$ \mu $ premeasure and continousness - proof verification

Question

Let be $ \mu $ a content on a Ring $R$

I want to prove following $$ (a) \leftrightarrow (b) \rightarrow (c) \leftrightarrow (d) $$ with

a) $ \mu $ is a premeasure

b) $\mu $ is continous "from below " , means $ A_n \uparrow A, A \in R$ and $\mu (A_n) \rightarrow \mu (A) $

c)$\mu $ is continous "from above " , means $ A_n \downarrow A, A \in R$ and $\mu (A_n) \rightarrow \mu (A) $

d) $ \mu $ is $ \emptyset $- continous, means there is a sequence $ (A_n) \in \mathbb{R}$ with $ A_n \downarrow \emptyset $ and $\mu (A_n) < \infty \forall n $ with $ \mu (A_n) \rightarrow 0 $

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here my proof: $ (a) \rightarrow (b) $ Let be $ A_n \uparrow A $ and define $A_0= \emptyset $

Then $B_n= A_n \backslash A_{n-1} $ is pairwise disjoint. Then you can write

$ A_n = \cup_{k=1}^n B_k$ and $ A= \cup_{k \in \mathbb{N}} B_k $

so is

$\mu (A_n)= \mu ( \cup_{k=1}^n B_k ) = \sum_{k=1}^n \mu (B_k) \rightarrow \sum_{k=1}^{ \infty} \mu (B_k)= \mu ( \cup_{k=1}^{ \infty } B_k) = \mu (A) $

$(a) \leftarrow (b) $

here the $ \sigma $- additivity needs to be show.

So let be $ B_1,...,B_n $ pairwise disjoint and $ A_n = \cup_{k=1}^n B_k $

then $ A_n \uparrow = \cup_{k=1} ^{ \infty} B_n $ hold

and you can write

$\mu (A_n) = \mu ( \cup_{k=1}^n B_k) = \sum_{k=1}^n \mu (B_k) \rightarrow \sum_{k=1}^{ \infty } \mu (B_n) $

and because of continousness from below it follows that : $ \mu (A_n) \rightarrow \mu (A) $

$ (c) \rightarrow (d) $

Set $ A= \emptyset $ then $ A_n \downarrow \emptyset $ and $ \mu (A_n) < \infty $ and $ \mu (A_n) \rightarrow \mu (A) = \mu ( \emptyset) = 0 $

$ (c) \leftarrow (d) $

let be $ \mu $ finite. Then $ A \backslash A_n \uparrow 0 $ holds and $ \mu (A) - lim_{n \rightarrow \infty} \mu (A_n) = lim_{ n \rightarrow \infty} \mu (A \backslash A_n ) = \mu ( \emptyset)= 0 $

$(b) \rightarrow (c) $

Is $ A_n \downarrow A $ then $ A_1 \backslash A_n \uparrow A_1 \backslash A $

and $ \mu (A) < \mu (A_n) < \infty $

so $ \mu (A_1)- \mu (A_n) = \mu (A_1 \backslash A_n) \rightarrow \mu ( A_1 \backslash A) = \mu (A_1 ) - \mu (A) $

Is this proof correct so far? Can I leave it like that or woud you add some adjustments? especially i'm not sure with case b) to c) Thanks in advance !!