this equality is true in the simple function case and then let me know if my thoughts regarding generalizing that case seem correct. As far as limit theorems go I only have the monotone, bounded, and dominated convergence theorems at my disposal as well as Fatous lemma.
Yes your approach is exactly right. In fact, the formula holds when $(X, F, \mu)$ is an arbitrary measure space and $g \colon X \to [0, \infty]$ is measurable. Let $\phi \colon X \to [0, \infty)$ be an arbitrary simple function. Let $\{a_1, \dots, a_n\} = \phi(X) \setminus \{0\}$. We have $$\phi = \sum_{i = 1}^{N}a_i\chi_{\phi^{-1}(\{a_i\})} = \sum_{i = 1}^{N}a_i\chi_{E_i},$$ where $E_i = \phi^{-1}(\{a_i\})$. Note that $E_1, E_2, \dots, E_N$ are disjoint and each $a_i > 0$. By rearranging the $E_i$s, we may assume that $0 < a_1 < a_2 < \dots < a_N$. Now you can compute a piecewise formula for $S(a) = \{x \in X : \phi(x) > a\}$. Then you can compute $\mu(S(a))$ and integrate it to verify the identity. Then you can use the monotone convergence theorem to prove the identity for measurable $g \colon X \to [0, \infty]$ (this is where it becomes evident why we have $\{g > a\}$ and not $\{g \geq a\}$ in the formula).
It is worth noting that the motivation for the formula comes from applying Tonelli's theorem to $$\int_X g(x)\,dx = \int_X \int_{\mathbb{R}} I(0 \leq a < g(x))\,da\,dx.$$ However, a proof using Tonelli's theorem would require sigma-finiteness of the measure space $(X, F, \mu)$ and proof of the measurability of the integrand $(x, a) \mapsto I(0 \leq a < g(x))$.