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Is the following true for outer measure?

$\forall j\in\mathbb{N},A_j\subset \mathbb{R}^n$.Then $$m^*\left(\bigcup_{j\in \mathbb{N}}A_j\right)=\lim_{N\to\infty}m^*\left(\bigcup_{j=0}^{N}A_j\right)$$

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I don't think this holds for outer measures, but on the other hand I don't have any counter examples in mind. If you look at the proof when we're dealing with a measure, it uses countable additivity, which does not hold for outer measures in general. – T. Eskin Apr 20 '12 at 8:27
@Thomas E. You will never find a counter example,because leo has found the proof in Zygmund & Wheeden Measure and Integral,theorem 3.27 :-) – Luqing Ye Apr 21 '12 at 4:24
That proof fails in the assumption that for each $k\in\mathbb{N}$ there exists a measurable cover $H_{k}$ of $E_{k}$, i.e. the first verse. This would require $m^{*}$ to be regular. In the Lebesgue case it holds since the Lebesgue outer measure is even Borel regular (by choosing a $G_{\delta}$ cover...). If you read the textbook, it is very explicit that this proof is for Lebesgue outer measure. – T. Eskin Apr 21 '12 at 20:46
I posted a counter-example now, see my reply. – T. Eskin Apr 21 '12 at 22:10

2 Answers 2

up vote 3 down vote accepted

It's false in general for outer measures. I decided to dig this counter-example because there seemed to be an accepted misunderstanding in this topic.

This was found from Paul R. Halmos, Measure Theory, 1974, page 53, exercise (2). He states that it is false for non-regular outer measures in exercise (4), and gives this type of construction as a hint from exercise (2).

We say that $A\subset \mathbb{R}$ has infinite condensation if there are uncountably many points of $A$ outside of every bounded interval. Denote $2^{\mathbb{R}}$ as the powerset of $\mathbb{R}$.

Let $m^{*}:2^{\mathbb{R}}\to [0,\infty]$ be such that:

  • $m^{*}(A)=0$ if $A$ is countable.

  • $m^{*}(A)=1$ if $A$ is uncountable but does not have infinite condensation.

  • $m^{*}(A)=\infty$ if $A$ has infinite condensation.

Then $m^{*}$ is an outer-measure and the only $m^{*}$-measurable sets are countable sets and sets with countable complements. You may check that for example $m^{*}([0,1])=1$ and that there exists no sets containing $[0,1]$ with countable complement and no infinite condensation. Hence $m^{*}$ is not regular.

Now let $A_{i}=[-i,i]$ for every $i\in \mathbb{N}$ whence $A_{1}\subset ... \subset A_{i}\subset A_{i+1}\subset ...\subset \mathbb{R}$. Each $A_{i}$ is uncountable and does not have infinite condensation so $m^{*}(A_{i})=1$ for each $i\in\mathbb{N}$. But $\mathbb{R}=\bigcup_{i=1}^{\infty}A_{i}$ and $m^{*}(\mathbb{R})=\infty$. Hence: \begin{align*}\lim_{i\to\infty}m^{*}(A_{i})=1\neq \infty =m^{*}(\mathbb{R})=m^{*}(\bigcup_{i=1}^{\infty}A_{i}) \end{align*} As required.

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My friend says that you are right.I need time to study your example.Thank you for your example! – Luqing Ye Apr 22 '12 at 4:17

Yes, it is true for Lebesgue outer measure. The proof is from Zygmund & Wheeden Measure and Integral:

Theorem 3.27 from Zygmund

(3.26) is just the desired result for measurable sets.

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Thanks a lot,leo! – Luqing Ye Apr 21 '12 at 4:16
This proof is ofcourse valid only for regular outer measures, such as Lebesgue outer measure. But does not hold in the general case for every outer measure $m^{*}$... – T. Eskin Apr 21 '12 at 20:47
@ThomasE. Thanks for point that out. Indeed, up to this point, the book just deal with the Lebesgue (outer) measure in $\mathbb{R}^n$ – leo Apr 22 '12 at 4:11

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