Show, that $\tilde{\mathcal{R}}$ is an algebra, $\tilde{\mu}$ is a pre-measure 

Let $\mathcal{R}$ be a ring on the set $\Omega$ and $\mu$ a pre-measure on $\mathcal{R}$. Show that
    $$
\tilde{\mathcal{R}}:=\left\{A\subset\Omega|\forall~R\in\mathcal{R}: R\cap A\in\mathcal{R}\right\}
$$
    is an algebra on $\Omega$ with $\mathcal{R}\subset\tilde{\mathcal{R}}$ and that
    $$
\tilde{\mu}(A):=\sup\left\{\mu(R)|R\subset A, R\in\mathcal{R}\right\}
$$
    defines a premeasure on $\tilde{\mathcal{R}}$ which continues $\mu$.


Hello, everybody! At first, I give you my proof, that $\tilde{\mathcal{R}}$ is an algebra on $\Omega$ with $\mathcal{R}\subset\tilde{\mathcal{R}}$. To show this, I have to show, that it is a ring which contains $\Omega$ (this are (1) - (4)) and of course $\mathcal{R}\subset\tilde{\mathcal{R}}$ (this is (5) below)).
ad (1): $\forall R\in\mathcal{R}$ it is $R\cap\emptyset=\emptyset$ and because $\mathcal{R}$ is a ring, it is $\emptyset\in\mathcal{R}$. So $\emptyset\in\tilde{\mathcal{R}}$.
ad (2): Let $A,B\in\tilde{\mathcal{R}}$, i.e. for all $R\in\mathcal{R}$ it is $R\cap A, R\cap B\in\mathcal{R}$. Then
$$
R\cap (A\cup B)=(R\cap A)\cup (R\cap B)\in\mathcal{R},
$$
because $\mathcal{R}$ is a ring and unions of two elements in $\mathcal{R}$ are in $\mathcal{R}$.
ad (3): Let $A,B\in\tilde{\mathcal{R}}$, i.e. for all $R\in\mathcal{R}$ it is $R\cap A, R\cap B\in\mathcal{R}$. It is $\forall R\in\mathcal{R}$:
$$
R\cap (A\setminus B)=((R\cap A)\cup (R\cap B))\setminus (R\cap B).
$$
Because of ad (2) and because of the fact, that $\mathcal{R}$ is a ring, this is in $\mathcal{R}$.
ad (4): For all $R\in\mathcal{R}$ it is $R\cap\Omega=R$. So $\Omega\in\tilde{\mathcal{R}}$.
ad (5): Let $R\in\mathcal{R}$. For all $R'\in\mathcal{R}$ it is
$$
R'\cap R=(R\cup R')\setminus (R\Delta R').
$$
This was, to my opinion, part 1 of the task. 
It remains to show, that $\tilde{\mu}$ is a pre-measure ((6)-(7) below) on $\tilde{\mathcal{R}}$ and that $\tilde{\mu}$, restricted to $\mathcal{R}$, is identical with $\mu$ (point (8) below).
ad (6): $\tilde{\mu}(\emptyset)=\sup\left\{\mu(R)|R\subset\emptyset, R\in\mathcal{R}\right\}=\sup\left\{\mu(\emptyset)\right\}=\sup\left\{0\right\}=0$, because $\emptyset$ is the only subset of $\emptyset$ that is in $\mathcal{R}$.
ad (7): Consider disjoint $(A_n)\in\tilde{\mathcal{R}}^{\mathbb{N}}$ with $A:=\biguplus_{i\geq 1}A_i\in\tilde{\mathcal{R}}$. Let $R\in\mathcal{R}$ with $R\subset A$. Because the $A_i$ are disjoint, $R$ can be written as a disjoint union $R=\biguplus_{i\geq 1}R\cap A_i$. Then, because $\mu$ is $\sigma$-additive, it is
$$
\mu(R)=\mu(\biguplus_{i\geq 1}R\cap A_i)=\sum_{i\geq 1}\mu(R\cap A_i)\leq\sum_{i\geq 1}\sup\left\{\mu(R)|R\subset A_i, R\in\mathcal{R}\right\}\\=\sum_{i\geq 1}\tilde{\mu}(A_i)\\\Rightarrow \underbrace{\sup\left\{\mu(R)|R\subset A, R\in\mathcal{R}\right\}}_{=\tilde{\mu}(A)}\leq\sum_{i\geq 1}\tilde{\mu}(A_i)
$$
In order to show the other direction, consider any $R\in\mathcal{R}$ with $R\subset\biguplus_{n\geq 1}A_n$. Again one can write $R$ as $R=\biguplus_{n\geq 1}(R\cap A_n)$. It is
$$
\sum_{n\geq 1}\mu(R\cap A_n)=\mu(R)\leq\tilde{\mu}(A).
