Integrade with disjoint sets

**tttcomrader** · Nov 9th 2009, 03:40 PM

Suppose that f is an integrable non-negative function mapping from a set X to all reals.

Let $E_n$ be disjoint sets in X, why is $\int _{ \cup E_n } f = \sum \int _{E_n} f$ ?

Thanks.

**Drexel28** · Nov 9th 2009, 03:43 PM

Originally Posted by tttcomrader

Suppose that f is an integrable non-negative function mapping from a set X to all reals.

Let $E_n$ be disjoint sets in X, why is $\int _{ \cup E_n } f = \sum \int _{E_n} f$ ?

Thanks.

I don't think I have ever seen you post your work. What have you done up to this point?

**Moo** · Nov 10th 2009, 12:15 PM

Oh yes he did at least several times

Spoiler:

**Drexel28** · Nov 10th 2009, 02:39 PM

Originally Posted by Moo

Oh yes he did at least several times

Spoiler:

tttcommander,transgalatic same thing.

**Moo** · Nov 10th 2009, 02:43 PM

Originally Posted by Drexel28

tttcommander,transgalatic same thing.

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 $\neq$ 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

**tttcomrader** · Nov 11th 2009, 06:44 AM

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 $1_{ \cup E_n } = \sum 1_{ E_n }$

Proof.

If $1_{ \cup E_n } (x) = 1$ , that means $x \in \cup E_n$ , implies that $x \in E_n$ for some n, and that is the only one since all En are disjoint.

Then $\sum 1_{ E_n } (x) = 1$ , and the arguement is the same the other way around.

If $1_{ \cup E_n } (x) = 0$ , then x is not in any of the E_n, so of course $\sum 1_{ E_n } (x) = 0$

So now, with that established, I need to think about:

$\int _{ \cup E_n } f = sup \{ \int _{ \cup E_n } s : 0 \leq s \leq f , s \ simple \ \}$

$\sum \int _{ E_n } f = \sum sup \{ \int _{ E_n } s : 0 \leq s \leq f , s \ simple \ \}$

Let $s = \sum a_k 1_k (x)$

And $\int _ { \cup E_n} s = \sum _k a_k m( \cup E_n)$

$\sum \int _ {E_n} s= \sum _n \sum _k a_k m( E_n )$

Now, can I say that $\sum _n m (E_n) = m( \cup E_n )$ to conclude the proof?

**Moo** · Nov 11th 2009, 07:32 AM

No,

Once you've proved (well) the preliminary thing, just note that :

$\int_{\cup E_n} f =\int f \cdot 1_{\cup E_n}=\int \sum_{n} f \cdot 1_{E_n}=\sum \int f \cdot 1_{E_n}=\sum \int_{E_n} f$

and of course, justify the inversion of sum and integral

(in that case it's simple since f is positive)

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