4
$\begingroup$

I understand that Lebesgue outer measure on $\mathbb R$ is not countably additive. But if there are two disjoint sets, does the outer measure of their union equal the sum of their outer measure? Can someone give me a counterexample?

$\endgroup$
3
$\begingroup$

Use transfinite induction to construct $2$ (or $2^{\aleph_0}$) disjoint subsets $A_i$ of the unit interval $[0,1]$, such that each $A_i$ has nonempty intersection with every uncountable closed subset of $[0,1]$. (These are called Bernstein sets.) Then each $A_i$ has outer measure $1$, as does the union of all the $A_i$.

$\endgroup$
0
$\begingroup$

Take $\Omega = \{a, b\}$. Define $$ \mu^*(A) = \left\{ \begin{array}{ll} 0, & A = \emptyset \\ 1, & A \neq \emptyset \end{array} \right. $$ This is an outer measure.

In this case, $$ \mu^*(\{a,b\}) = 1 \neq 2 = \mu^*(\{a\}) + \mu^*(\{b\}). $$

$\endgroup$
  • $\begingroup$ Maybe I wasn't specific enough. I'm only considering Lebesgue outer measure regarding subsets of real numbers. Do you have counterexamples on the real line? $\endgroup$ – Badoe Sep 30 '13 at 3:08
  • $\begingroup$ @Badoe: Maybe you can try something like this... Consider the Lebesgue outer measure in $[0,1]$. Take a non measurable set $A$. Since it is non measurable, it's inner measure must not be equal its outer measure. But maybe it means that $\lambda^*(A) + \lambda^*(A^c) \neq 1$. I am not sure... $\endgroup$ – André Caldas Sep 30 '13 at 3:22
  • $\begingroup$ I guess you have to be more specific about what you admit as sets. There's a lot of riff-raff in $\mathcal{P}(\mathbb{R})$ that's really not worth considering, whereas if you admit docile (measurable) sets $E$ which satisfy $m_*(A) = m_*(A \cap E) + m_*(A \cap E^c)$ you get disjoint additivity immediately. Your counter-example will require a choice, Mr. Anderson. $\endgroup$ – snar Sep 30 '13 at 4:16
  • $\begingroup$ @snarski: I don't think I have to be that specific about what is a set when writing a comment!!! What is the problem with a "choice"? We make choices all the time. You, for instance, have made the particular choice of making a comment to my comment! :-P $\endgroup$ – André Caldas Sep 30 '13 at 23:48
  • $\begingroup$ @AndréCaldas Sorry, I was trying to be funny. What I mean is this: very roughly, you have two types of $E \subset \mathbb{R}$: measurable w.r.t. "Lebesgue" outer measure and not measurable. Thus, when you say "two disjoint sets" it is either trivial or impossible that $m_*(A \cup B) = m_*(A) + m_*(B)$. The pun about "choice" is that I think you need (something equivalent to) the axiom of choice to construct these sets. Maybe you know all this? Does this address your question at all? $\endgroup$ – snar Oct 1 '13 at 2:35
-1
$\begingroup$

Countably additive holds only If the sets $\{ E_n \} $ are measurable and pairwise disjoint: That is. If $\{ E_n \}_{n\geq 1} $ are measurable, then

$$ \mu^{*} ( \bigcup E_n ) = \sum \mu^*(E_n) $$

$\endgroup$
  • $\begingroup$ I am not asking for countably additivity. Does the equality hold for just 2 disjoint sets (whether measurable or not)? If not, what is a possible counterexample? Thank you. $\endgroup$ – Badoe Sep 30 '13 at 2:45
  • $\begingroup$ THe answer is NO. But I cannot think of a counter example right now $\endgroup$ – ILoveMath Sep 30 '13 at 2:45
-1
$\begingroup$

For a non-measurable set, the outer measure will necessarily be positive. Therefore it suffices to consider a non-Lebesgue measurable subset $A$ of the circle, and let $B$ be its complement in the circle. Then additivity must fail for this pair of sets for the following example.

Consider a set $A$ of representatives for equivalence classes under the action of rational rotations on the circle (the usual example of a non-measurable set). Recall that the circle is a countable union of the sets $A_\alpha$ obtained by rotating $A$ by a rational angles $\alpha$.

Let $\epsilon=\mu(A)>0$ where $\mu$ is the outer measure. Let $N=1+\lfloor\frac{1}{\epsilon}\rfloor$. Now take $N$ rational translates of $A$, namely $A_1,A_2,\ldots,A_N$. These are all disjoint by construction. If $\mu$ satisifed finite additivity, then the total outer measure would be greater than $1$: $$ \mu(A_1\cup\cdots\cup A_N)=N\cdot\epsilon>1, $$ yielding the desired contradiction. This shows that the outer measure is not even finitely additive.

$\endgroup$
  • $\begingroup$ This makes no sense. If the set is not measurable, it has NO Lebesgue measure. Therefore, it cannot have an outer measure greater then something that does not exist. $\endgroup$ – André Caldas Oct 2 '13 at 15:46
  • $\begingroup$ Thanks, this was written in too much of a hurry. I clarified what I had in mind above. $\endgroup$ – Mikhail Katz Oct 3 '13 at 8:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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