Equivalently, it seems to me that
By some argument of monotone convergence?
]]>but
as one can see by considering what happens when .
]]>One way to think of a measurable function is as the pointwise limit of simple functions (finite linear combinations of indicator functions of measurable sets). This ties in with the underlying intuition of Lebesgue integration as coming from “horizontally” slicing up the graph of a function (as opposed to Riemann integration, which is focused instead on “vertically” slicing up the graph).
Strengthening the measurability requirement by placing the Lebesgue sigma algebra, rather than the Borel sigma algebra, on the range leads to some pathologies, for instance the very natural function from to is now a nonmeasurable function! (if is a nonmeasurable subset of , then is a null set and is thus Lebesgue measurable, but not Borel measurable in .) One can achieve similar pathologies in onedimension by using an essentially bijective map between, say, the unit interval and the Cantor set.
]]>For doing integration of functions , why is it enough to use Borel measure in the target space?
]]>If I don’t understand it wrong, the “uniqueness of the Lebesgue integral” means the following:
Let be a operator such that
 when is simple
 for any measurable functions f and g


Show that for any Lebesgue measurable functions.
is immediately true when is simple. How can one past the case of simple functions the to general cases?
]]>I don’t see what strategy one should use in
http://terrytao.wordpress.com/2010/10/21/245aproblemsolvingstrategies/
It seems that one can take the “finite measure support” out from (4) in Lemma 7. Are simple functions with finite measure support of special interest in application? Isn’t “uniform convergence” preferable?
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