Prob. 23, Chap. 2, in Royden's REAL ANALYSIS: How is $m^{***}(A)=\sup\{m^*(F) | F \subset A, F \mbox{ closed} \}$ related to the outer measure?

Question

Here is Prob. 23, Chap. 2, in the book Real Analysis by H.L. Royden and P.M. Fitzpatrick, 4th edition:

For any set $A$, define $m^{***}(A) \in [0, \infty]$ by $$ m^{***}(A) = \sup \left\{ m^*(F) \, | \, F \subseteq A, F \mbox{ closed} \right\}. $$ How is this set function $m^{***}$ related to $m^*$?

My Attempt:

If $F$ is any (closed) set such that $F \subseteq A$, then by the monotonicity of the outer measure, we must have $m^*(F) \leq m^*(A)$, and thus $m^*(A)$ is an upper bound for the set $$ \left\{ m^*(F) \, | \, F \subseteq A, F \mbox{ closed} \right\}. \tag{1} $$ Therefore we must have $$ m^{***}(A) \leq m^*(A). \tag{2} $$

How to establish the reverse inequality?

Here our set can either be measurable or not, and also either $m^*(A) = \infty$ or $m^*(A) < \infty$.

How to show in each one of the above four cases that $$ m^*(A) \leq m^{***}(A)? $$

PS:

Here is my attempted proof of the reverse inequality of (2) above.

If $m^{***}(A) = \infty$, then we trivially have $m^*(A) \leq m^{***}(A)$. So let us assume that $m^{***}(A) < \infty$. Then, for any closed set $F$ such that $F \subseteq A$, we must have $m^*(F) \leq m^{***}(A) < \infty$. Thus we have $m^*(F) < \infty$ for any closed set $F$ contained in $A$.

If $A$ is measurable, then by Theorem 11 (iii), Chap. 2, in Royden, for any real number $\epsilon > 0$, there exists a closed set $F_\epsilon$ contained in $A$ for which $$ m^* \left( A \setminus F_\epsilon \right) < \epsilon. \tag{3} $$ But as $F_\epsilon$ is a closed set, so $F_\epsilon$ is measurable, and as $F_\epsilon \subseteq A$, so we also have $m^*\left( F_\epsilon \right) < \infty$. Therefore (3) together with the excision property yields $$ m^*(A) - m^* \left( F_\epsilon \right) = m^* \left( A \setminus F_\epsilon \right) < \epsilon, $$ which implies
$$ m^*(A) < m^* \left( F_\epsilon \right) + \epsilon \leq m^{***}(A) + \epsilon, $$ which in turn implies $$ m^*(A) \leq m^{***}(A) + \epsilon $$ for any real number $\epsilon > 0$, and therefore we can conclude that $$ m^*(A) \leq m^{***}(A). \tag{4} $$

If $A$ is not measurable, then by Prob. Prob. 17, Chap. 2, in Royden, there exists a real number $\epsilon_0 > 0$ such that, for any open set $O$ and for any closed set $F$ such that $F \subseteq A \subset O$, we have $$ m^*(O \setminus F ) \geq \epsilon_0. \tag{5} $$ But any closed set is measurable, and in particular any closed set $F$ contained in $A$ is measurable with finite outer measure; we therefore have $$ m^*(O \setminus F) = m^*(O) - m^*(F), $$ by the excision property, and the last identity together with (5) gives $$ m^*(O) - m*(F) \geq \epsilon_0, $$ and thus we have $$ m^*(O) \geq m^*(F) + \epsilon_0. \tag{5} $$

What next? How to proceed from here and prove that (4) above holds in the case when $A$ is not measurable?

Answer 1

$m^{***}$ is the called the Lebesgue inner measure, often denoted by $m_*$ Your proof of $m_* \le m^*$ looks good.

We want to show $m_*(A) = m^*(A)$ when $A$ is measurable. After that, we show $m_*(A) < m^*(A)$ when $A$ is not measurable and $m^*(A)$ is finite. It is possible that $m_*(A) = m^*(A) = +\infty$ but $A$ is not measurable. Can you construct an example using Vitali set in $[0, 1]$? To begin, we first prove the outer and inner regularity of $m^*$.

Outer regularity:

Let $A$ be measurable. Then for all $\varepsilon > 0$ we can find open $G$ with $A \subseteq G$ such that $m^*(G \setminus A) < \varepsilon$.

