3 added 3429 characters in body
Sup.lim_nsup.lim_n\ C_n:= \cap_n(\cup_{h\geq n} C_h)$: \$A\Delta B= \widetilde{A}\Delta \widetilde{B})$follow that this sequence converging to$\cup_n(\cap_{h\geq n} \widetilde{C_h})$then applying the (uniform) map$C \mapsto \widetilde{C}$follow that the sequence$(C_n)_n)$converging to the superior limit$sup.lim_n\ C_n:= \cap_n(\cup_{h\geq n} C_h)$too, and this sequence is equivalent (i.e. has the some limit) the increasing sequence$(\cap_{n\leq k} C_k)_n$, and if$(C_n)_n)$is a increasing$Chauchy$sequence then it converging to its union$C=\cup_n C_n$.\\ \ If$\mu \in o_fMis(X)$for subadditivity we have$|\mu(A)-\mu(B)|\leq \mu(A\Delta B)$for$ A,B\in [\mu < \infty]$then$\mu: [\mu < \infty]\to [0, \infty[$is uniformly continuous. If$\mu \in oMis(X)$then from$(1)$follow that$\mu: \mathcal{P}(X)\to [0,\infty]$is continuous: it's enough show that for$\mu(C)=\infty$and$C_n$increasing sequence with union$C$then$sup_n (C_n)=\infty$: we have that$lim_{n\to\infty}\mu(C\setminus C_n)=0$then follow from$\mu(C) \leq \mu(C\setminus C_n)+\mu(C_n)$.\\ \C_n)+\mu(C_n)$.