$$
Because $R$ is arbitrary, it is
$$
\sup\limits_{R\in\mathcal{R},R\subset A}\left\{\sum_{n\geq 1}\mu(R\cap A_n)\right\}\leq\tilde{\mu}(A).
$$
Because of 
$$
\sum_{n\geq 1}\mu(R\cap A_n)\leq\sum_{n\geq 1}\tilde{\mu}(A_n)
$$
for any $R\in\mathcal{R}$ with $R\subset A$ it is
$$
\sup\limits_{R\in\mathcal{R},R\subset A}\left\{\sum_{n\geq 1}\mu(R\cap A_n)\right\}=\sum_{n\geq 1}\tilde{\mu}(A_n)
$$
which shows that
$$
\sum_{n\geq 1}\tilde{\mu}(A_n)\leq\tilde{\mu}(A).
$$
All in all, it is
$$
\tilde{\mu}(A)=\sum_{n\geq 1}\tilde{\mu}(A_n).
$$
ad (8): Consider any $R\in\mathcal{R}$. Consider
$$
\tilde{\mu}(R)=\sup\left\{\mu(R')|R'\subset R, R'\in\mathcal{R}\right\}.
$$
Because $\mu$ is monotonous, it is $\mu(R')\leq\mu(R)$ for all subsets $R'\in\mathcal{R}$ of $R$. If $R'\neq R$, i.e. $R'\varsubsetneq R$, it is $\mu(R')<\mu(R)$. For $R'=R$, it is $\mu(R')=\mu(R)$, so considering the supremum gives
$$
\tilde{\mu}(R)=\sup\left\{\mu(R')| R'\subset R, R'\in\mathcal{R}\right\}=\mu(R)
$$
I think, that's it!
Would be great if you gave me a feedback!
By the way: Does the finit additvity of $\tilde{\mu}$ follow from the shown $\sigma$-additivity?
 A: The only step in your proof I do not understand is the claim
$$
\sup\limits_{R\in\mathcal{R},R\subset A}\left\{\sum_{n\geq 1}\mu(R\cap A_n)\right\}=\sum_{n\geq 1}\tilde{\mu}(A_n)
$$
from which your derive that
$$
\sum\limits_{n\geq 1}\tilde{\mu}(A_n)\leq\tilde{\mu}(A)\tag{*}
$$
So I'll give my own proof of $(*)$. 
Fix $\varepsilon>0$. From definition of $\tilde{\mu}$ for each $n\in\mathbb{N}$ we have $R_n\in\mathcal{R}$ such $\tilde{\mu}(A_n)\leq\mu(R_n)+\varepsilon 2^{-n}$ and $R_n\subset A_n$. Since $(A_n)$ is a family of disjoint sets so does $(R_n)$. Denote $R=\biguplus_{n\geq 1} R_n$, then clearly $R\subset\biguplus_{n\geq 1}A_n=A$. Now we have
$$
\sum\limits_{n\geq 1}\tilde{\mu}(A_n)
\leq\sum\limits_{n\geq 1}(\mu(R_n)+\varepsilon 2^{-n})
\leq\mu\left(\biguplus\nolimits_{n\geq 1}R_n\right)+\varepsilon
=\mu(R)+\varepsilon
\leq\tilde{\mu}(A)+\varepsilon
$$
Since $\varepsilon>0$ is arbitrtary $(*)$ is proved.
P.S.
To prove finite additivity from $\sigma$-additivity just add to finite system of sets infinite family of empty sets.
