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If $X = [0,1]$ and we let $m$ be the Lebesgue measure on $[0,1]$ and $\nu$ be the counting measure on $[0,1]$, then are the following statements true?

  1. The only subset $S \subset [0,1]$ that is of measure zero relative to the counting measure is $\{\}$. Therefore the counting measure is complete on $[0,1]$.

  2. The Lebesgue Measure is not complete on $[0,1]$, for there exist pathological subsets of null sets in $[0,1]$ that are nevertheless not Lebesgue measurable.

  3. The counting measure is not $\sigma$-finite on $[0,1]$ since no countable number of finite subsets of $[0,1]$ could possibly cover $[0,1]$ (which has uncountably many points).

  4. The Lebesgue measure is $\sigma$-finite on $[0,1]$ since $m([0,1]) = 1$ implies $m$ is finite on $X$ and hence $m$ is trivially $\sigma$-finite on $X$ as well.

Are these all correct?

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    $\begingroup$ What are your thoughts on the topic? $\endgroup$
    – Asaf Karagila ♦
    Dec 20, 2012 at 21:45
  • 1
    $\begingroup$ Which do you think are correct? $\endgroup$
    – Alex Becker
    Dec 20, 2012 at 21:54
  • $\begingroup$ I think all of them are correct but wanted to check my understanding. $\endgroup$
    – user53843
    Dec 20, 2012 at 21:55
  • $\begingroup$ It's nonsense to speak of a measure being complete without declaring what sigma-algebra it's defined on. Lebesgue measure is in fact complete on the Lebesgue sigma-algebra, more or less by definition. $\endgroup$
    – Chris Eagle
    Dec 21, 2012 at 0:31

1 Answer 1

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The arguments given in 1, 3, and 4 are certainly correct.

However, if I were your teacher, I'd tell you that the argument in 2 is incomplete without giving a specific null set $X\subset[0,1]$ and non-Lebesgue-measurable subset $Y\subset X$. Why not try to find one?

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