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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 ?

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    You wanted to show that all sets of zero measure are measurable2012-12-15

2 Answers 2

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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.

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

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    @Amr TA ballsed up.2012-12-15
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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 $\nu(E) = \nu ( E \cap A^c)$.

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    All with your help :)2012-12-15