Let $\mathcal{R}_L=${$\mathcal{R}(\mu)=${$A\subset X | \forall \epsilon \geq 0 \exists R\in \mathcal{ R } : \mu (A\Delta R)\leq \epsilon$} the topological closure of $\mathcal{R}$, it is also the Cauchy completion, then is still a (boolean topological) ring ; and from last observation on $(1)$ the ring $\mathcal{R}_L$ \mathcal{R}(\mu)$is a$\sigma $-Ring and the continuous extension$\mu:\mathcal{R}_L\to[0, \mu:\mathcal{R}(\mu)\to[0, \infty]$is still addictive, and in particular continuous for the increasing sequences, then it is$\sigma $-additive. EDIT: Let$ \mathcal{R}\subset \mathcal{P}(X)$a ring and$\mu: \mathcal{R} \to [0, \infty[$a measure (then$\sigma$-addittive ), and suppose that$X$is a countable unions of elements$\mathcal{R}$. Define the Lebesgue extension$µ_L\in oMis(X)$:$ \mu_L (A):=inf_{\ A\subset \cup_n R_n} \sum_n \mu(R_n)$where the 'Inf' is on the countable families$(R_n)_n$such that$A\subset \cup_n R_n$. Let$Mis(\mu_L)$the class of the Caratheodory$\mu_L$-measurable sets, this is a$\sigma$-ring that containing$\mathcal{R}$and$\mu_L$is strong-regular: for any$A\subset X$there exist a measurable set$E_A\in Mis(\mu_L)$such that$A\subset E_A$and$\mu_L(A)=\mu_L(E_A)$. d) Let$\mu^\star\in ofMis(X)$, let$\mu^\star$additive and finite on a subsets algebra$\mathcal{R}\subset \mathcal{P}(X)$. For$A\subset X$considering the following property: i)$\mu^\star (X)= \mu^\star (A) + \mu^\star (X\setminus A)$ii)$A\in Mis(\mu^\star)$iii)$A\in \mathcal{R}(\mu^\star)$. Then$(ii)\Rightarrow (i)$and$(iii)\Rightarrow (i)$; and$(i)\Leftrightarrow (ii)$if$\mu$is strong-regular and$\mu^\star (A)<\infty$. These are all equivalent if$\mu^\star= µ_L$(where$\mu $is$\sigma $-addittive and$X\sigma $-finite).\DIM.$(ii)\Rightarrow(i)$: Obvious.$(iii)\Rightarrow(i):$If$\mu\star(X)=\infty$follow by subaddittivity, otherwise for$\epsilon >0$let$ R\in \mathcal{R}$with$\mu^\star(A\Delta R)=\mu^\star((X\setminus A)\Delta(X\setminus R))<\epsilon $; we have that$\mu^\star(A)\leq \mu^\star(R)+\epsilon$(from$ A\subset A\Delta R \cup R $) and$ \mu^\star(X\setminus A)\leq \mu^\star(X\setminus R)+\epsilon $and follow that$ \mu^(X) - \mu^\star(X\setminus A) \geq \mu^\star(X) - \mu^\star(X\setminus R)-\epsilon \geq \mu^\star(A) -2\epsilon $the last follow from$\mu^\star(X)+\epsilon \geq \mu^\star(X)-\mu^\star(R)+ \mu^\star(A)=\mu^\star(X\setminus R)+ \mu^\star(A)$.$(i)\Rightarrow (ii):$We have$\mu^\star(X)= \mu^\star (A)+ \mu^\star (X\setminus A)$and let$A \subset M\in Mis(\mu^\star)$with$\mu^\star (A)= \mu^\star (M)$, from$\mu^\star (A) = \mu^\star (A\cap M) + \mu^\star (M\setminus A)= \mu^\star (A)+ \mu^\star (M\setminus A)$follow$\mu^\star (M\setminus A)=0$for$E\subset X$we have$\mu^\star (E\cap M) \leq \mu^\star (E\cap A)+ \mu^\star (E\cap (M\setminus A)) = \mu^\star (E\cap A)$then$\mu^\star (E \cap M)= \mu^\star (E \cap A)$and from$E\setminus M \subset E\setminus A=(E\setminus M) \cup (E \cap (M\setminus A))$follow$\mu^\star (E\setminus M) = \mu^\star (E\setminus A)$then$\mu^\star (E)= \mu^\star (E \cap M)+ \mu^\star (E\setminus M)= \mu^\star (E \cap A)+ \mu^\star (E\setminus A)$.$(ii)\Rightarrow(iii):$For$\epsilon >0$let$A\subset \cup_n B_n$with$B_n\in \mathcal{R}$with$\sum_n \mu(B_n)< \mu^\star(A)+\epsilon /2$and let$N>0$a integer such that$\sum_{n>N} \mu (B_n)<+\epsilon /2$, let$F:=\cup_{1\leq k\leq n } B_n$, then$A\setminus F \subset \cup_{ n>N } B_n$and$\mu^\star(A\setminus F)<\epsilon /2$, from$F\setminus A \subset \cup_n \ B_n\setminus A$follow$\mu^\star (F\setminus A)\leq \mu^\star (\cup_n B_n\setminus A) =\ ^{\mu^\star\ is\ \sigma-addittive\ on\ measurables}$=$= \mu^\star (\cup_ n\ B_n)-\mu^\star (A)\leq \sum_{1\leq i\leq n } \ \mu(B_i)\ -\ \mu^\star(A)< \epsilon /2$then$\mu^\star (A\Delta F) <\epsilon $. 2 improve latex; deleted 4 characters in body I did this many time ago (as a student), I did it alone (teacher dont care a lot of about), I hope all work well: 0) DEFINITION.