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I was reading Terry Tao's second volume on Analysis and he makes the following remark on page 586. I was hoping that someone could explain why this is the case since he does not give a proof.

"The unit interval has outer measure 1, but the unit interval $\{(x,0) : 0 \leq x \leq 1\}$ in $R^2$ has two dimensional outer measure 0. . . . Note that the above remarks and countable additivity imply that the entire x-axis has two-dimensional outer measure 0, despite the fact that $R$ has infinite one-dimensional measure"

I am not sure if I am understanding the terminology correctly. I assume he is talking about the two dimensional outer measure of the unit interval and not the two dimensional outer measure of a unit box.

Given that outer measure is defined as follows for some set $B$ that is covered by an interval $[a,b]$

$$ m^*_n(B) := \prod^n_{i=1}(b_i - a_i) $$

So the one dimensional outer measure$(m^*_1(\cdot))$ of the unit interval would be:

$$ m^*_1([0,1]) = (1-0) = 1 $$

When he says the two dimensional outer measure of the unit interval, is he talking about the two dimensional measure in the sense of $\{(0,1), (0,0)\}$ meaning ?:

$$ m^*_2([0,1]) = (1-0)(0-0) = 0 $$

Let me know if I am looking at this incorrectly. Thanks.

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    $\begingroup$ Your definition of outer measure is very informal to the point of being incorrect. However, yes: it is sufficient to say that the unit interval is "covered" by a rectangle of height zero. $\endgroup$ – Omnomnomnom Nov 30 '15 at 20:53
  • $\begingroup$ Yep, sorry I should have included the more detailed definition as in the response by @anomaly below. $\endgroup$ – krishnab Nov 30 '15 at 23:51
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The definition of $m^*$ you give only allows for a single rectangle. With that, the disc of radius $r$ would have measure $4r^2$, for example. Instead, you want something like $m_n^*(E) = \inf \sum_i \lambda(B_i)$, where \begin{align*} \lambda([a_1, b_1] \times \cdots \times [a_n, b_n]) = \prod_i (b_i - a_i) \end{align*} and the infimum runs over finite sets of $n$-dimensional rectangles that cover $E$. (We can also consider products of open intervals, etc.)

With that, we can show that $m_2^*([0, 1]) = m_2^*([0, 1]\times \{0\})$ vanishes by considering the cover with a single rectangle $[0, 1] \times [-\epsilon, \epsilon]$ for arbitrarily small $\epsilon > 0$. Then $m_2^*([0, 1]) \leq 2\epsilon$ by definition; and since clearly $m_2^*([0, 1]) \geq 0$, we have $m_2^*([0, 1]) = 0$.

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  • $\begingroup$ Okay right. Yes, thanks for your more complete answer and for validating my original intuition. You are correct, I should have included the more detailed definition of outer measure. I did not mean to leave out the important detail. $\endgroup$ – krishnab Nov 30 '15 at 23:50

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