Suppose that f is an integrable non-negative function mapping from a set X to all reals.
Let $\displaystyle E_n $ be disjoint sets in X, why is $\displaystyle \int _{ \cup E_n } f = \sum \int _{E_n} f $?
Thanks.
Oh yes he did at least several times
http://www.mathhelpforum.com/math-he...w-measure.html
http://www.mathhelpforum.com/math-he...rsections.html
http://www.mathhelpforum.com/math-he...t-algebra.html
http://www.mathhelpforum.com/math-he...open-sets.html
http://www.mathhelpforum.com/math-he...-infinity.html
http://www.mathhelpforum.com/math-he...ous-below.html
http://www.mathhelpforum.com/math-he...ng-unions.html
http://www.mathhelpforum.com/math-he...ric-space.html
http://www.mathhelpforum.com/math-he...sure-sums.html
Do you want more ?
I think it's rather scarce to see someone detailing this way, with good latex, what he's done so far, eh ?
So I guess you'd better check before saying such a thing.
Hint :
Spoiler:
Not at all !
They certainly don't have the same syntax. One speaks a correct English, the other one is obviously a non native speaker !
The use of the LaTeX is not the same. And the level of the questions asked in the forums is completely different from one to another.
And I showed you 9 proofs that tttcomrader $\displaystyle \neq$ transgalactic.
Please check before saying that.
Coming from someone who's only been a member for 10 days and who apparently wasn't in this forum before that, it's quite inappropriate
lol, sorry about the lack of the work that I showed in this thread, this question is actually part of a proof that I read from the current chapter. I didn't type the whole proof out since it isn't really my original work and I only have question on that particular part.
Anyway, thank you and I appreicate the help, and again I apologize for the confusion.
So my work following the hint:
Prove that $\displaystyle 1_{ \cup E_n } = \sum 1_{ E_n } $
Proof.
If $\displaystyle 1_{ \cup E_n } (x) = 1 $, that means $\displaystyle x \in \cup E_n $, implies that $\displaystyle x \in E_n $ for some n, and that is the only one since all En are disjoint.
Then $\displaystyle \sum 1_{ E_n } (x) = 1$, and the arguement is the same the other way around.
If $\displaystyle 1_{ \cup E_n } (x) = 0 $, then x is not in any of the E_n, so of course $\displaystyle \sum 1_{ E_n } (x) = 0 $
So now, with that established, I need to think about:
$\displaystyle \int _{ \cup E_n } f = sup \{ \int _{ \cup E_n } s : 0 \leq s \leq f , s \ simple \ \} $
$\displaystyle \sum \int _{ E_n } f = \sum sup \{ \int _{ E_n } s : 0 \leq s \leq f , s \ simple \ \} $
Let $\displaystyle s = \sum a_k 1_k (x) $
And $\displaystyle \int _ { \cup E_n} s = \sum _k a_k m( \cup E_n) $
$\displaystyle \sum \int _ {E_n} s= \sum _n \sum _k a_k m( E_n ) $
Now, can I say that $\displaystyle \sum _n m (E_n) = m( \cup E_n ) $ to conclude the proof?
No,
Once you've proved (well) the preliminary thing, just note that :
$\displaystyle \int_{\cup E_n} f =\int f \cdot 1_{\cup E_n}=\int \sum_{n} f \cdot 1_{E_n}=\sum \int f \cdot 1_{E_n}=\sum \int_{E_n} f$
and of course, justify the inversion of sum and integral (in that case it's simple since f is positive)