Let $(X, \mu)$ be a measure space. Let $X = A_1 \cup\cdots\cup A_k (A_i \cap A_j = \emptyset$ for $i \neq j)$, where each $A_i$ is measurable. We say $\pi = \{A_1,\dots,A_k\}$ is a finite measurable partition of $X$. We denote by $\mathfrak{P}$ the set of such partitions of $X$. Let $f\colon X \rightarrow [0, \infty]$ be a measurable function on $X$. Let $\pi = \{A_1,\dots,A_k\} \in \mathfrak{P}$. Let $a_i = \inf\{f(x); x \in A_i\}$. Let $b_i = \sup\{f(x); x \in A_i\}$.
We denote $\sum_i a_i\mu(A_i)$ by $\xi(f,\pi)$.
We denote $\sum_i b_i\mu(A_i)$ by $\Xi(f,\pi)$.
We denote $\sup\{\xi(f,\pi); \pi \in \mathfrak{P}\}$ by $\xi(f)$.
We denote $\inf\{\Xi(f,\pi); \pi \in \mathfrak{P}\}$ by $\Xi(f)$.
Is the following proposition true? If yes, how would you prove this?
Proposition
(1) $\xi(f) = \int_X f d\mu$.
(2) If $0 \le f \le M$, where $M$ is a constant such that $0 < M < \infty$, Then $\Xi(f) = \int_X f d\mu$.