Lebesgue and Jordan measure and some example

Question

Let $S$ be semiring with identity element $E$ and $m$ is $\sigma$-additive measure on $S$.

Theorem: Suppose $S$ - semiring. Then minimal ring, containing $S$ is the following set: $$R(S)=\left\{\bigsqcup \limits_{i=1}^{n}A_i:A_i\in S\right\}.$$ In other words, $R(S)$ is the family of all finite disjoint unions of sets in $S$.

Definition 1: If $A\subset E$, then we define outer Jordan measure in the following way: $$\mu_J^*(A)=\inf \left\{\sum \limits_{i=1}^{n}m(A_i): A_1,\dots, A_n\in S, A\subseteq \bigcup\limits_{i=1}^{n}A_i\right\}.$$

Definition 2: If $A\subset E$, then we define outer Lebesgue measure in the following way: $$\mu^*(A)=\inf \left\{\sum \limits_{i=1}^{\infty}m(A_i): A_1, A_2,\dots\in S, A\subseteq \bigcup\limits_{i=1}^{\infty}A_i\right\}.$$

Remark 1: It is not so difficult to show that for any $A\subseteq E$ we have $\mu^*(A)\leq \mu^*_J(A)$.

Remark 2: ("Triangle inequality" for measure) If $A, B\subseteq E$ then the following estimate holds: $$|\mu^*(A)-\mu^*(B)|\leq \mu^*(A\triangle B)$$

Also the same "triangle inequality" is true for outer Jordan measure.

Definition: We say that $A\subseteq E$ is Lebesgue measurable (Jordan measurable), if for any $\varepsilon>0$ $\exists A_{\varepsilon}\in R(S)$ such that $\mu^*(A\triangle A_{\varepsilon})<\varepsilon$ ($\mu_J^*(A\triangle A_{\varepsilon})<\varepsilon$). Denote by $M$ and $M_J$ the family of all subsets of $E$ which are Lebesgue measurable and Jordan measurable, respectively.

Then by remark 1 it follows that $M_J\subseteq M$. Also, note that $R(S)\subseteq M_J\subseteq M$. But the below example will show that $M_J\neq M$.

Example: Denote by $\{a,b\}$ the following intervals: $[a,b], [a,b), (a,b], (a,b)$. Consider the semiring $S=\{\{a,b\}:\{a,b\}\subseteq [0,1]\}$ with identity $E=[0,1]$. Consider the set $E=\mathbb{Q}_{[0,1]}\subset [0,1]$.

Then we can show that $\mu^*(E)=0$ and $\mu^*_J(E)=1$. I can show that $E\in M$.

But how to show that $E\notin M_J$?

I have tried to prove by contradiction and also tries to use remark $2$ but no results. I would be very thankful if anyojne can show the solution?

Answer 1

Similiarly to how you showed that $\mu_J^*(E) = 1$, you should be able to show that every dense subset of $[0,1]$ has outer Jordan measure 1. Now show that for any set $B \in R(S)$, the set $E \triangle B$ is dense. Hence you cannot find any $B \in R(S)$ for which $\mu_J^*(E \triangle B)$ is small.

To show $E \triangle B$ is dense, let $(a,b) \subset [0,1]$ be an arbitrary interval. We have to show that $(a,b)$ contains a point of $E \triangle B$. If $(a,b)$ contains a point of $E \setminus B$ we are done. If not, then $B$ contains all the rationals in $(a,b)$. But since $B$ has to look like a finite union of intervals, show that in this case, $B$ contains some irrationals from $(a,b)$ (you fill in the details), so that $(a,b)$ contains a point of $B \setminus E$.

Answer 2

Let $Q= \mathbb{Q} \cap [0,1]$. Suppose $I \subset [0,1]$ is an interval, then $\mu_J^* (Q \cap I) = \mu_J^* I$. Similarly, $\mu_J^*(I \setminus Q)= \mu_J^* I$.

If $I_k \in S$ are pairwise disjoint, then $\mu_J^* (A \cap (I_1 \cup \cdots \cup I_l)) = \mu_J^* (A \cap I_1) + \cdots + \mu_J^* ( A \cap I_l)$ (for any $A \subset [0,1]$).

Note that $[0,1] \in S$ so for any $B \in S$ we have $B^c \in S$.

Suppose $Q \subset A_1\cup\cdots \cup A_n$ where $A_i \in S$. Without loss of generality, we can assume the $A_i$ are pairwise disjoint. We can also write $(A_1\cup\cdots \cup A_n)^c = B_1\cup\cdots \cup B_m$, where the $B_j \in S$ are pairwise disjoint.

Hence $Q \triangle (A_1\cup\cdots \cup A_n) = (Q \cap B_1) \cup \cdots \cup (Q \cap B_m) \cup (A_1 \setminus Q) \cup \cdots \cup (A_n \setminus Q)$, and so $\mu_J^* (Q \triangle (A_1\cup\cdots \cup A_n)) = \mu_J^* B_1 + \cdots + \mu_J^* B_m + \mu_J^* A_1 + \cdots + \mu_J^* A_n = 1$.

The reason this works for $\mu^*$ is because we can find a countable collection of $A_i \in S$ such that $Q \subset \cup_i A_i$ and $\cup_i A_i$ has arbitrarily small $\mu^*$ measure. Hence $Q \triangle \cup_i A_i = \cup_i (A_i \setminus Q) \subset \cup_i A_i$.

Addendum: To see why $\mu_J^* (Q \cap I) = \mu_J^* I$: Note that $\mu_J^* (Q \cap I) \le \mu_J^* I$ follows by definition. Suppose $Q \cap I \subset A_1 \cup \cdots \cup A_n$. with $A_i \in S$. Then $I \setminus (A_1 \cup \cdots \cup A_n)$ must be finite and only the endpoints of the intervals can be missing). Then we can replace the $A_i$ by $\overline{A_i}$ which have the same $m$ measure and $I \subset \overline{A_1} \cup \cdots \cup \overline{A_n}$. Hence, taking $\inf$s we have $\mu_J^* (Q \cap I) = \mu_J^* I$.