Let $(X, \mathcal M, \mu)$ be a measure space. Let $g: X \rightarrow [0, \infty]$ be a non-negative extended real-valued function. We call $g$ an elementary function if $g$ is measurable and $g(X)$ is countable. We define $\int g \ \mathsf d\mu$ in the obvious manner. Let $f: X \rightarrow [0, \infty]$ be a non-negative extended real-valued function. Define $$\int^* f \ \mathsf d\mu = \inf \left\{\int g\ \mathsf d\mu: f \le g, g\text{ elementary} \right\}.$$ We call $\int^* f \ \mathsf d\mu$ the upper integral of $f$. Suppose $f$ is measurable. It can be proved by using a high-profile theorem in the theory of Lebesgue integration that $\int f \ \mathsf d\mu = \int^* f \ \mathsf d\mu$. See Definition of upper integral.

A function $s: X\rightarrow \mathbb R$ is called a simple function if $s$ is measurable and $s(X)$ is a finite set. Let $f: X \rightarrow [0, \infty]$ be a measurable function. I would like to prove the following fact without using the knowledge of Lebesgue integration theory . The shorter the better.

$$\int^* f \ \mathsf d\mu = \sup \left\{\int s \ \mathsf d\mu: 0\le s \le f, s\text{ simple} \right\}.$$

  • $\begingroup$ @EricWofsey The OP does not use the definition of Lebesgue integral in his question. $\endgroup$ – JHW Aug 6 '15 at 1:15
  • $\begingroup$ I think you could reframe your question and use the Lebesgue monotone convergence theorem. $\endgroup$ – user247608 Aug 6 '15 at 4:41
  • $\begingroup$ @user247608 Maybe you misunderstand my point. I came up with a new definition of Lebesgue integrals. I would like to develop an integration theory upon my definition from scratch. Which means proving basic theorems on the integration without using prior knowledges for Lebesgue integration theory. The current question is the first step toward my goal. $\endgroup$ – Deep Aug 6 '15 at 4:56

Let $f: X \rightarrow [0, \infty]$ be a measurable function. For integers $n\ge 1, k \ge 0$, let $$E_{n,k} = \{x\in X: k/2^n \le f(x) \lt (k+1)/2^n\}.$$ Let $$E_{\infty} = \{x\in X: f(x) = \infty\}.$$ We denote by $\chi_{n, k}$ the characteristic function of $E_{n,k}$ and by $\chi_{\infty}$ the characteristic function of $E_{\infty}$. Define $$g_n = \sum_{k=1}^{\infty} \frac k{2^n} \chi_{n, k} + n\chi_{\infty}$$ and $$h_n = \sum_{k=0}^{\infty} \frac{k+1}{2^n} \chi_{n, k} + \infty\chi_{\infty}.$$ Then $0 \le g_1 \le g_2\le \cdots$ and $h_1\ge h_2\ge \cdots$ and $g_n(x) \le f(x) \le h_n(x)$ for all $n$ and $x\in X$.

By definition,

$$\int g_n\ \mathsf d\mu = \sum_{k=1}^{\infty} \frac k{2^n}\mu(E_{n,k}) + n\mu(E_{\infty}),$$

$$\int h_n\ \mathsf d\mu = \sum_{k=0}^{\infty} \frac{k+1}{2^n}\mu(E_{n,k}) + \infty\mu(E_{\infty}).$$

Lemma. Let $f: X \rightarrow [0, \infty]$ be a measurable function. Let $g_n, h_n$ be as defined above. Then $$\sup \left\{\int g_n\ \mathsf d\mu : n = 1, 2, \cdots\right\} = \inf \left\{\int h_n\ \mathsf d\mu: n = 1, 2, \cdots\right\}.$$

Proof: If $\mu(E_{\infty}) \gt 0$, then $\sup \left\{\int g_n\ \mathsf d\mu: n = 1,2, \cdots\right\} = \infty$ and $\int h_n\ \mathsf d\mu = \infty$ for all $n$. Hence we may assume $\mu(E_{\infty}) = 0$.

