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?
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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 $[0,1]$.
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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.
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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).
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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?
Best Answer
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?