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Background Information:

In the theorem here we fix a complete Lebesgue-Stiltjes measure $\mu$ on $\mathbb{R}$ associated to the increasing right continuous function $F$, and we denote by $M_{\mu}$ the domain of $\mu$. Thus, for any $E\in M_{\mu}$ $$\mu(E) = \inf\{\sum_{1}^{\infty}[F(b_j) - F(a_j)]:E\subset \bigcup_{1}^{\infty}(a_j,b_j]\} = \inf\{\sum_{1}^{\infty}\mu((a_j,b_j]):E\subset \bigcup_{1}^{\infty}(a_j,b_j]\}$$

Theorem 1.18 - If $E\in M_{\mu}$, then \begin{align*} \mu(E) &= \inf\{\mu(U): E\subset U, U \ \text{open}\}\\ &= \sup\{\mu(K):K\subset E, K \ \text{compact}\} \end{align*}

Theorem 1.19 - If $E\subset \mathbb{R}$, the following are equivalent

a.) $E\in M_{\mu}$

b.) $E = V\setminus N_1$ where $V$ is a $G_{\delta}$ set and $\mu(N_1) = 0$

c.) $E = H\cup N_2$ where $H$ is a $F_{\sigma}$ set and $\mu(N_2) = 0$

Lebesgue measure $m^n$ on $\mathbb{R}^n$ is the completion of the $n$-fold product of Lebesgue measure on $\mathbb{R}$ with itself, that is, the completion of $m\times \ldots \times m$ on $B_{\mathbb{R}}\otimes \ldots \otimes B_{\mathbb{R}} = B_{\mathbb{R}^n}$, or equivalently the completion of $m\times \ldots \times m$ on $L\otimes \ldots \otimes L$.

The domain $L^n$ of $m^n$ is the class of Lebesgue measurable sets in $\mathbb{R}^n$.

In what follows, if $E = \prod_{1}^{n}E_j$ is a rectangle in $\mathbb{R}^n$, we shall refer to the sets $E_j\subset \mathbb{R}$ as the sides of $E$.

Question:

Suppose $E\in L^n$

a.) $m(E) = \inf\{m(U):E\subset U, U \ \text{open}\} = \sup\{m(K):K\subset E, K \ \text{compact}\}$.

b.) $E = A_1\cup N_1 = A_2\setminus N_2$ where $A_1$ is a $F_{\sigma}$ set, $A_2$ is a $G_{\delta}$ set, and $m(N_1) = m(N_2) = 0$.

c.) If $m(E) < \infty$, for any $\epsilon > 0$ there is a finite collection $\{R_j\}_{1}^{n}$ of disjoint rectangles whose sides are intervals such that $m(E \ \triangle \ \bigcup_{1}^{n}R_j) < \epsilon$.

Proof a.) - Let $E\in L^n$ and $\epsilon > 0$ then by definition of product measures there exists a countable family $\{T_j\}$ of rectangles such that $E\subset \bigcup_{1}^{\infty}T_j$ and $\sum_{1}^{\infty}m(T_j) \leq m(E) + \epsilon$. For each $j$, applying theorem 1.18 to the sides of $R_j$ we can find a rectangle $T_j\subset U_j$ whose sides are open sets such that $m(U_j) < m(T_j) + \epsilon 2^{-j}$. Let $U = \bigcup_{1}^{\infty}$, then $U$ is open and $$m(U) \leq \sum_{1}^{\infty}m(U_j) \leq m(E) + 2\epsilon$$ thus we can define $m(E) = \inf\{m(U):E\subset U, U \ \text{open}\}$.

I am not sure how to show the second part.

Proof b.) Not sure

Proof c.) Let $m(E) < \infty$ and $\epsilon >0$. So we know that $m(U_j) < \infty$ for all $j$. Now since the sides of $U_j$ are countable unions of open intrvals, take a sufficient number of subunions to obtain rectangles $V_j\subset U_j$ whose sides are finite unions of intervals such that $m(U_j) \leq m(V_j) + \epsilon 2^{-j}$. Take $N$ to be sufficiently large,then $$m(E\setminus \bigcup_{1}^{N}V_j) \leq m(\bigcup_{1}^{N}U_j\setminus V_j) + m(\bigcup_{N+1}^{\infty}U_j) < 2\epsilon$$ and $$m(\bigcup_{1}^{N}V_j\setminus E) \leq m(\bigcup_{1}^{\infty}U_j\setminus E) < \epsilon$$ thus we have $$m(E\triangle \bigcup_{1}^{N}U_j) < 3\epsilon$$

I am not sure if more details needed to be added to make this more clear. Any suggestions is grealy appreciated.

