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I am struggling with proving that the following measure is not necessary complete. I just can't find the way to grab it properly, missing an initial idea. Suppose we got a set function denoted by $$\mu(A)=\int_A f(x) \; d\lambda(x), $$ where $\lambda$ is a Lebesgue measure on $\mathcal{A}$, $f$ is a nonnegative measurable function and $A\in\mathcal{A}$. I have already proven that $\mu$ is measure on $\mathcal{A}$, but I need to make a comment about completeness. My teacher said that it's not necessary complete. How to grab it?

When I think of negligible set of Lebesgue measure (which is complete), I think of countable sets. But Lebesgue integrals over countable set (set of singletons) will always be zero, which should make measure $\mu$ complete. Another thing could be that $f\equiv 0$ on some interval $I$, that means $\mu(I)=0$, but subset of that interval will also have measure zero.

MatEZ
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  • what is $a$ Lebesgue measure? – blamethelag Apr 08 '22 at 13:21
  • There are uncountable sets of Lebesgue measure 0. – eyeballfrog Apr 08 '22 at 13:21
  • @eyeballfrog: Yes, for example Cantor set, I forgot about those sets. – MatEZ Apr 08 '22 at 13:22
  • @blamethelag completion of interval-like measure from borel sets to lebesgue-measurable sets? – MatEZ Apr 08 '22 at 13:24
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    So $\lambda$ is $the$ Lebesgue measure on $\mathbb R$ and $\mathcal A = \mathcal L(\mathbb R)$ is $the$ Lebesgue sigma algebra. If you take $f = 0$ your measure is certainly not complete. You can generalize it with $f$ null on a set of positive measure. – blamethelag Apr 08 '22 at 13:26
  • OK, so suppose I have a set $A\in\mathcal{L}(\mathbb{R})$ for which $\lambda(A)>0$. If $f=0 ; \forall x\in A$, that means $\mu(A)=0$. But how do I find subset of $A$ which is $\mu$-non-measurable? Anyway, thanks for the answer – MatEZ Apr 08 '22 at 13:37
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    Try to use the fact that any set with positive measure contains a non Lebesgue measurable set – blamethelag Apr 08 '22 at 14:19
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    See here https://math.stackexchange.com/questions/1287212/non-measurable-subset-of-a-positive-measure-set for a proof of the property mentioned by @blamethelag. – PhoemueX Apr 09 '22 at 07:41

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