Lebesgue measurable homework problem Let $X \subseteq \mathbb{R}$. A subset $E \subseteq \mathbb{R}$ is called a hull of $X$ if


*

*$E$ is measurable

*$X \subseteq E$

*If $F$ is any measurable set such that $X \subseteq F$, then $E$\ $F$ is a zero set
A hull of $X$ should be thought of as a sort of "smallest" measurable set containing $X$.
This is made precise in the problem.
(a) Prove that every subset of $\mathbb{R}$ has a hull
(b) Prove that if $E_{1}$ and $E_{2}$ are both hulls of $X$, then $E_{1}$\ $E_{2}$ and $E_{2}$\ $E_{1}$ are zero sets. Thus the hull of $X$ is unique up to zero sets 
(c) Let E be a hull of X. Prove that $m(E)=m^{*}(X)$
The hint says to consider the case $m^{*}(X)< \infty$, then consider the case let $R_i=(i,i+1)$ for $i \in \mathbb{Z}$. Let $X_i=X \cap R_i$. Then $X=\cup_{i \in \mathbb{Z}} X_{i} \cup \mathbb{Z}$. I need help to prove this.
 A: a): Hulls occur very often in differents fields of mathematics. Usually, you have some set $M$ which doesn't fullfill a particular property (measureable, in this case), and you're looking for the *smallest (in some sense) set which includes $M$ and does have the desired property. For example, in a vector space, a natural definition for the hull of a set of vectors $M$ is the smallest subspace with contains $M$. Now, if the propertery in question (being a subspace, in the case of vector spaces) is preserved under intersections, then this is easy - just define $$
  \textrm{hull } M := \bigcap_{X \supset M, \textrm{$X$ has desired property}} X \text{.}
$$
The problem in your case is that it's not true that the intersection of arbitrary measurable sets is measureable. But the intersection of arbitrarily many closed sets is closed, and every closed set is measurable. So a natural idea for finding the measurable hull of an arbitrary set $X \subset \mathbb{R}$ is to set $$
  \textrm{hull } X := \bigcap_{X \subset T \subset \mathbb{R}, \textrm{$T$ closed}} T \text{.}
$$
It's clear that $\textrm{hull } X$ is closed, and hence measurable. What remains to show is that if $F \supset X$ is measurable, $\mu(\textrm{hull } X \setminus F) = 0$. For that, you'll need to use that the closed sets in some way (which, exactly?) generate all the measurable sets.
c): If you can show that for the special hull constructed in (a) that $m^*(\textrm{hull } X) = m^*(X)$, then the same must hold for any hull $E$ - just invoke (b).
A: Let $X\subset \mathbb{R}$ with finite measure, then there exist open sets $A_n\supset X$ such that
$$m^*(A_n) - m^*(X) \leq \frac{1}{n}.$$ 
Define $A = \cap A_n$, then $A$ is measurable and $A\supset X$. To show (c)
$$m^*(A) - m^*(X) \leq m^*(A_n) - m^*(X) \leq \frac{1}{n}$$
for each $n$, we have that $m^*(A) = m^* (X)$. 
Let $F\supset X$ be measurable, then
$$m^*(A) = m^*(A\cap F) + m^*(A\cap F^c) \geq m^*(X) + m^*(A\cap F^c)$$
which gives
$$0 \geq m^*(A\cap F^c).$$
For part (b), use the same argument as above, take $F$ to be another Hull of $X$, all the inequality would still work.
For  $X\subset\mathbb{R}$ with $m^*(X) = \infty$, define $X_i = X\cap([-i, -i+1)\cup [i-1, i))$, then $X = \cup_{i=1}^\infty X_i.$ By previous arguement, there exists $A_i$ for each $X_i$, with all $A_i$ disjoint. Define $A = \cup_i A_i$, we see that
$$m^*(A) = \sum_i m^*(A_i) = \sum_i m^*(X_i) \geq m^*(X) = \infty.$$
A: I know this is an old question, but I keep waking up in the middle of the night with the sense that I've only dreamed the existence of measurable hulls and eventually find my way to this thread, so I would like to provide a more general answer so that I can finally get some sleep.
Let $(X,\Sigma,\mu)$ be a finite measure space, i.e. $\mu(X)<\infty$ (please note that I have made $X$ the space here, not the set we want the hull of, because $X$ is the typical symbol for a general measure space; I hope this doesn't create any confusion). Let $A\subset X$ be any subset, measurable or non-measurable. We want to find a hull of $A$, i.e. a measurable set $E$ containing $A$ with property 3 in the question.
Since $\mu(X)<\infty$,
$$
M:= \inf_{S \in \Sigma,\ A\subset S} \mu(S) < \infty.
$$
Then we can find measurable set $S_1, S_2,\dots$ containing $A$ such that $\mu(S_n)$ decreases to $M$. Let
$$
E = \bigcap_n S_n.
$$
Then $E$ is measurable, $A\subset E$, and since this measure space is finite, we can see that $\mu(E)$ is equal to the infimum. Lastly, suppose $F\supset A$ is also measurable. Then $E\cap F = E \setminus (E\setminus F)$ is also a measurable set containing $A,$ and is also a subset of $E$. Thus $M \leq \mu(E\cap F) \leq \mu(E) = M,$ so it follows that $\mu(E\setminus F) = 0$. Therefore, $E$ satisfies all the desired properties of a measurable hull.
This can be extended to $\sigma$-finite spaces, which are measure spaces $(X,\Sigma,\mu)$ in which $X$ written as a countable union of sets $Y_n$ of finite measure, even if $\mu(X)=\infty$. To find a hull in this case, we simply intersect $A$ with each $Y_n$ $X$, and a hull for the intersection in each $Y_n$, and then take the union. To check that property 3 is satisfied, remember that a countable union of sets of measure $0$ has measure $0$.
$\mathbb{R}$ with the Lebesgue measure is $\sigma$-finite (think along the lines of the hint in the problem), so this concludes the proof. Notice, though, that I never mentioned the Lebesgue outer measure. This might be disconcerting, because with a measure constructed from an outer measure, it seems just as good to define your hull to be a measurable set $E$ containing $A$ such that $\mu^*(A) = \mu(E)$, and with the properties above (in this case, property 3 is needed only to deal with sets of infinite outer measure). However, consider the way in which the hull I found and the outer measure are both defined using infima. This should show you how to prove that the definitions in this case are equivalent.
