On Daniell integral and the notion of measurability Recently, I decided to learn Daniell integration and after a couple of months on it I like to think that I got the notions right. I also understood the Daniell-Stone theorem that established that, under certain regularity conditions, Lebesgue and Daniell integration are the same thing.
However, there is one notion that is still fuzzy for me, i.e., the notion of measurability, for at least two reasons.
1) I do not understand why necessarily one wants to delve into measurability once we already have the integral, unless it is to determine on the spot the integrability of certain functions.
2) This is my real question. I noticed in different notes/textbooks/papers that there seems to be two notions of measurability:


*

*(a) One says that a nonnegative function f on X is measurable iff $\phi \wedge f $ is integrable for all $\phi $ in the class of elementary functions.

*(b) The second definition says that a function $f $ is measurable if there exists a sequence of elementary functions $\{\phi_n: n \in N\} $ such that $\phi_n \to f $ a.e. on $X. $


I cannot prove that these two notions are equivalent and I am not even sure they are, in fact, equivalent.  I wonder if someone with more mathematical ability and/or more experience on this topic could give me a few pointers.  Thank you in advance to everyone for his or her kindness
Maurice
 A: Daniell's approach to integrability gives exactly the same integrable functions as Lebesgue-Charatheodory's approach.
The notion of measurability in Daniell's approach is based on Littlewood's principles:
Definition:
A real-valued function  $f$ is Daniell-measurable if for any integrable set $A$ and $\varepsilon>0$ there exists an integrable set $A_0$ contained in $A$ on which $f$ is the uniform limit of elementary functions.
A set $B$ is measurable if the indicator $\mathbb{1}_B$ is measurable.
This means that a function is measurable if it is "smooth" in large integrable sets.
The Lebesgue--Charatheodory measurability starts with a measure $\mu$ on a measurable space $(\Omega,\mathscr{F})$ and uses the outer measure $\mu^*$ built from $\mu$.
Definition:
A set $B$ is measurable if 
$$ \mu^*(A)=\mu^*(A\cap B) + \mu^*(A\setminus B) $$
for any subset $A$ of $\Omega$.
A real--valued function $f$ is measurable if $\{f>r\}$ is measurable for any $r\in\mathbb{R}$.
It turns out that both approaches to integration give exactly the same set of measurable functions. 
The space of all real--valued Daniell-measurable functions $\mathscr{M}_\mathbb{R}$ is an algebra lattice containing the constant function $\mathbb{1}$. It is also sequentially closed, that is, if the sequence $(f_n)\subset\mathscr{M}_{\mathbb{R}}$ converges a.s. to a real valued function $f$, then $f\in\mathscr{M}_\mathbb{R}$. The last statement is part of Egorov's theorem.
The following statement shows some equivalence statemnts for Daniell measurability
Proposition
The following statements are equivalent


*

* (a) A function $f$ is Daniell measurable 

* (b) $\{f>r\}$ is Daniell measurable for any $r\in\mathbb{R}$. 

* (c) $f$ is the limit of a sequence of simple Daniell-measurable functions. 

* (d) $(-n)\vee(f\wedge n)$ is Daniell-measurable for any $n\in\mathbb{N}$.


A modern approach to Daniell integration, that starts with a Stone-lattice $\mathcal{E}$
 of bounded functions and a positive $\sigma$--additive elementary integral $I$ and then constructs upper Daniell integral $I^*$ followd by a Daniell mean $\|f\|^*=I^*(|f|)$
is the book of K. Bichteller "Integration: A functional approach".
