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I have found this in a book but I have no idea how to prove it, does anyone know how to?

Let $R$ be a ring in $\Omega$ and $\mu$ a premeasure on $R$. Define: $B_R=\{A \subset \Omega\, :\, A \cap X \in R\text{ for all }X \in R\}$.

$\mu^*(A) := \sup${$μ(R) : X \in R, X \subset A$}, for all $A \in B_R$

How can I prove that $B_R$ is an algebra in $\Omega$ that contains $R?$ Solved.

And that $\mu^*$ is a premeasure on $B_R$ that extends $\mu$?

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By definition, a ring $\mathcal A$ is an algebra if and only if $\Omega\in\mathcal A$. It is clear that $\Omega\in B_R$. By the identity $(A\cap B)\cap X=A\cap(B\cap X)$ and the fact that $R$ is a ring, it is clear that $B_R$ is closed under intersection. By the identity $(A\setminus B)\cap X=(A\cap X)\setminus(B\cap X)$ and the fact that $R$ is a ring it follows that $B_R$ is closed under set difference. Since $R$ is a ring, $R\subseteq B_R$.

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    I don't see how do you use the identity $(A\cap B)\cap X=A\cap(B\cap X)$ to prove that $B_R$ is closed under intersection... – 2017-02-13
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    @TylerKant Suppose $A,B\in B_R$. This means that, for all $X\in R$, $B\cap X\in R$. But then $A\cap(B\cap X)\in R$. Therefore, $(A\cap B)\cap X\in R$ for all $X\in R$. – 2017-02-13
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    If you like it better, you can also use $(A\cap B)\cap X=(A\cap X)\cap(B\cap X)$. – 2017-02-13
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    Ok now I see it thanks. And $R\subseteq B_R$ is just proved by definition? – 2017-02-13
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    @TylerKant $R$ is intersection-closed, so all the elements of $R$ satisfy the defining property of $B_R$. – 2017-02-13