Ytukyg

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: $$R(S)=left{bigsqcup
limits_{i=1}^{n}A_i:A_iin Sright}.$$ In other words, $R(S)$ is the
family of all finite disjoint unions of sets in $S$.

Definition 1: If $Asubset E$, then we define outer Jordan measure in the following way: $$mu_J^*(A)=inf left{sum
limits_{i=1}^{n}m(A_i): A_1,dots, A_nin S, Asubseteq
bigcuplimits_{i=1}^{n}A_iright}.$$

Definition 2: If $Asubset E$, then we define outer Lebesgue measure in the following way: $$mu^*(A)=inf left{sum
limits_{i=1}^{infty}m(A_i): A_1, A_2,dotsin S, Asubseteq
bigcuplimits_{i=1}^{infty}A_iright}.$$

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

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

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

Definition: We say that $Asubseteq E$ is Lebesgue measurable (Jordan measurable), if for any $varepsilon>0$ $exists
A_{varepsilon}in R(S)$ such that $mu^*(Atriangle
A_{varepsilon})<varepsilon$ ($mu_J^*(Atriangle
A_{varepsilon})<varepsilon$). 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 $M_Jsubseteq M$. Also, note that $R(S)subseteq M_Jsubseteq M$. But the below example will show that $M_Jneq M$.

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 $Ein M$.

But how to show that $Enotin M_J$?

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?

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 $B setminus E$.

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 $Q subset A_1cupcdots cup A_n$ where $A_i in S$. Without loss of generality, we can assume the $A_i$ are pairwise disjoint. We can also write
$(A_1cupcdots cup A_n)^c = B_1cupcdots cup B_m$, where the $B_j in S$ are
pairwise disjoint.

Hence $Q triangle (A_1cupcdots cup A_n) = (Q cap B_1) cup cdots cup (Q cap B_m) cup (A_1 setminus Q) cup cdots cup (A_n setminus Q)$, and so
$mu_J^* (Q triangle (A_1cupcdots cup A_n)) = mu_J^* B_1 + cdots + mu_J^* B_m + mu_J^* A_1 + cdots + mu_J^* A_n = 1$.

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 $mu_J^* (Q cap I) = mu_J^* I$.