# Is an integrable map from a measure space to a Banach space always measurable?

Is every integrable mapping defined in a general measure space to a Banach space measurable?

The answer is yes if it is function (real valued). The answer is yes if it is a mapping into a Banach space but defined on euclidean space.

The monograph by J. Mikisuinski says that you must add this fact as an axiom for general Bochner integral. Serge Lang in his book "real analysis" defines Lebesgue integral for mappings into a Banach space and specifically insists on measurability but definition makes sense if a mapping f is not measurable but | f| is |f| is a real valued function . So why should we force measurability?

As far as I know, nobody has given a specific mapping say on a locally compact space to a Banach space which is integrable and of course |f| measurable but f not measurable.Can somebody furnish an actual example? Will it involve transfinite induction (axiom of choice in some form)? Will it involve a cardinal greater than the cardinal number of real line? a transfinite induction in some sense of higher order? These questions are not at all addressed.

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If possible, could you please edit your post to fix the spelling and capitalization? –  Carl Mummert Jun 10 '11 at 11:46
Perhaps you will need to provide a definition of "integrable" as well. –  Gerald Edgar Jun 10 '11 at 13:55
also: "measurable" wrto which sigma-algebra on the Banach space? –  Pietro Majer Jun 10 '11 at 16:54
Wikipedia has definitions of the Bochner integral (en.wikipedia.org/wiki/Bochner_integral), Bochner space (en.wikipedia.org/wiki/Bochner_space) and Bochner measurability (en.wikipedia.org/wiki/Bochner_measurable_function). I assume these are the intended definitions. –  George Lowther Jun 13 '11 at 3:39
Maybe the question is asking whether Bochner measurable functions are measurable in the usual sense with respect to the Borel sigma-algebra. But I'm not sure. –  George Lowther Jun 13 '11 at 3:57