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It seems obvious but I am having a hard time explaining to myself why that is. Considering that in general, a subset of a measurable set need not be measurable. For instance, the Vitali subset of $[0, 1]$ is not measurable. But it seems that there’s an exception when the subset in question is a subset of a set of measure zero.

Does it follow from the definition of measure or from a Theorem perhaps?

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This depends on exactly how you define “measure zero”. For subsets of the real line, I’d define “$X$ has measure zero” to mean that, for every $\varepsilon>0$, $X$ can be covered by open intervals of lengths adding up to at most $\varepsilon$. That makes subsets of measure-zero sets have measure zero trivially, because anything that covers $X$ also covers all its subsets.

If you want to limit yourself to the real line

$$\text{The set $\;X\;$ has measure zero}\;\implies\;\forall\;\epsilon>0\;\exists\;\text{intervals}\;\{I_n\}_{n\in\Bbb N}\;\;s.t.$$

$$ X\subset \bigcup_{m=1}^\infty I_n\;\;\text{and}\;\;\sum_{n=1}^\infty|I_n|<\epsilon$$

If we have a subset $\;Y\subset X\;$ then for any $\;\epsilon>0\;$ you can take *exactly* the same sequence of intervals as above…

A subset of a null set need not be measurable. If the measure is complete, then all subsets of null sets are also null sets.

The Lebesgue measure is complete. This follows, for example, from the construction using an outer measure since any covering of a set is also a covering of a subset.

If $N$ is a null set and $A \subset N$ is also measurable, then since the measure is additive, we have $\mu N = \mu A + \mu (N\setminus A)$. Since the measure is non-negative, we have $\mu N \ge \mu A \ge 0$, and since $\mu N = 0$, we have $\mu A = 0$.

If a measurable set $A$ is a subset of a measurable set $B$ then $B \setminus A$ is also measurable (by axioms of the $\sigma$-algebras) and since $A$ and $B\setminus A$ are disjoint, by additivity of measure $\mu(A) + \mu(B \setminus A) = \mu(B)$, so that $\mu(A) \leq \mu(B)$. In particular, when $B$ has measure $0$, so does $A$.

On the other hand, not every subset of a measurable set needs to be measurable. But for the subsets of measure zero there’s no harm in introducing another measure on an extension of the original $\sigma$-algebra, a completion of $\mu$ which has the property that every subset of a measure $0$ set is measurable (and so has measure $0$) while it agrees with the original measure on the sets belonging to the original $\sigma$-algebra.

It is a matter of personal preference whether one works with complete measures or not. It makes some proofs cleaner but at the cost of introducing many new sets into the $\sigma$-algebra. E.g. completion of a Borel $\sigma$-algebra is the Lebesque $\sigma$-algebra which has strictly greater cardinality (at least for real line — I am not sure in general).

It is not necessary that every subset of a measurable set of measure $0$ is itself measurable. If such a subset is measurable, it has measure $0$, of course. Furthermore, it is always possible to extend the $\sigma$-algebra of measurable subsets by adding all subsets of measure $0$ to it. That is, given a $\sigma$-algebra $\mathcal F$ and a measure $\mu$ on $\mathcal F$, form a new $\sigma$-algebra $\mathcal F’$ by $$\mathcal F’ = \{ X \cup Z \; \mid \; X\in {\mathcal F} \land \exists Y \in {\mathcal F} [ Z \subseteq Y \land \mu(Y) = 0 ] \}$$ and extend $\mu$ to this new $\sigma$-algebra.

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