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

$(A_1\setminus A_2) \Delta (B_1\setminus B_2)= [(A_1\setminus B_1)\cap(B_2\setminus A_2)]\cup[(B_1\setminus A_1)\cap(A_2\setminus B_2)]\subset$

$\subset [(A_1\setminus B_1)\cup(B_1\setminus A_1)]\cap[(A_2\setminus B_2)\cup(B_2\setminus A_2)]=(A_1\Delta B_1)\setminus (A_2\Delta B_2)$

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:

$(\cap_{k\geq m} F_k) \Delta F_m= F_m\setminus(\cap_{k\geq m} F_k)= (\cup_{ k>m } F_m\setminus F_k)\subset$

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

And

$E\Delta(\cap_{k\geq m} F_k)= E\setminus(\cap_{k\geq m}F_k) =$

$\bigcup_n[(\cap_{h\geq n} F_h)\setminus( \cap_{ k\geq m }F_k)]=$

$= \bigcup_ n[(\cap_{h\geq n} F_h)\setminus (\cap_{n>k\geq m} F_k) \bigcap (\cap_{h\geq n} F_h)]=$

$=\bigcup_{ n>m } [(\cap_{h>n} F_h)\setminus( \cap_{n>k\geq m} F_k)] =$

$= \bigcup_{ n>m } [(\cap_{h>n} F_h)\setminus F_m \cup (\cap_{h>n} F_h)\setminus F_{ m+1})\ldots \cup (\cap_{h>n} F_h)\setminus F_n)]\subset$

$\subset \bigcup_{ n>m } (F_{ m+1} \setminus F_m) \cup (F_{ m+2} \setminus F_{ m+1}) \ldots \cup (F_{ n+1} \setminus F_n)$.

Then

$E\Delta F_m= (E\Delta (\cap_{k\geq m} F_k)) \Delta ((\cap_{k\geq m} F_k) \Delta F_m)\subset$

$\subset (E\Delta (\cap_{k\geq m} F_k)) \bigcup (( \cap_{k\geq m} F_k) \Delta F_m)\subset$

$\subset (F_m\Delta F_{ m+1}) \bigcup ( F_{ m+1} \Delta F_{ m+2})\cup\ldots$

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 $\mathcal{R}(\mu)=${$ 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 $\mathcal{R}(\mu)$ is a $\sigma $-Ring and the continuous extension $\mu:\mathcal{R}(\mu)\to[0, \infty]$ is still additive, 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 $.