2
$\begingroup$

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 ?

  • 0
    How about this counterexample (I hope I am not wrong this time)2012-12-15
  • 0
    @Amr - but whats wrong with $\{0\}$ being not $\nu$-measurable ?2012-12-15
  • 0
    $\{0\}$ has zero measure2012-12-15
  • 0
    @Amr - the empty set need to be $\nu$ measurable, but there is no need for all the sets to be $\nu$ measurable2012-12-15
  • 0
    I didnt get your last comment2012-12-15
  • 0
    You wanted to show that all sets of zero measure are measurable2012-12-15

2 Answers 2

2

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.

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

  • 0
    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.2012-12-15
  • 0
    How did the book have such an exercise? I think that the book was asking for a proof or a counterexample.2012-12-15
  • 0
    At the LHS you should have the set $E$ which you tooked to be $\{0\}$2012-12-15
  • 0
    I have proved that the space is $\nu$ measurable without assuming that $\nu$ is monotone, I don't think this is a counterexample2012-12-15
  • 0
    Yes . You are right.2012-12-15
  • 0
    Yes, I think this example is correct2012-12-15
  • 0
    So Why did the exercise ask for a proof ?2012-12-15
  • 0
    @Amr TA ballsed up.2012-12-15
1

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 $\nu(E) = \nu ( E \cap A^c)$.

  • 0
    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: $\dots$".2012-12-15
  • 0
    Thanks for the answer, but I didn't manage to prove that $\nu$ is monotone on sets that are not in $M$2012-12-15
  • 0
    Regarding your comment - thats a mistake, so I need to prove that $\nu$ is monotone for that to or everything I wrote is wrong2012-12-15
  • 0
    @Belgi Ok. But are you sure that what you have there is not an [outer measure](http://en.wikipedia.org/wiki/Outer_measure)? Where does your definition come from?2012-12-15
  • 0
    Those are the definitions given in the beginning of the exercise. I may have miss translated and wrote "external" instead of "outer"2012-12-15
  • 0
    @Belgi If it is an outer measure it is also monotone and your proof is correct. If on the other hand it is not, I will delete my answer.2012-12-15
  • 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 mistake2012-12-15
  • 1
    The TA sais that he forogt about it and I should assume that $\nu$ is also monotone, thanks!2012-12-15
  • 0
    @Belgi I'm glad to hear. Well done spotting it!2012-12-15
  • 0
    All with your help :)2012-12-15