\ Let$X\in Set$and$\mathcal{P}(X)$the set of parts of$X$. Let$o_fMis(X)$the class of finite-outer-measure on$X$i.e. the maps$\mu: \mathcal{P}(X)\to [0, \infty] $with:$\mu(\emptyset)=0,\ \mu(A) \leq \mu(B) \ for \ A \subset B$,$\mu(\cup_I A_i) \leq \sum_{i\in I } \mu(A_i)\ for\ I\ finite $, if in the last property$I$is countable then$\mu$is said a outer-measure, and these make the subclass$oMis(X)\subset o_fMis(X)$. For$\mu \in o_fMis(X)$define$d_\mu: \mathcal{P}(X)\times \mathcal{P}(X) \to [0, \infty]$as$d_\mu(A, B):=\mu(A\Delta B)$(where$A\Delta B:= (A\setminus B)\cup(B \setminus A) $) and$\rho_\mu:\mathcal{P}(X)\times \mathcal{P} (X)\to [0,1]$as$\rho_\mu(A,B):= d_\mu(A,B)/(1+ d_\mu(A,B))$(let$\infty/\infty:=1$) this is a is a pseudo-metric and from$d(A\Delta S,A\Delta T)=d(S, T)$this pseudo-metric is additive. Further, from$(A_1\setminus A_2) \Delta (B_1\setminus B_2)= [(A_1\setminus B_1)\cap(B_2\setminus A_2)]\cup[(B_1\setminus A_1)\cap(A_2\setminus B_2)] \subset B_2)]\subset\subset [(A_1\setminus B_1)\cup(B_1\setminus A_1)]\cap[(A_2\setminus B_2)\cup(B_2\setminus A_2)]=(A_1\Delta B_1)\setminus (A_2\Delta B_2)$follow that the map$(A,B) \mapsto A\setminus B$is uniformly continuous, then there are also the maps:$(A,B) \mapsto A\cap B=A\setminus (A\setminus B)$,$(A,B) \mapsto A\cup B=X\setminus (X\setminus A \cap X\setminus B)$,$(A,B) \mapsto A\Delta B=A\cup B\setminus (A\cap B)$; then the (boolean) ring$(\mathcal{P}(X), \Delta,\cup, 0,1)$is a uniformly ring. In the the subspace$[\mu< \infty]:={A\subset X| \mu(A)<\infty}$we have the pseudo-metric$d_µ$equivalent to the restriction of$\rho_\mu$. 1) Let$\mu\in oMis(X)$and fixed the pseudo-metric$\rho_\mu$. We prove that a$Cauchy$-sequence$ (C_n)_n$converging to the inferior limit:$inf.lim_nC_n:=\cup_n(\cap_{h\geq n} C_h)$and to the superior limit$ Sup.lim_n\ C_n:= \cap_n(\cup_{h\geq n} C_h)$: \ For$\epsilon >0$let$ N(\epsilon )>0$such that$ d(C_n,C_m)<\epsilon /2\ for\ n,m\geq N(\epsilon )$; let$F_n:= C_{ N(1/2^n)}$and put$E:= \cup_n(\cap_{h\geq n} F_h)$. We have that:$(\cap_{k\geq m} F_k) \Delta F_m= F_m\setminus(\cap_{k\geq m} F_k)= (\cup_{ k>m } F_m\setminus F_k)\subset F_k)\subset\subset (\cup_{ k>m } (F_m\setminus F_{ m-1} \cup\ldots\cup F_{k+1} \setminus F_k)$. And$E\Delta(\cap_{k\geq m} F_k)= E\setminus(\cap_{k\geq m}F_k) =\bigcup_n[(\cap_{h\geq \bigcup_n[(\cap_{h\geq n} F_h)\setminus( \cap_{ k\geq m }F_k)]= F_k)]== \bigcup_ n[(\cap_{h\geq n} F_h)\setminus (\cap_{n>k\geq m} F_k) \bigcap (\cap_{h\geq n} F_h)]= \bigcup_{ F_h)]==\bigcup_{ n>m } [(\cap_{h>n} F_h)\setminus( \cap_{n>k\geq m} F_k)] == \bigcup_{ n>m } [(\cap_{h>n} F_h)\setminus F_m \cup (\cap_{h>n} F_h)\setminus F_{ m+1})\ldots \cup (\cap_{h>n} F_h)\setminus F_n)] \subset F_n)]\subset\subset \bigcup_{ n>m } (F_{ m+1} \setminus F_m) \cup (F_{ m+2} \setminus F_{ m+1}) \ldots \cup (F_{ n+1} \setminus F_n)$. Then$E\Delta F_m= (E\Delta (\cap_{k\geq m} F_k)) \Delta ((\cap_{k\geq m} F_k) \Delta F_m)\subset F_m)\subset\subset (E\Delta (\cap_{k\geq m} F_k)) \bigcup (( \cap_{k\geq m} F_k) \Delta F_m)\subset\ F_m)\subset\subset (F_m\Delta F_{ m+1}) \bigcup ( F_{ m+1} \Delta F_{ m+2})\cup\ldots$By countable subadditivity follow that the sequence$(F_n)_n$(and then the sequence$(C_n)_n)$) converging to$E=inf.lim_n C_n$. Applying this to the sequence$(\widetilde{C_n})_n)$(where put$\widetilde{C}:=X\setminus A$) from$A\Delta B= \widetilde{A}\Delta \widetilde{B})$follow that this sequence converging to$\cup_n(\cap_{h\geq n} \widetilde{C_h})$then applying the (uniform) map$C \mapsto \widetilde{C}$follow that the sequence$(C_n)_n)$converging to the superior limit$sup.lim_n\ C_n:= \cap_n(\cup_{h\geq n} C_h)$too, and this sequence is equivalent (i.e. has the some limit) the increasing sequence$(\cap_{n\leq k} C_k)_n$, and if$(C_n)_n)$is a increasing$Chauchy$sequence then it converging to its union$C=\cup_n C_n$.\\ \ 2) If$\mu \in o_fMis(X)$for subadditivity we have$|\mu(A)-\mu(B)|\leq \mu(A\Delta B)$for$ A,B\in [\mu < \infty]$then$\mu: [\mu < \infty]\to [0, \infty[$is uniformly continuous. If$\mu \in oMis(X)$then from$(1)$follow that$\mu: \mathcal{P}(X)\to [0,\infty]$is continuous: it's enough show that for$\mu(C)=\infty$and$C_n$increasing sequence with union$C$then$sup_n (C_n)=\infty$: we have that$lim_{n\to\infty}\mu(C\setminus C_n)=0$then follow from$\mu(C) \leq \mu(C\setminus C_n)+\mu(C_n)$.\\ \ 3) Let$\mu\in oMis(X)$and$\mathcal{ R }$a ring of subset of$X$with$\mu: \mathcal{ R } \to[0,\infty]$additive. Let$\mathcal{R}_L=${$ \mathcal{R}_L=${$ A\subset X | \forall \epsilon \geq 0 \exists R\in \mathcal{ R } : \mu (A\Delta R)\leq \epsilon $} the topological closure of$\mathcal{R}$, it is also the Cauchy completion, then is still a (boolean topological) ring ; and from last observation on$(1)$the ring$L(\mathcal{R})$\mathcal{R}_L$ is a $\sigma$-Ring and the continuous extension $\mu:L(\mathcal{ R} )\to[0, \mu:\mathcal{R}_L\to[0, \infty]$ is still addictive, and in particular continuous for the increasing sequences, then it is $\sigma$-additive.

