I do not know if I get correctly your question. If you have some kind of coordinate processes on a Banach space then you can reduce measurability to R^n measurability. If for some reason you have to define a measurability structure on a Banach space, then you must define an appropriate topology and then a Borel algebra induced by it.

Have a look also at, Laurent Schwartz, Radon measures on arbitrary topological spaces and cylindrical measures. Oxford University Press, 1973 

1. What is necessary to consider and what is not depends on the circle of problems that one studies.

2. When one speaks about Banach space valued functions there is no sense in avoiding concepts that depend on the axiom of choice, because in the Banach space theory everything is based on the Hahn-Banach theorem, which depends on the axiom of choice. So without this axiom you can do nothing. Moreover, on the axiom of choice depends the existence of non-measurable functions, the existence of measurable ones does not depend. So what is the problem?

3. The more tools you know to use, the more results you can obtain.  There are several different (not equivalent) concepts of measurability for Banach space X valued function f. Scalar measurability (composition of f with any bounded linear functional x* is measurable), usual measurability (preimages of Borel sets are measurable) and strong measurability (approximability a.e. by a sequence of simple functions) are the most important ones. The interplay between these types of measurability is an important tool in Banach space theory. For example it is used in the proof of the fact that the dual space to a separable Banach space has the Radon-Nikodym property if and only if this dual space is separable.

4. An advantage of measurability is that it is preserved by many operations that do not preserve integrability. Say, you may manipulate with functions and don't care if the Bochner integrability is preserved on each step: you just need to check measurability at each step (which is usually evident)  and if what you get finally has an integrable majorant, then it is integrable. So breaking  integrability in two parts (strong measurability and existence of an integrable majorant) is very useful. 

respected sir let me clarify subtelkities of my question.

each absolutely integrable function is measurable but an absolutely integrable mapping may not be measurable ( i reserve term function exclusively for real valued mappings following Serge Lang.

Now axiom of choice is needed for blanket staements like dual of abanch space is nonempoty , or maximal ideals ecxist. in any specific cases one can explicitly compute dual etc without using axiom of choice.

true existence of nomeasurable functions only depends upon axiom of choice. but herein lies crux. the scalar measurabiltry ie | f| is measurable which automatically follows if | f| raised to p > 1 is integrable. 

the relevance of the question is in gauge integral that is henstock kurzweil integral by definition the integral of a mapping is approximated by step mappings .

in lebesgue integration one has density theorem that each  integrbale function and measurable integrbale mapping in lp spaces can be approximated in p norm by integral of simple mappings (step mappings ( in case domain is euclideaand one technin) which enables us to prove results about integrals of mappings from integrals of simple maps.

 in hk theory also this resulot holds but is unnecessary as explained in gthe prvious para. so one just assumes pth power of  \f\ is integrable 

Dear Professor Pedgaonkar,

Maybe to make the conversation fruitful we must fix the notation.

1. If one forgets the measurability, then maybe the definition of absolutely integrable function must be the following: a function f: \Omega \to (-\infty, \infty) such that |f| is an integrable function. Do you mean this definition?  If so, then the statement that ``each absolutely integrable function is measurable'' is not true. Just consider a function that equals 1 on a non-measurable subset A and equals -1 on the complement to A. The modulus of f equals 1 in each point, so |f| is integrable (and every power |f|p is integrable as well), but  f  itself is not measurable.

2. I cannot agree with you that the consequences of the axiom of choice like Hahn-Banach theorem or Krein-Milman theorem can be avoided in concrete spaces, because in these spaces we have  descriptions of corresponding dual spaces.  Say, we know that the dual to $\ell_infty$ is the space of all finitely additive measures, but how this helps to prove the existence of a multiplicative functional on $\ell_infty$, different from coordinate functionals? Or the existence of shift-invariant generalized limit (so-called Banach limit)? There is a number of other examples, when we use duality argument, bipolar theorem or weak compactness in concrete spaces, and all this technique needs multiple use of the axiom of choice. For example, the duality is widely used in analysis of very concrete operators on concrete function spaces, and if we reject all these tools, we make our life much more complicated. As for me, there is a lot of hard mathematical problems, so there is no need to make artificially the simple things harder.

All the best, Vladimir

i do not avoid axiom of choice. we fix notation real valued mappings aand only these are termed as functions  all others are called as mappppings .  a function is real valued 

 but it seems to me absolutely integrable mapping is a mapping f such that f and | f \ are bith integrable.. if oine carefully studies lang analysis II or mikisuinski bocner integral, or henstock kurzweil integral an integrable mapping may not be measurable and for hk integrable absolutely integrable mapping f may not be measurable.

 technically all lp spaces can be defined in terms of scalar measurability that is |f | is measurable.

 of course in case of a cauchy sequence if  mappings are measurable then limit is measurable..nonmeasurable functions only exist theoratically . so issue of mappings which are integrable but not measurable is a purely theoratical issue.

 so why not avoid this issue in defining lp spaces. ( rember mappings not functions.) even for functions i can consider l2 wherdee |f/\ square is integrable but f may not be measurable. very fine i get bigger l2 space and i can rim it if there is neeed

As far as i know, a mapping f from a measure space X (endowed with a positive measure), into a a topological space Y, is called measurable if and only if f ^(-1)(V) is a measurable set in X, for any open subset V in Y. In the particular case when Y=R (or C), we obtain the concept of measurable real (respectively complex) function. The spaces L^p, 1

Yes but ibprefer tobeena r lang alsooneneed not insist on measurability there canbea sqare integrable mapping but not measurable i meanmappinginto Banach space not a complex valued  function if one consider hence talk or  kurzweil integral then absolute integrable mapping may not b measurable but just scalarly measurable also generalized ito integral called skorhood integral is defined for non measurable stochastic processes

Waltetvrudins book is not preferred for measure theory but royden is lucid lang analysis  ii is more general