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