Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am not sure I am using the standard definitions so I will open by defining what I need:

  1. Let $X$ be a set, $\nu:\, \mathscr{P}(X)\to[0,\infty]$ will be called an external measure if $\nu(\emptyset)=0$ and for any $\{A_{i}\}_{i=1}^{\infty}\subseteq\mathscr{P}(X)$ (not neccaseraly disjoint) it holds that $\nu(\cup_{i=1}^{\infty}A_{i})\leq\sum_{i=1}^{\infty}\nu(A_{i})$

  2. Let $\nu$ be an external measure on a set $X$ then we say that a set $A$ is $\nu$ measurable if for any $E\subseteq X$: $\nu(E)=\nu(E\cap A)+\nu(E\cap A^{c})$

I have an exersice that asks me to prove that if $\nu(A)=0$ then $A$ is $\nu$- measurable.

Note: I have already proved that the set of all $\nu$-measurable sets, denoted by $M$, is a $\sigma$-algebra.

What I tried:

For any $E\subseteq X$:

From containment: $$\nu(E\cap A^{c})\leq\nu(E)$$

But $E=(E\cap A)\cup(E\cap A^{c})$ thus $$\nu(E)=\nu((E\cap A)\cup(E\cap A^{c}))\leq\nu(E\cap A)+\nu(E\cap A^{c})$$

This is the part I want to say that since $E\cap A\subseteq A$ and $\nu(A)=0$ then $\nu(E\cap A)=0$ and so $$\nu(E)=\nu(E\cap A^{c}) $$

thus

$$\nu(E)=0+\nu(E\cap A^{c})=\nu(E\cap A)+\nu(E\cap A^{c})$$

but the problem is that I do not know that $\nu$ is monotone (I can argue that its monotone on sets in $M$ but $E\cap A^{c}$and the other sets here need not be in $M$).

Can someone please help me to prove that $\nu$ is monotone, or suggest another approach ?

share|improve this question
    
How about this counterexample (I hope I am not wrong this time) –  Amr Dec 15 '12 at 14:38
    
@Amr - but whats wrong with $\{0\}$ being not $\nu$-measurable ? –  Belgi Dec 15 '12 at 14:49
    
$\{0\}$ has zero measure –  Amr Dec 15 '12 at 14:49
    
@Amr - the empty set need to be $\nu$ measurable, but there is no need for all the sets to be $\nu$ measurable –  Belgi Dec 15 '12 at 14:51
    
I didnt get your last comment –  Amr Dec 15 '12 at 14:52
show 1 more comment

2 Answers

up vote 2 down vote accepted

Let $X=\{0,1\}$. Let $\nu(\emptyset)=0,\nu(\{0,1\})=1$ and $\nu(\{0\})=0,\nu(\{1\})=2.$ It is easy to verify that $\nu$ is an external measure yet $\{0\}$ is not $\nu-$measurable.

$\nu(\{0,1\})\not=\nu(\{0,1\}\cap\{0\})+\nu(\{0,1\}-\{0\})$

share|improve this answer
    
+1. I guess the definition has to include monotonicity. –  Matt N. Dec 15 '12 at 14:21
    
As far as I understood the OP's question the definition of external measure does not include monoticity. Thus, the OP was looking for someone to prove monoticity so that the result that all sets of zero measure would follow easily. –  Amr Dec 15 '12 at 14:23
    
How did the book have such an exercise? I think that the book was asking for a proof or a counterexample. –  Amr Dec 15 '12 at 14:24
    
At the LHS you should have the set $E$ which you tooked to be $\{0\}$ –  Belgi Dec 15 '12 at 14:25
    
I have proved that the space is $\nu$ measurable without assuming that $\nu$ is monotone, I don't think this is a counterexample –  Belgi Dec 15 '12 at 14:27
show 4 more comments

But as far as I can tell you are done and your proof is correct:

You want to show that for $E \subseteq X$ it holds that $\nu(E) = \nu (E \cap A) + \nu (E \cap A^c)$.

As you correctly observed, since $\nu$ is monotone and $E \cap A \subseteq A$ you have that $\nu(E \cap A ) = 0$.

Hence the proof boils down to showing that $\nu(E) = \nu (E \cap A^c)$. Again by monotonicity of $\nu$ and the inclusion $E \cap A^c \subseteq E$ you have $\nu(E \cap A^c) \le \nu ( E )$.

Using $\nu(E)=\nu((E\cap A)\cup(E\cap A^{c}))\leq\nu(E\cap A)+\nu(E\cap A^{c}) = \nu(E \cap A^c)$ you get the missing inequality so that you have $\nu(E) = \nu ( E \cap A^c)$.

share|improve this answer
    
Though, reading your question a second time I am slightly confused: your definition of $\nu$ does not say that $A \subset B$ implies $\nu (A) \le \nu (B)$ and yet, you use it on the first line where you write "From containment: $\dots$". –  Matt N. Dec 15 '12 at 14:05
    
Thanks for the answer, but I didn't manage to prove that $\nu$ is monotone on sets that are not in $M$ –  Belgi Dec 15 '12 at 14:06
    
Regarding your comment - thats a mistake, so I need to prove that $\nu$ is monotone for that to or everything I wrote is wrong –  Belgi Dec 15 '12 at 14:07
1  
I think I miss-translated into English and that it is called an outer measure. But being monotone its not part of the definition in the question which seems like a mistake. I sent an email to the TA to check if this was a mistake –  Belgi Dec 15 '12 at 14:20
1  
The TA sais that he forogt about it and I should assume that $\nu$ is also monotone, thanks! –  Belgi Dec 15 '12 at 20:39
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.