A finite additive measure $\displaystyle \mu $ is a measure iff it is continuous from above. [sorry, I made a mistake in the title, should be "from above"]

I finished the proof from finite additive to continuous from above, but stuck on the other one...

Proof so far.

Suppose that $\displaystyle \mu : \mathbb {M} \rightarrow [0, \infty ] $ is continuous from above, that is, if $\displaystyle \{ E_j \} ^ \infty _{j=1} \subset \mathbb {M} $ with $\displaystyle E_{j+1} \subset E_j $, then I have $\displaystyle \lim _{j \rightarrow \infty } \mu (E_j)= \mu ( \bigcap ^ \infty _{j=1} E_j ) $.

I will show that $\displaystyle \mu $ is a measure.

Let $\displaystyle \{ E_j \} ^n _{j=1} \subset \mathbb {M} $ be disjoint sets.

Claim: $\displaystyle \mu ( \bigcup ^n _{j=1} E_j) = \sum _{j=1}^n \mu (E_j) $

Let $\displaystyle F_n = \bigcup ^ \infty _{j=1}E_j \ \bigcup ^n _{i=1} E_i$, note that $\displaystyle F_j \subset F_{j+1} $ and $\displaystyle \bigcup ^n_{j=1}F_j$

Now, since $\displaystyle \mu $ is continuous from above, I have $\displaystyle \lim _{n \rightarrow \infty } \mu (F_n) = \mu ( \bigcup ^ \infty _{j=1} F_j ) $

I'm defining the sequence as such base on how I work the other direction, but I can't seem to be able to break the left hand side into a sum of $\displaystyle \mu (E_1) + . . . + \mu (E_n) $.

Any hints? Thank you

Hello,

Suppose that $\displaystyle \mu : \mathbb {M} \rightarrow [0, \infty ] $ is continuous from above, that is, if $\displaystyle \{ E_j \} ^ \infty _{j=1} \subset \mathbb {M} $ with $\displaystyle E_{j+1} \subset E_j $, then I have $\displaystyle \lim _{j \rightarrow \infty } \mu (E_j)= \mu ( \bigcap ^ \infty _{j=1} E_j ) $.

This limit holds only if at least one of the $\displaystyle E_j$ has a finite measure. If they're all of infinite measure, I think it should be easy to finish your proof by using (*) below. So now assume we can use this formula.

I will show that $\displaystyle \mu $ is a measure.

Since $\displaystyle \mu$ is finite-additive, the first axiom of a measure :
$\displaystyle \mu(\emptyset)=0$
is verified.
And the positivity too.

Let $\displaystyle \{ E_j \} ^n _{j=1} \subset \mathbb {M}$ be disjoint sets.

Claim: $\displaystyle \mu ( \bigcup ^n _{j=1} E_j) = \sum _{j=1}^n \mu (E_j) $

And we want to prove that for any pairwise disjoint sequence $\displaystyle \{F_j\}$, we have :
$\displaystyle \mu\left(\bigcup_{j=1}^\infty F_j\right)=\sum_{j=1}^\infty \mu(F_j)$ (*)

Let $\displaystyle F_n = \bigcup ^ \infty _{j=1}E_j \backslash \bigcup ^n _{i=1} E_i$, note that $\displaystyle F_j \subset F_{j+1} $ and $\displaystyle \bigcup ^n_{j=1}F_j$

The backslash is not \ which gives a space in latex, but \backslash

Now, since $\displaystyle \mu $ is continuous from above, I have $\displaystyle \lim _{n \rightarrow \infty } \mu (F_n) = \mu ( \bigcup ^ \infty _{j=1} F_j ) $

I guess you're being confused with below and above...
The above continuity deals with an intersection, not an union, and with $\displaystyle F_{j+1}\subset F_j$ (which you actually have)


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To sum up... (not too formally) :
We know that $\displaystyle E_{j+1}\subset E_j$, $\displaystyle \lim_{j\to\infty} \mu(E_j)=\mu\left(\bigcap_{j=1}^\infty E_j\right)$ (continuous from above)
We know that $\displaystyle E_j$ pairwise disjoint, $\displaystyle \mu\left(\bigcup_{j=1}^n E_j\right)=\sum_{j=1}^n \mu(E_j)$ (finite additivity)

