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Let $(X,\mathfrak M,\mu)$ be a measure space (Here $X$ is the set, $\mathfrak M$ is the $\sigma$-algebra, and $\mu$ is the measure). Let $N\in\mathfrak M$ with $\mu(N)=0$. It may happen that $N$ may have a subset $E$ which is not a member of $\mathfrak M$.

I am having a hard time grasping this concept. I would appreciate it if anyone could give me some examples to illustrate this.

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  • $\begingroup$ you can define a trivial measure with $\mu (E) = 0 $ for any $E$ in the $\sigma$ algebra $\endgroup$ – k76u4vkweek547v7 Sep 28 '14 at 6:23
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Every measure can be completed, so that subsets of measure zero sets are measurable. This suggests that the problem is with $\mathfrak{M}$, not $\mu$. You can then build tons of examples: Consider Dirac's delta in $\mathbb{R}$ (so that $\mu(A)=1$ if $0\in A$ and $\mu(A)=0$ otherwise) on the sigma-algebra $\mathfrak{M}=\{ \mathbb{R}, \emptyset, \{0\}, \mathbb{R}\setminus\{0\}\}$.

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Let $\mu$ be a Borel classical measure on $[0,1]$(i.e., the restriction of the linear Lebesgue measure on $B[0,1]$, where $B[0,1]$ denotes Borel $\sigma$-algebra of subsets of $[0,1]$). Let $C$ be a Cantor set. It is clear that $\mu(C)=0$ and $card(C)=2^{\aleph_0}$, where $\aleph_0$ denotes the cardinality of all natural numbers. Notice also that $card(B[0,1])=2^{\aleph_0}$.

Now we claim that there is a subset of $C$ which does not belong to $\mbox{dom}(\mu)=B[0,1]$.

Indeed, if assume the contrary then $\{Y:Y \subseteq C\} \subset B[0,1]$. Since the cardinality of the class $\{Y:Y \subseteq C\}$ is equal to $2^{c}$(here $c=2^{\aleph_0}$), we claim that the cardinality of $B[0,1]$ is bigger or equal to $2^c$, which is a contradiction because $card(B[0,1])=2^{\aleph_0}=c$.

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