Integrade with disjoint sets

**tttcomrader** · November 9th 2009, 04:40 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** · November 9th 2009, 04:43 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** · November 10th 2009, 01:15 PM

Oh yes he did at least several times

Spoiler:

**Drexel28** · November 10th 2009, 03:39 PM

Originally Posted by Moo

Oh yes he did at least several times

Spoiler:

tttcommander,transgalatic same thing.

**Moo** · November 10th 2009, 03:43 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** · November 11th 2009, 07:44 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** · November 11th 2009, 08:32 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)

Thread: Integrade with disjoint sets

LinkBack

Thread Tools

Display

Integrade with disjoint sets

Similar Math Help Forum Discussions

Disjoint sets and Axiom of choice.

Connected Sets in the Plane which are not Disjoint

analysis help (countable union of disjoint sets)

Disjoint sets/basis

Disjoint sets

Search Tags