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This question is related to What's the quickest way to see that the subset of a set of measure zero has measure zero? But I'm specifically concerned about Lebesgue measure, $m$, on a real interval, $X=[a,b]$, and specifically about why a subset, $A$, of zero measure set, $E$, is measurable? Is it by the construction of the Lebesgue measure?

I can understand that the outer measure of $A$ is zero, i.e. $m^*(A)=0$, since any open cover of $E$ also covers $A$. But to claim that $A$ is measurable, I think (by definition) we need to show that $A$ is a countable union of finitely $m$-measurable sets. It's not clear to me how this may be done? I'd appreciate some help.

BTW, according to Rudin's Principles of Mathematical Analysis, a set $B$ is finitely $m$-measurable, if there is a sequence $\{B_n\}$ of elementary sets such that $B_n \to B$.

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    $\begingroup$ A measure space is said to be complete if every subset of a zero set is measurable. See, e.g. math.stackexchange.com/questions/791021/…, for some hints about showing the Lebesgue measure is complete $\endgroup$ – eepperly16 May 16 '17 at 5:40
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    $\begingroup$ Lebesgue measure is defined as the completion of another measure (at least in Folland, not sure if there are alternate definitions). So the subset of a measure zero set is measurable by definition. $\endgroup$ – mathematician May 16 '17 at 5:46
  • $\begingroup$ @syeh I do not know which definitions you are using, I am writing an answer assuming a particular definition of measurability used in Rudin. $\endgroup$ – Juanito May 16 '17 at 6:03
  • $\begingroup$ @Juanito Thanks a lot! I'm reading Rudin's Principles of Mathematical Analysis. He defines measurability on page 305, which is also listed at the end of the question. $\endgroup$ – syeh_106 May 16 '17 at 6:08
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Suppose $\lambda(E)=0$, and that $A\subseteq E$. Then as you said, $\lambda^*(A) = E$. In order for $A$ to be measurable, it must be the case that for each $B\subseteq [a,b]$, we have that $$\lambda^*(A) = \lambda^*(A\cap B)+\lambda^*(A\cap B')$$ where $B' = [a,b]\backslash B$. Indeed, since $A\cap B, A\cap B' \subseteq E$, both of these sets have outer measure zero, and so the equality is satisfied. Thus, $A$ is measurable.

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  • $\begingroup$ Thanks a lot, @florence! This proof is quite simple. The only thing that I'm not certain about is whether this definition of measurability is equivalent to Rudin's, as given at the end of the question in the construction of Lebesgue measure. $\endgroup$ – syeh_106 May 16 '17 at 5:59
  • $\begingroup$ When you say $m$-measurable, does that refer to Jordan measure? $\endgroup$ – florence May 16 '17 at 6:06
  • $\begingroup$ I'm not familiar with Jordan measure. According to Rudin's book (PMA), $m([x,y])=y-x$, for $x\le y$. $\endgroup$ – syeh_106 May 16 '17 at 6:16
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I will use the following well known result: If $ N \subseteq \mathbb{R}^n$ and $m^*(N)=0$ then $N$ is Lebesgue measurable.

Also, remember that the Lebesgue measure $m$ is the restriction of $m^*$ to the Lebesgue measurable set $\mathscr{L}(\mathbb{R}^n)$: $m=m^*\big|_{\mathscr{L}(\mathbb{R}^n)}$. It means $m(M)=m^*(M),\forall M \in \mathscr{L}(\mathbb{R}^n)$.

Now, turning to the problem, suppose $M \in \mathscr{L}(\mathbb{R}^n):m(M)=0$ and let $E\subseteq M $. Then $0 \leq m^*(E)\leq m^*(M)$ by monotonicity of outer measures. Since $m=m^*\big|_{\mathscr{L}(\mathbb{R}^n)}$, we have that $0=m(M)=m^*(M)$. Therefore, using the result of the first line above, we conclude that $E \in \mathscr{L}(\mathbb{R}^n)$. It means that the set $E$ is Lebesgue measurable.

So, this is an easy way to see the completeness of $(\mathbb{R}^n,\mathscr{L}(\mathbb{R}^n),m)$ if you know those basic concepts of the Lebesgue measure.

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Let us use the Caratheodory criterion.

If E is measurable, then for any B, $\mu(B)=\mu(B\cap E)+\mu(B\cap E^c)$

Now, as E is measure zero $\mu(B\cap A)\leq\mu(B\cap E)\leq \mu( E)=0$, and as measure is non-negative all measures must be zero.

Therefore, $\mu(B)=\mu(B\cap A)+\mu(B\cap E^c)\leq \mu(B\cap A)+\mu(B\cap A^c)$

But outer measure is sub additive, hence

$\mu(B)=\mu(B\cap A)+\mu(B\cap A^c)$

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