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I would like to show that it's possible to associate to a measure a monotone increasing right-continuous function s.t.: $\mu(\left(a,b\right])=F(b)-F(a)$.

How can I prove that a function like

$$F(x)=\mu(\left(0,x\right]) ~~ x>0$$

$$F(x)=0 ~~ x=0$$

$$F(x)=-\mu(\left(x,0\right])~~ x<0$$

is right-continuous? I thought about considering a sequence $x_n$ decreasing to $x$ such that $\lim(\left(x,x_n\right])=\emptyset$, $\lim \>\mu(\left(x,x_n\right])=0$ and then directly express all in terms of $F$ . My doubts concern the fact that there is an hint for the exercise saying: consider the countable additivity property of the measure in order to show that the function is right continuous.

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Your idea of proof is right. I think the hint is here to help you show $\lim \mu((x,x_n]) = 0$ as $\lim (x,x_n] = \emptyset$. BTW, I think there should be a condition that $\mu((a,b]) < \infty$ for all $(a,b],a,b < \infty$ –  zeno tsang Apr 29 at 12:05
    
I think I got the solution...it's enough to use the fact that we have a decreasing succession of sets convergent to the intersection (the empty set) so that the measure of the succession tends to the measure of the limit (with the hypothesis the $\mu(\left(a,b\right))<\infty$). –  user73793 Apr 29 at 13:04

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