We want to prove that $\displaystyle E_j$ pairwise disjoint, $\displaystyle \mu\left(\bigcup_{j=1}^\infty E_j\right)=\sum_{j=1}^\infty \mu(E_j)$


With the information we have, this is equivalent to proving $\displaystyle \mu\left(\bigcup_{j=1}^\infty E_j\right)=\lim_{n\to\infty} \mu\left(\bigcup_{j=1}^n E_j\right)$

Then I recommend you read this : http://www.mathhelpforum.com/math-he...ection-ei.html (the second post).

So base on the $\displaystyle F_n$ that I define, I have the following:

1. $\displaystyle \lim _{j \rightarrow \infty } \mu (F_j) = \mu ( \bigcap _{j=1} ^ \infty F_j) $

2. $\displaystyle \bigcup ^ \infty _{j=1} E_j = \bigcap _{j=1} ^ \infty F_j $

3. $\displaystyle F_j = \bigcup ^ \infty _{j=1} E_j \backslash \bigcup _{i=1} ^n E_i $

So I have: $\displaystyle \mu ( \bigcup ^ \infty _{j=1} E_j ) = \mu ( \bigcap _{j=1} ^ \infty F_j ) = \lim _{j \rightarrow \infty } \mu (F_j) = \lim _{ j \rightarrow \infty } ( \bigcup ^ \infty _{j=1} E_j \backslash \bigcup _{i=1} ^n E_i ) $


What I'm trying to do is get the last term to be $\displaystyle \sum _{j=1}^ \infty \mu (E_j) $, but I'm suspecting that the F_n that I define is not really help...

Will this get me to the right track? Thanks.

Okay, I confused some signs, so the reasoning I was pointing you at was not correct... Sorry

Indeed you defined badly $\displaystyle F_n$. I believe I wrote it above

Let's start it again.


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We know that $\displaystyle E_{j+1}\subset E_j, \lim_{j\to\infty} \mu(E_j)=\mu\left(\bigcap_{j=1}^\infty E_j\right)$ (continuous from above)
We know that $\displaystyle E_j$ pairwise disjoint, $\displaystyle \mu\left(\bigcup_{j=1}^n E_j\right)=\sum_{j=1}^n \mu(E_j)$ (finite additivity)

We want to prove that $\displaystyle E_j$ pairwise disjoint, $\displaystyle \mu\left(\bigcup_{j=1}^\infty E_j\right)=\sum_{j=1}^\infty \mu(E_j)$


With the information we have, this is equivalent to proving $\displaystyle \mu\left(\bigcup_{j=1}^\infty E_j\right)=\lim_{n\to\infty} \mu\left(\bigcup_{j=1}^n E_j\right)$, where the Ej are pairwise disjoint << this is important. It uses the finite additivity. Do you understand this point ?


Then, assuming the measure of the whole space is finite (if it's infinite, it may work too but it ain't no fun)...

Let's take a sequence Ej of pairwise disjoint measurable sets.

Define $\displaystyle F_n=\bigcap_{j=1}^n \{E\backslash E_j\}$
- why did I choose the intersection ? Because we need a decreasing sequence for using the continuity from above.
- why did I take the complement ? Because it makes no sense taking the intersection of disjoint sets ! And because of de Moivre's formula, which will help us transform the intersection into an union.

So by the continuity from above applied to the sequence $\displaystyle F_n$, we have $\displaystyle \lim_{n\to\infty} \mu\left(\bigcap_{j=1}^n \{E\backslash E_j\}\right)=\mu\left(\bigcap_{j=1}^\infty\{E\back slash E_j\}\right)$

$\displaystyle \lim_{n\to\infty} \mu(E)-\mu\left(\bigcup_{j=1}^n E_j\right)=\mu(E)-\mu\left(\bigcup_{j=1}^\infty E_j\right)$

and finally $\displaystyle \lim_{n\to\infty}\mu\left(\bigcup_{j=1}^n E_j\right)=\mu\left(\bigcup_{j=1}^\infty E_j\right)$


Does it look somehow clear ?