Let $\mu^{*}(X) < \infty $ and suppose $g: X \rightarrow \mathbb{R} $ is a measurable and a bounded function. Show that $g$ is an upper function.
A function $g: X \rightarrow \mathbb{R}$ is a upper functionn if there exists a sequence of step functions, $ \{ \phi_{n} \} $ such that $$(1). \phi_{n} \uparrow g $$ $$(2). lim \int \phi_n < \infty$$. Recall the Lebesgue integral is the following defintion: $$\int \phi_n = \sum_{k = 1}^{n} a_k \mu^*(A_k)$$
Since $g$ is a bounded function, there is a $N \in \mathbb{R}$ such that $-N \leq f(x) \leq N \ \forall x \in X$. Since $g$ is a measurable function, there is a sequence of simple functions, {\phi_n } such that $lim_{n \rightarrow \infty } \phi_n = g$ So I was going to define the sequence $\beta_n = -N \chi_{X} \wedge N \chi_{X}$. I am not sure if this sequence is correct. Is there another way to approach this problem? Thank you for all your help!