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 non-measurable function! (if is a non-measurable subset of , then is a null set and is thus Lebesgue measurable, but not Borel measurable in .) One can achieve similar pathologies in one-dimension 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/2010/10/21/245a-problem-solving-strategies/

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?

]]>the set is equal to

Since , it suffices to show that

One direction is obvious:

since

for all . But the other direction is not:

only implies that

.

Where do I do wrong?

]]>for and

, where are the non-negative rationals. ”

——————————

In the first formula, if one uses instead of , then it seems that the formula works for also.

In the second formula, why doesn’t it work when ? (the LHS is an empty set and the RHS is an empty union…)

*[The formulae work at these endpoints also. -T.]*