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Suppose $E\in \mathcal{L}^n$

a.) $m(E) = \inf\{m(U):E\subset U, U \ \text{open}\} = \sup\{m(K):K\subset E, K \ \text{compact}\}$.

b.) $E = A_1\cup N_1 = A_2\setminus N_2$ where $A_1$ is a $F_{\sigma}$ set, $A_2$ is a $G_{\delta}$ set, and $m(N_1) = m(N_2) = 0$.

c.) If $m(E) < \infty$, for any $\epsilon > 0$ there is a finite collection $\{R_j\}_{1}^{n}$ of disjoint rectangles whose sides are intervals such that $m(E \ \triangle \ \bigcup_{1}^{n}R_j) < \epsilon$.

Proof a.)

Part 1: Let $E\in L^n$ and $\epsilon > 0$ then by definition of product measures there exists a countable family $\{T_j\}$ of rectangles such that $E\subset \bigcup_{1}^{\infty}T_j$ and $\sum_{1}^{\infty}m(T_j) \leq m(E) + \epsilon$. For each $j$, applying theorem 1.18 to the sides of $T_j$ we can find a rectangle $U_j \supset T_j$ whose sides are open sets such that $m(U_j) < m(T_j) + \epsilon 2^{-j}$. Let $U = \bigcup_{1}^{\infty}U_j$, then $U$ is open and $$m(E)\leq m(U) \leq \sum_{1}^{\infty}m(U_j) \leq \sum_{1}^{\infty}m(T_j) +\epsilon\leq m(E) + 2\epsilon \tag{1}$$ thus we have proved $m(E) = \inf\{m(U):E\subset U, U \ \text{open}\}$.

Part 2: Now, let us prove $\mu(E) =\sup\{\mu(K):K\subset E, K \ \text{compact}\}$.

First suppose $E$ is bounded. Then there is $k\in \mathbb{N}$ such that $E\subset [-k,k]^n$. Then, $[-k,k]\setminus E \in \mathcal{L}^n$ and then, by part 1, $$\mu([-k,k]^n\setminus E) = \inf\{\mu(U):([-k,k]^n\setminus E)\subset U, U \ \text{open}\}$$ So , for any $\varepsilon>0$, there is $U$ such that $([-k,k]^n\setminus E)\subset U$, $U$ is open and $$\mu([-k,k]^n\setminus E)\leq \mu(U) \leq \mu([-k,k]^n\setminus E) + \epsilon$$ So we have, for any $\varepsilon>0$, there is $U$ such that $([-k,k]^n\setminus U)\subset E$, $U$ is open and $$\mu(E)=\mu([-k,k]^n)-\mu[-k,k]^n\setminus E)\geq \mu([-k,k]^n)-\mu(U) \geq \\ \geq \mu([-k,k]^n)-\mu([-k,k]^n\setminus E) -\varepsilon=\mu(E)-\varepsilon$$ Note that $([-k,k]^n\setminus U)$ is closed and bounded, so it is compact and $\mu([-k,k]^n\setminus U)=\mu([-k,k]^n)-\mu(U)$. So we proved that for any $\varepsilon>0$, there is $K$ (take $K=[-k,k]^n\setminus U$) such that $K\subset E$, $K$ is compact and $$\mu(E)\geq \mu(K) \geq \mu(E)-\varepsilon$$ So we have proved that, if $E\in \mathcal{L}^n$ is bounded then $$\mu(E) =\sup\{\mu(K):K\subset E, K \ \text{compact}\}$$

If $E$ is not bounded, since for all $K\subset E$, $\mu(K)\leq \mu(E)$, we $$\sup\{\mu(K):K\subset E, K \ \text{compact}\} \leq \mu(E)$$

Define, for $j\in \mathbb{N}$,$j>1$ $E_j=E\cap [-j,j]^n$. Note that $\{E_j\}_j$ is a monotone non-decreasing sequence of measurable sets and $$E=\bigcup_jE_j$$ So, since $\mu$ is continuous from below, we have $\lim_{j\to \infty}\mu(E_j)=\mu(E)$.

