Is the set $\{1,\frac12,\frac13,...\}$ of measure zero? Why?
Measure of $\{1,\frac12,\frac13,...\}$
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0What's the definition of the measure of a sequence? – 2012-12-06
3 Answers
Using Lebesgue measure in $\mathbb R$, any countable set is of measure zero.
Take any $\,\epsilon>0\,$ and take the open intervals
$I_n:=\left(\frac{1}{n}-\frac{\epsilon}{2^n}\,,\,\frac{1}{n}+\frac{\epsilon}{2^n}\right)\Longrightarrow \left\{\frac{1}{n}\right\}_{n\in\Bbb N}\subset \bigcup_{n\in\Bbb N}I_n\,\,,\,\;\;\sum_{n=1}^\infty|I_n|\leq\sum_{n=1}^\infty\frac{\epsilon}{2^{n-1}}<\epsilon$
Here is a reason that wouldn't work for an arbitrary countable set, but works in this case (or for other sets of terms of a convergent sequence).
For all $\varepsilon>0$,
$\left\{1,\frac12,\frac13,\frac14,\ldots\right\}\subset(-\varepsilon,\varepsilon)\cup\text{(a finite set)},$ which implies that the measure is at most
$2\varepsilon+\text{(the measure of a finite set)}.$
If you can show that the measure of an arbitrary finite set is $0$ (perhaps a good warmup for the countable case), then you can conclude that the measure must be $0$, because it is smaller than each positive number.