Find the Lebesgue measure of the following sets:

i) $A=\left (\cup_{n=1}^\infty [2^n, 2^n + \frac{1}{2^n}) \right)$ \ $\mathbb{Z}$

ii) $B=\left(\cup_{n=1}^\infty (n^n, n^n + \frac{1}{2^n})\right)$ $\cap$ $\mathbb{Q}$.

For i), I believe that $\mu(\cup_{n=1}^\infty [2^n, 2^n + \frac{1}{2^n}))$=$\sum_{n=1}^\infty \mu[2^n, 2^n + \frac{1}{2^n}))$. $\mu[2^n, 2^n + \frac{1}{2^n})=(2^n + \frac{1}{2^n}-2^n)=\frac{1}{2^n}$.

Therefore $\sum_{n=1}^\infty \mu[2^n, 2^n + \frac{1}{2^n}))$=$\sum_{n=1}^\infty \frac{1}{2^n} = 1$.

For ii), do I need to transform the closed interval into a half open one? Can I say that: $(n^n, n^n + \frac{1}{2^n}) \subset [n^n-\epsilon, n^n + \frac{1}{2^n})$ and then use this to find the Lebesgue measure.

I can't seem to wrap my head around this and feel like I'm heading in the complete wrong direction. Any help would be greatly appreciated! Thank you.

  • 2
    $\begingroup$ $B$ is a subset of $\mathbb{Q}$. What is the measure of $\mathbb{Q}$? $\endgroup$
    – user296602
    May 16, 2016 at 3:14
  • $\begingroup$ The measure of $\mathbb{Q}$ =0. So am I right in saying that since B is a subset of $\mathbb{Q}$ we are able to say that the measure of B is also zero? $\endgroup$
    – locorolo
    May 16, 2016 at 3:24

1 Answer 1


For i), your belief can be made rigorous in the following way. First, if $N$ is a set whose Lebesgue measure is equal to $0$, then $\lambda\left(A\setminus N\right)=\lambda\left(A\right)$. Second, we can argue that the collection $\left(\left[2^n,2^n+2^{-n}\right)\right)_{n\geqslant 1}$ is a collection of pairwise disjoint sets and use $\sigma$-additivity.

For ii), if the Lebesgue measure was defined for half open interval, the we can show that singletons have measure $0$, as $\left\{x\right\}\subset \left[x,x+1/n\right)$ for any positive integer $n$. So the fact that the intervals are not half-open is not a problem. However, as pointed out by user296602, $B$ is a subset of $\mathbb Q$ which have Lebesgue measure $0$.


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