I am new to measure theory. This is my understanding of the Lebesgue integral for positive measurable functions:
Let $f:X\to\Bbb R^+$ be a positive Borel measurable function. Take the image $f(X)$ and partition it into $P=\{y_0\lt y_1\lt y_2\lt y_3\lt\cdots\lt y_N\}$ for some finite $N$, of course such that $y\in f(X)\iff \exists n:y\in[y_{n+1},y_n]$, and define $A_n=f^{-1}([y_{n+1},y_n])$ for $n\in\{0,1,\cdots,N-1\}$. Then define the simple function $h_P(x)=\sum_{n=0}^{N-1}y_n\cdot\chi_{A_n}$, where $\chi_{A_n}$ is the characteristic function of $A_n$. It is easily shown that $h_P$ is also a Borel measurable function from $X\to\Bbb R^+$, and by construction $h(x)\le f(x)$ for all $x\in X$. By the properties of the Lebesgue integral for simple functions, the natural generalisation of the integral to arbitrary $f$ keeps the monotonicity property. As the supremum will always exist if you allow $\infty$, then the definition:$$\int_X f(x)\,\mathrm{d}\mu=\sup_{n\in\Bbb N}\left\{\int_X h_P(x)\,\mathrm{d}\mu:P\text{ is a partition, order $n$, of $f^{-1}(X)$}\right\}$$Is well-defined and natural.
This is my own phrasing of what I've learned so far. I am deeply curious to know if the following definition that I just thought of is valid:
Call the integral defined as above by $L(f)$. Redefine $h_P$ as $\sum_{n=1}^Ny_n\cdot\chi_{A_{n-1}}$, so that $f(x)\le h_P(x)$. With this $h_P$, define: $$U(f)=\inf_{n\in\Bbb N}\left\{\int_X h_P(x)\,\mathrm{d}\mu:P\text{ is a partition, order $n$, of $f^{-1}(X)$}\right\}$$In the same spirit as the definition of the upper and lower Darboux integrals.
My question: Is $U(f)=L(f)=\int_Xf(x)\,\mathrm{d}\mu$ always? If so, can I safely not care whether I choose the infimum of upper bound simple integrals or the supremum of lower bound simple functions? If not, why, and when?
Best Answer
Let $S^+(X)$ denote the space of nonnegative simple functions on a measurable space $(X, \mathcal{F})$ (a simple function is a measurable function $\phi : X \to \mathbb{C}$ with $\phi(X)$ a finite set). Let $\mu$ be a measure on $(X, \mathcal{F})$. Given measurable $f \colon X \to [0, \infty]$, we are interested in determining whether $$A(f) := \sup\{\int_X \phi\,d\mu : \phi \in S^+(X), \phi \leq f\} = B(f) := \inf\{\int_{X}\phi\,d\mu : \phi \in S^+(X), \phi \geq f\}.$$ Clearly $A(f) \leq B(f)$. This inequality can be strict, e.g. when $f$ is unbounded and $A(f) < \infty$ since if $f$ is unbounded, then $\{\phi \in S^+(X) : \phi \geq f\} = \emptyset$, so $B(f) = \infty$.