I what to prove this statement
But it seems to me that is part from the definition, here is the definition:
I mean; if $a<b$ then $A=(a,b] \in B(\mathbb{R})$ then $\mu_g(A)=\mu_g((a,b])=g(b)-g(a)$. So there is something I do not get? because for example what is the continuous from the right part for?
Edit:
I understand that they refer this example, but this uses the outer measure but is not necesay in this case, isn't it?
I'll illustrate with an example: Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $$ g(x) = z $$ for all $x \in [z,z+1)$, where $z \in \mathbb{Z}$. Note that $$ \bigcup_{z \in \mathbb{Z}} [z,z+1) = \mathbb{R}, $$ and it is a disjoint union, so it is a partition of $\mathbb{R}$. So $g$ is well defined.
The continuous from the right part is satisfied, but it is not continuous from the left. So $$ \mu_g((z,z+1]) = \lim_{h \rightarrow 0} g(z+1-h) - \lim_{h \rightarrow 0} g(z+h) = z - g(z) = z - z = 0, $$ for every interval of the partition, for this definition of $g$. Do you get where the continuous from the right part is used?
Disclaimer: if you see any mistake, edit it, I'm responding via cellphone.