1

I did this many time ago (as a student), I did it alone (teacher dont care a lot of about), I hope all work well:

0) DEFINITION.\ Let $X\in Set$ and $\mathcal{P}(X)$ the set of parts of $X$. Let $o_fMis(X)$ the class of finite-outer-measure on $X$ i.e. the maps $\mu: \mathcal{P}(X)\to [0, \infty]$ with: $\mu(\emptyset)=0,\ \mu(A) \leq \mu(B) \ for \ A \subset B$ , $\mu(\cup_I A_i) \leq \sum_{i\in I } \mu(A_i)\ for\ I\ finite$, if in the last property $I$ is countable then $\mu$ is said a outer-measure, and these make the subclass $oMis(X)\subset o_fMis(X)$.

For $\mu \in o_fMis(X)$ define $d_\mu: \mathcal{P}(X)\times \mathcal{P}(X) \to [0, \infty]$ as $d_\mu(A, B):=\mu(A\Delta B)$ (where $A\Delta B:= (A\setminus B)\cup(B \setminus A)$) and $\rho_\mu:\mathcal{P}(X)\times \mathcal{P} (X)\to [0,1]$ as $\rho_\mu(A,B):= d_\mu(A,B)/(1+ d_\mu(A,B))$ (let $\infty/\infty:=1$) this is a is a pseudo-metric and from $d(A\Delta S,A\Delta T)=d(S, T)$ this pseudo-metric is additive. Further, from

$(A_1\setminus A_2) \Delta (B_1\setminus B_2)= [(A_1\setminus B_1)\cap(B_2\setminus A_2)]\cup[(B_1\setminus A_1)\cap(A_2\setminus B_2)] \subset [(A_1\setminus B_1)\cup(B_1\setminus A_1)]\cap[(A_2\setminus B_2)\cup(B_2\setminus A_2)]=(A_1\Delta B_1)\setminus (A_2\Delta B_2)$ follow that the map $(A,B) \mapsto A\setminus B$ is uniformly continuous, then there are also the maps: $(A,B) \mapsto A\cap B=A\setminus (A\setminus B)$, $(A,B) \mapsto A\cup B=X\setminus (X\setminus A \cap X\setminus B)$, $(A,B) \mapsto A\Delta B=A\cup B\setminus (A\cap B)$ ; then the (boolean) ring $(\mathcal{P}(X), \Delta,\cup, 0,1)$ is a uniformly ring. In the the subspace $[\mu< \infty]:={A\subset X| \mu(A)<\infty}$ we have the pseudo-metric $d_µ$ equivalent to the restriction of $\rho_\mu$.

1) Let $\mu\in oMis(X)$ and fixed the pseudo-metric $\rho_\mu$.