Suppose $\text{inf} \{\int h_n\ \mathsf d\mu: n = 1, 2, \cdots\} = \infty$. Then $$\int h_n\ \mathsf d\mu = \sum_{k=0}^{\infty} \frac{k+1}{2^n}\mu(E_{n,k}) = \infty$$ for all $n$. Choose an integer $n \ge 1$. Let $$E = \{x\in X: f(x) \lt \infty\},$$ then $$E = \bigcup_{k=0}^{\infty} E_{n,k}.$$ Suppose $$\mu(E) = \sum_{k=0}^{\infty} \mu(E_{n,k}) \lt \infty.$$ Since $$\int h_n\ \ \mathsf d\mu = \sum_{k=0}^{\infty} \frac k{2^n}\mu(E_{n,k}) + \frac1{2^n} \left(\sum_{k=0}^{\infty} \mu(E_{n,k})\right) = \infty,$$ we have $$\sum_{k=0}^{\infty} \frac k{2^n}\mu(E_{n,k}) = \infty.$$ Hence $\int g_n\, d\mu = \infty$. Thus the assertion is proved in this case.

Suppose $$\mu(E) = \sum_{k=0}^{\infty} \mu(E_{n,k}) = \infty.$$ Since $$E = \bigcup_{k=0}^{\infty} E_{n,k} = \bigcup_{m=1}^{\infty} \bigcup_{k=1}^{\infty} E_{n+m,k},$$ we have $$\lim_{m\rightarrow \infty} \sum_{k=1}^{\infty} \mu(E_{n+m,k}) = \infty.$$ Since $$\sum_{k=1}^{\infty} \mu(E_{n+m,k})\le \int g_{n+m}\ \mathsf d\mu,$$ we have $$\lim_{m\rightarrow \infty} \int g_{n+m}\ \ \mathsf d\mu = \infty.$$ Hence $$\sup\left\{\int g_n\ \ \mathsf d\mu: n = 1, 2, \cdots\right\} = \infty.$$ Thus the assertion is also proved in this case.

Suppose $$\inf\left\{\int h_n\ \mathsf d\mu: n = 1, 2, \cdots\right\} \lt \infty.$$ There exists $n_0$ such that $$\int h_{n_0}\ \ \mathsf d\mu \lt \infty.$$ Let $n$ be any integer such that $n\ge n_0$. Since $g_n \le h_n \le h_{n_0}$, $$\int g_n\ \mathsf d\mu \le \int h_n\ \mathsf d\mu \le \int h_{n_0}\ \mathsf d\mu \lt \infty.$$ Then $$\int h_n\ \mathsf d\mu - \int g_n\ \mathsf d\mu = \sum_{k=0}^{\infty} \frac1{2^n}\mu(E_{n,k}) = \frac1{2^n} \sum_{k=0}^{\infty} \mu(E_{n,k}).$$ Since $\sum_{k=0}^{\infty} \mu(E_{n,k}) \lt \infty$, letting $n\rightarrow \infty$ we are done. $\Box$

Now we prove your assertion $$\int^* f\ \mathsf d\mu = \sup\left\{\int s\ \mathsf d\mu: 0\le s\le f, s \text{ simple }\right\}.$$ For integers $n,m \ge 1$, define $$s_{n,m} = \sum_{k=1}^m \frac k{2^n} \chi_{n, k} + n\chi_{\infty}.$$ Then $s_{n,m}$ are simple functions and for each $n$, $\{s_{n,m}\}$ is an increasing sequence with respect to $m$. It is clear that $$\lim_{m\rightarrow \infty} \int s_{n,m}\ \mathsf d\mu = \int g_n\ \mathsf d\mu.$$ Hence by the lemma $$\sup\left\{\int s_{n,m}\ \mathsf d\mu: n, m \ge 1\right\} = \sup\{\int g_n\ \mathsf d\mu: n \ge 1\} = \inf\left\{\int h_n\ \mathsf d\mu: n \ge 1\right\}.$$

Let \begin{align} \alpha &= \inf\left\{\int h_n\ \mathsf d\mu: n \ge 1\right\},\\ \beta &= \sup\left\{\int s\ \mathsf d\mu: 0\le s\le f, s \text{ simple }\right\},\\ \beta' &= \sup\left\{\int s_{n,m}\ \mathsf d\mu: n, m \ge 1\right\}. \end{align} Then $\beta' \le \beta \le \alpha$. Since $\beta' = \alpha$, $\beta = \alpha$. On the other hand, $\beta \le \int^* f\ \mathsf d\mu \le \alpha$. Hence $$\beta = \int^* f\ \mathsf d\mu.$$ $\Box$

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