A finite additive measure that is continuous from below is a measure

**tttcomrader** · Sep 26th 2009, 09:14 PM

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

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

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

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

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

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

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 $\mu (E_1) + . . . + \mu (E_n)$ .

Any hints? Thank you

**Moo** · Sep 27th 2009, 02:07 AM

Hello,

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

This limit holds only if at least one of the $E_j$ 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 $\mu$ is a measure.

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

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

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

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

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

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

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

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

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

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

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

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

**tttcomrader** · Sep 27th 2009, 10:35 AM

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

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

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

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

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

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

Will this get me to the right track? Thanks.

**Moo** · Sep 27th 2009, 11:13 AM

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

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

Let's start it again.

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

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

With the information we have, this is equivalent to proving $\mu\left(\bigcup_{j=1}^\infty E_j\right)=\lim_{n\to\infty} \mu\left(\bigcup_{j=1}^n E_j\right)$ , 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 $F_n=\bigcap_{j=1}^n \{E\backslash E_j\}$
- 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 $F_n$ , we have $\lim_{n\to\infty} \mu\left(\bigcap_{j=1}^n \{E\backslash E_j\}\right)=\mu\left(\bigcap_{j=1}^\infty\{E\back slash E_j\}\right)$

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

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

Does it look somehow clear ?

Thread: A finite additive measure that is continuous from below is a measure

LinkBack

Thread Tools

Display

A finite additive measure that is continuous from below is a measure

Similar Math Help Forum Discussions

Showing that a specific outer measure is a measure

continuity+finite additivity of measure

finite measure space

Finite additive measure is a measure if it is continuous from below

Finite field subtraction, i.e. additive inverse of a geometric point

Search tags for this page

continuous from above

content

Search Tags