Now, given any $\varepsilon>0$, since each $E_j$ is in $\mathcal{L}^n$ and is bounded, there is $K_j$ compact, such that $K_j\subset E_j$ and $$\mu(E_j)-\frac{\varepsilon}{j}\leq \mu(K_j) \leq \mu(E_j)$$ So we have $\lim_{j\to \infty}\mu(K_j)=\mu(E)$. So we have $$\mu(E) =\sup\{\mu(K):K\subset E, K \ \text{compact}\}$$

So we have proved that, if $E\in \mathcal{L}^n$ (bounded or not) then $$\mu(E) =\sup\{\mu(K):K\subset E, K \ \text{compact}\}$$

Proof b.)

It follows exactly as the proof of the Theorem 1.19. See my answer in

Real Analysis, Folland Theorem 1.19 Borel Measures

See in my answer there, the part General Case, item "a.) implies b.)" and item "a.) implies c.)".

Proof c.)

Let $m(E) < \infty$ and $\epsilon >0$. Let $\{U_j\}_j$ be the family of open rectangles as in item a.) part 1 above. From $(1)$, we have that $m(U_j) < \infty$ for all $j$.

Now since the sides of $U_j$ are countable unions of open intervals, then for each $j$, take a suitable subunion to obtain a rectangle $V_j\subset U_j$ whose sides are finite unions of intervals such that $m(V_j) \leq m(U_j) \leq m(V_j) + \epsilon 2^{-j}$. So, we have,

$$m(E)\leq m(U) \leq \sum_{1}^{\infty}m(U_j) \leq \sum_{1}^{\infty}m(V_j) +\epsilon \leq \sum_{1}^{\infty}m(U_j) +\epsilon \leq m(E) + 2\epsilon $$

Take $N$ to be sufficiently large, then

$$m(E\setminus \bigcup_{1}^{N}V_j) \leq m(\bigcup_{1}^{N}U_j\setminus V_j) + m(\bigcup_{N+1}^{\infty}U_j) < 2\epsilon$$ and $$m(\bigcup_{1}^{N}V_j\setminus E) \leq m(\bigcup_{1}^{\infty}U_j\setminus E) < \epsilon$$ thus we have $$m(E\triangle \bigcup_{1}^{N}V_j) < 3\epsilon$$

Since $\bigcup_{1}^{N}V_j$ can be expressed as a finite disjoint union of rectangles whose sides are intervals, we have proved c.).

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  • $\begingroup$ For part a.) we have $$\mu([-k,k]^n\setminus E) = \inf\{\mu(U):([-k,k]^n\setminus E)\subset U, U \ \text{open}\}$$ So for any $\epsilon > 0 $ there is a $U$ such that $([-k,k]^n\setminus E)\subset U$, $U$ is open and this is what I don't seem to understand: $$\mu([-k,k]^n) - \mu([-k,k]^n\setminus E) \geq \mu([-k,k]^n) - \mu(U)$$ Is it because $-\mu([-k,k]^n\setminus E)$ will be the supremum? I understand the next step since $([-k,k]^n\setminus E)\subset U$. $\endgroup$ – Wolfy Jul 8 '16 at 2:14
  • $\begingroup$ Also, for the case where $E$ is not bounded you define $j\in\mathbb{N}$, $j > 1$, $E_j = E\cap [-j,j]^n$ how is $E_j$ bounded? i.e. How is $E_j$ bounded from below and bounded from above? $\endgroup$ – Wolfy Jul 8 '16 at 2:36
  • $\begingroup$ For b. from the link the general case states that $\mu$ is $\sigma$-finite. I remember we discussed that was the case for section 1.5 does that apply as well for section 2.6? $\endgroup$ – Wolfy Jul 8 '16 at 3:28
  • $\begingroup$ I still don't understand the logic here, to me the last inequality works since we are subtracting i.e. we have $$\mu([-k,k]^n\setminus E) \leq \mu(U)$$ so then $$-m([-k,k]^n\setminus U) \geq -\mu(U)$$ is that correct? $\endgroup$ – Wolfy Jul 8 '16 at 3:37
  • $\begingroup$ @Wolfy , For part a.) : Since $$([-k,k]^n\setminus U)\subset E$$ then $$([-k,k]^n\setminus E)\subset U$$ So $$\mu([-k,k]^n\setminus E)\leq \mu(U)$$ So $$\mu([-k,k]^n)-\mu[-k,k]^n\setminus E)\geq \mu([-k,k]^n)-\mu(U) $$ $\endgroup$ – Ramiro Jul 8 '16 at 3:42

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