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This is an example from Royden's Real Analysis 4th edition that I am having a lot of trouble understanding.

A countable set has outer measure zero. Let $C$ be a countable set enumerated as $C=\{c_k\}_{k=1}^{\infty}$. Let $\epsilon > 0$. For each natural number $k$, define $I_k=(c_k-\epsilon/2^{k+1}, c_k+\epsilon/2^{k+1})$. The countable collection of open intervals $\{I_k\}_{\infty}^{k=1}$ covers $C$. Therefore:

$$ 0 \leq m^*(C)\leq \sum^{\infty}_{k=1} \ell(I_k)=\sum^{\infty}_{k=1}\epsilon/2^k = \epsilon$$

This inequality holds for each $\epsilon > 0$. Hence $m^*(E) = 0$.

What I understand along with questions:

  • A countable set is a set like the natural numbers.
  • $C=\{c_k\}_{k=1}^{\infty}$ is just a way to describe said countable set using indexing for all of the elements in the set.
  • $I_k=(c_k-\epsilon/2^{k+1}, c_k+\epsilon/2^{k+1})$ I kind of understand. This is defining an open cover over the countable set (I think).
  • Why the $\epsilon/2^{k+1}$ instead of just $\epsilon$? It looks like the interval is continually shrinking as n goes to infinity.
  • The equation in the middle is summing up all of the intervals. The outer measure is simply the sum of all of the open intervals in the set C. $m^*(C)$ is countably subadditive, hence the $\leq$ sign here: $m^*(C)\leq \sum^{\infty}_{k=1} \ell(I_k)$
  • How is the last summation producing epsilon?
  • How is this summing to zero?

Thank you for your help.

-Idle

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  • $\begingroup$ Why $\epsilon /2^{k+1}$ rather than $\epsilon$? $\sum_{k=1}^\infty\epsilon$ diverges, which isn't very useful. $\endgroup$ Jan 17 '18 at 6:14
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6) $\displaystyle\sum_{k=1}^{\infty}\frac{\epsilon}{2^{k}}=\epsilon\sum_{k=1}^{\infty}\dfrac{1}{2^{k}}=\epsilon\cdot\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}=\epsilon$.

7) No, the sum is $\epsilon$, but now we have $0\leq m^{\ast}(C)<\epsilon$ for every $\epsilon>0$, if $m^{\ast}(C)>0$, then put $\epsilon$ to be $m^{\ast}(C)$, then we arrive to $m^{\ast}(C)<m^{\ast}(C)$, a contradiction.

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  • $\begingroup$ Thank you. Any advice for how to get better at this? I study hard but this is even more difficult, conceptually, than real analysis. $\endgroup$ Jan 17 '18 at 14:20
  • $\begingroup$ Maybe you can consult to a more elementary book to build up the foundation, I would suggest that, Theory of Measure and Integration, J Yeh, this book actually does not seem to be easier, but at least, the author writes out every detail of the argument, which leads the reader for easy reading. $\endgroup$
    – user284331
    Jan 17 '18 at 19:06
  • $\begingroup$ I should have said "classical real analysis" in my last comment. Of course, measure theory / lebesgue integration is real analysis just at a higher level. $\endgroup$ Jan 17 '18 at 22:20
  • $\begingroup$ Okay, then try A First Course in Mathematical Analysis, David Alexander Brannan, this book builds up the essential feeling of mathematical analysis at its best. $\endgroup$
    – user284331
    Jan 17 '18 at 22:24

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