Definition of upper integral

Question

Answering this question it occurred to me that the OP's definition of integral is unsatisfactory in the following sense. He defines it using the usual Lebesgue integral. I think it would be far more satisfactory if we could define the integral without using Lebesgue integral. In other words, it would be far more satisfactory if we could define the upper integral without using Lebesgue integral. Here is my effort to define the upper integral along this line.

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 $\int g \, d\mu$ in the obvious manner. Let $f: X \rightarrow [0, \infty]$ be a non-negative extended real-valued function. Define $\int^* f \,d\mu = \inf \left\{\int g \, d\mu; f \le g, g:\text{ elementary} \right\}$.

My question is:

$$\int f \,d\mu = \int^* f \,d\mu\text{ if }f\text{ is measurable?}$$

Answer 1

Yes.

It's certainly true if $\int f\,d\mu=\infty$; assume then that $\int f\,d\mu<\infty$.

Just to simplify the notation we can assume that $f>0$ (consider $X'\subset X$). And since $f$ is integrable the set where $f=\infty$ has measure zero, so we can also assume $f<\infty$.

For $\lambda>1$ and $n\in\Bbb Z$ let $$E(\lambda,n)=\{x\,:\,\lambda^n\le f(x)<\lambda^{n+1}\}.$$ Let $$g_\lambda=\sum_{n\in\Bbb Z}\lambda^{n+1}\chi_{E(\lambda,n)}.$$Then $g_\lambda$ is elementary and $g_\lambda>f$. In fact we have $$f<g_\lambda\le\lambda f,$$so $g_\lambda\to f$ pointwise as $\lambda$ decreases to $1$.

And for $1<\lambda\le 2$ we have $g_\lambda\le\lambda f\le 2f$. Since $2f$ is integrable, dominated convergence shows that $\int g_\lambda\,d\mu\to\int f\,d\mu$ as $\lambda$ decreases to $1$.