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.]*

Could you give an example to explain your remark above? (What “problem” are you refering to?)

]]>