We prove that a $Cauchy$-sequence $(C_n)_n$ converging to the inferior limit: $inf.lim_nC_n:=\cup_n(\cap_{h\geq n} C_h)$ and to the superior limit $Sup.lim_n\ C_n:= \cap_n(\cup_{h\geq n} C_h)$: \ For $\epsilon >0$ let $N(\epsilon )>0$ such that $d(C_n,C_m)<\epsilon /2\ for\ n,m\geq N(\epsilon )$ ; let $F_n:= C_{ N(1/2^n)}$ and put $E:= \cup_n(\cap_{h\geq n} F_h)$. We have that:

$(\cap_{k\geq m} F_k) \Delta F_m= F_m\setminus(\cap_{k\geq m} F_k)= (\cup_{ k>m } F_m\setminus F_k)\subset (\cup_{ k>m } (F_m\setminus F_{ m-1} \cup\ldots\cup F_{k+1} \setminus F_k)$.

And

$E\Delta(\cap_{k\geq m} F_k)= E\setminus(\cap_{k\geq m}F_k) = \bigcup_n[(\cap_{h\geq n} F_h)\setminus( \cap_{ k\geq m }F_k)]= \bigcup_ n[(\cap_{h\geq n} F_h)\setminus (\cap_{n>k\geq m} F_k) \bigcap (\cap_{h\geq n} F_h)]= \bigcup_{ n>m } [(\cap_{h>n} F_h)\setminus( \cap_{n>k\geq m} F_k)] = \bigcup_{ n>m } [(\cap_{h>n} F_h)\setminus F_m \cup (\cap_{h>n} F_h)\setminus F_{ m+1})\ldots \cup (\cap_{h>n} F_h)\setminus F_n)] \subset \bigcup_{ n>m } (F_{ m+1} \setminus F_m) \cup (F_{ m+2} \setminus F_{ m+1}) \ldots \cup (F_{ n+1} \setminus F_n)$.

Then

$E\Delta F_m= (E\Delta (\cap_{k\geq m} F_k)) \Delta ((\cap_{k\geq m} F_k) \Delta F_m)\subset (E\Delta (\cap_{k\geq m} F_k)) \bigcup (( \cap_{k\geq m} F_k) \Delta F_m)\subset\ (F_m\Delta F_{ m+1}) \bigcup ( F_{ m+1} \Delta F_{ m+2})\cup\ldots$

By countable subadditivity follow that the sequence $(F_n)_n$ (and then the sequence $(C_n)_n)$) converging to $E=inf.lim_n C_n$. Applying this to the sequence $(\widetilde{C_n})_n)$ (where put $\widetilde{C}:=X\setminus A$) from $A\Delta B= \widetilde{A}\Delta \widetilde{B})$ follow that this sequence converging to $\cup_n(\cap_{h\geq n} \widetilde{C_h})$ then applying the (uniform) map $C \mapsto \widetilde{C}$ follow that the sequence $(C_n)_n)$ converging to the superior limit $sup.lim_n\ C_n:= \cap_n(\cup_{h\geq n} C_h)$ too, and this sequence is equivalent (i.e. has the some limit) the increasing sequence $(\cap_{n\leq k} C_k)_n$, and if $(C_n)_n)$ is a increasing $Chauchy$ sequence then it converging to its union $C=\cup_n C_n$ .\\ \

2) If $\mu \in o_fMis(X)$ for subadditivity we have $|\mu(A)-\mu(B)|\leq \mu(A\Delta B)$ for $A,B\in [\mu < \infty]$ then $\mu: [\mu < \infty]\to [0, \infty[$ is uniformly continuous. If $\mu \in oMis(X)$ then from $(1)$ follow that $\mu: \mathcal{P}(X)\to [0,\infty]$ is continuous: it's enough show that for $\mu(C)=\infty$ and $C_n$ increasing sequence with union $C$ then $sup_n (C_n)=\infty$: we have that $lim_{n\to\infty}\mu(C\setminus C_n)=0$ then follow from $\mu(C) \leq \mu(C\setminus C_n)+\mu(C_n)$.\\ \

3) Let $\mu\in oMis(X)$ and $\mathcal{ R }$ a ring of subset of $X$ with $\mu: \mathcal{ R } \to[0,\infty]$ additive. Let $\mathcal{R}_L=$ {$A\subset X | \forall \epsilon \geq 0 \exists R\in \mathcal{ R } : \mu (A\Delta R)\leq \epsilon$} the topological closure of $\mathcal{R}$, it is also the Cauchy completion, then is still a (boolean topological) ring ; and from last observation on $(1)$ the ring $L(\mathcal{R})$ is a $\sigma$-Ring and the continuous extension $\mu:L(\mathcal{ R} )\to[0, \infty]$ is still addictive, and in particular continuous for the increasing sequences, then it is $\sigma$-additive.