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I had a question about the additivity property of the outer measure.

Can someone provide an example of a disjoint union of sets which doesn't have an outer measure equal to the sum of the outer measure of each set (in $\mathbb{R}^n$)? That is:

$m_*(E_1\bigcup E_2)\neq m_*(E_1) + m_*(E_2)$, where $E_1\bigcap E_2=\emptyset$ and $d(E_1,E_2)$ possibly equal to zero.

The theorem states this holds in general if $d(E_1,E_2)> 0$, but I can't find a counterexample where it doesn't if $d(E_1,E_2)=0$.

Thanks!

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Let $B$ be a Bernstein set in $\mathbb R$. Then both $$ E_1 = [0,1] \cap B \qquad \text{and}\qquad E_2 = [0,1]\setminus B $$ have inner measure zero and outer measure one. But they are disjoint with union $[0,1]$. Now $0 + 0 \ne 1$ so inner measure is not additive, and $1+1 \ne 1$ so outer measure is not additive.

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First, a clarification on terminology. The property that you are mentioning is what makes an outer measure a metric outer measure. Lebesgue measure is a metric outer measure, so we know that $m_*(E_1\bigcup E_2)= m_*(E_1) + m_*(E_2)$, where $E_1\bigcap E_2=\emptyset$ must be true if $d(E_1,E_2)>0$.


Now, to your question. An outer measure is additive on the $\sigma$-algebra of measurable sets $\mathcal{M}$. In fact, if $E_1, E_2 \in \mathcal{M}$, we have $$ m_*(E_1\cup E_2)=m_*((E_1\cup E_2)\cap E_1)+m_*((E_1\cup E_2)\cap E_1^c)= m_*(E_1) + m_*(E_2), $$ using the definition of measurable set. So we need to look for a counterexample among non-measurable sets. Note that Solovay proved in 1970 that the existence of a non-measurable set is possible only if we assume the Axiom of Choice, so there won't be a trivial answer to your question.


The most famous construction of a non-measurable set is the Vitali set. As in this answer by JDH, we can construct a Vitali set $V$ in such a way that is contained in $[0,a]$, for any $a \in (0,1)$. Also, since the inner measure of $V$ is zero, we have that $m_*([0,a]\setminus V)=a$ and then $[0,a]\setminus V$ is a non-measurable set of measure $a$. Note that we must have $m_*(V)>0$, because a set is measurable if and only if its outer and inner measures coincide, and we know the latter to be zero. Then we have, by setting $E_1=V$ and $E_2= [0,a]\setminus V$, $$ a=m_*([0,a])=m_*(E_1 \cup E_2) < m_*(E_1)+m_*(E_2)=a + m_*(V). $$

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Let Continuum Hypothesis $2^{\omega}=\omega_1$ is valid. Then we have the following representation $[0,1]=(x_{\xi})_{\xi<\omega_1}$. We put $A=\{ (x_{\eta}, x_{\xi}): \eta <\xi <\omega_1\}$ and $B=\{ (x_{\eta}, x_{\xi}):\omega_1> \eta \ge \xi\}$. It is obvious that $A\cup B=[0,1]\times [0,1]$ and $A\cap B=\emptyset$. Then we get $1=m^*(A \cup B)\neq m^*(A)+m^*(B)=1+1=2$, where $m^*$ denotes an outher measure produced by Lebesgue measure $m$ in $R^2$. If we denote by $m_*$ an inner measure produced by $m$, then we get $1=m_*(A\cup B)\neq m_*(A)+m_*(B)=0+0=0$.

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  • $\begingroup$ Hint: For proof use Fubini theorem. $\endgroup$ – Gogi Pantsulaia Feb 27 '15 at 12:51

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