Proof:

Suppose $A$ has finite measure. Fix $\varepsilon > 0$. By the definition of $m^*$, we can take an open interval cover $(I_n)$ for $A$ such that $$ m^* \left( \bigcup_n I_n \right) - m^*(A) < \varepsilon $$ Now let $G = \bigcup_n I_n$ and apply the excision property to get the result. When $A$ has infinite measure, we make use of the $\sigma$-finiteness of $m^*$ by chopping $A$ into $A_n = A \cap [n, n + 1)$. Now each $A_n$ has finite measure and using the above we can find open $G_n$ with $A_n \subseteq G_n$ such that $$ m^*(G_n \setminus A_n) < \frac{\varepsilon}{2^{n+1}} $$ Let $G = \bigcup_n G_n$. We have $$ m^*(G \setminus A) \\ = m^*\left(\bigcup_n G_n \setminus \bigcup_n A_n\right) \\ = m^*\left(\bigcup_n (G_n \setminus A_n)\right) \\ \le \sum_n m^*(G_n \setminus A_n) \\ \le \sum_n \frac{\varepsilon}{2^{n+1}} \\ = \frac{\varepsilon}{2} \\ < \varepsilon $$

Inner regularity:

Let $A$ be measurable. Then for all $\varepsilon > 0$ we can find closed $F$ with $F \subseteq A$ such that $m^*(A \setminus F) < \varepsilon$.

Proof:

We use the proof in Approximation of finite Lebesgue measurable set by closed subset. Fix $\varepsilon > 0$. Using outer regularity, we can find open $G$ with $A^c \subseteq G$ such that $m^*(G \setminus A^c) < \varepsilon$. Let $F = G^c$. Then $F$ is closed with $F = G^c \subseteq A$ and $$ m^*(A \setminus F) \\ = m^*(A \setminus G^c) \\ = m^*(A \cap G) \\ = m^*(G \setminus A^c) \\ < \varepsilon $$

After proving the above lemmas, we next show $m_*(A) = m^*(A)$ when $A$ is measurable.

Fix $\varepsilon > 0$. By inner regularity, we can find closed $F$ with $F \subseteq A$ such that $m^*(A \setminus F) < \varepsilon$. When $m^*(A)$ is finite, $m^*(F)$ is also finite. By the excision property, we see that $m^*(A) - m^*(F) < \varepsilon$. Therefore, $$ m^*(A) < m^*(F) + \varepsilon \le m_*(A) + \varepsilon $$ As $\varepsilon$ can be arbitrarily small, we get $m^*(A) \le m_*(A)$. When $m^*(A)$ is infinite, $m^*(A) = m^*(A \setminus F) + m^*(F)$ implies $m^*(F)$ must also be infinite. Therefore, $m^*(A) = m_*(A)$.

Finally, We want to show ($A$ is not measurable and $m^*(A)$ is finite) $\implies m_*(A) < m^*(A)$. Note that this is logically equivalent to ($m_*(A) = m^*(A)$ and $m^*(A)$ is finite) $\implies$ $A$ is measurable.

We use the idea from Lebesgue measurable implies caratheodory measurable and adapt it to our case. Suppose $m_*(A) = m^*(A)$ with $m^*(A)$ being finite. Fix set $B$ and $\varepsilon > 0$. Then we can find closed $F$ with $F \subseteq A$ such that $m^*(A) - m^*(F) < \frac{\varepsilon}{4}$, and open $G$ with $A \subseteq G$ such that $m^*(G) - m^*(A) < \frac{\varepsilon}{4}$. Since $G \setminus F$ is measurable and $F \subseteq A$ must also have finite measure, by the excision property, $$ m^*(G \setminus F) \\ = m^*(G) - m^*(F) \\ = m^*(G) - m^*(A) + m^*(A) - m^*(F) \\ < \frac{\varepsilon}{4} + \frac{\varepsilon}{4} \\ = \frac{\varepsilon}{2} $$ This is known as regularity. Now $$ m^*(B) \\ \le m^*(B \cap A) + m^*(B \setminus A) \\ \le m^*(B \cap G) + m^*(B \setminus F) \\ = m^*(B \cap (A \cup (G \setminus F))) + m^*(B \setminus (A \setminus (G \setminus F))) \\ \le m^*(B \cap A) + m^*(B \cap (G \setminus F)) + m^*(B \setminus A) + m^*(B \cap (G \setminus F)) \\ = m^*(B \cap A) + m^*(B \setminus A) + 2 \cdot m^*(B \cap (G \setminus F)) \\ \le m^*(B \cap A) + m^*(B \setminus A) + 2 \cdot m^*(G \setminus F) \\ \le m^*(B \cap A) + m^*(B \setminus A) + \varepsilon $$ Since $\varepsilon$ can be arbitrarily small, $m^*(B) = m^*(B \cap A) + m^*(B \setminus A)$. Therefore, $A$ is measurable by definition.