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I want to calculate the measure of following intervals: $$ \mu^*((a,b)), \mu^*((a,b)], \mu^*([a,b]) $$ Therefore I consider $g: \mathbb{R}\rightarrow \mathbb{R}$ and $ \mu : F^1 \rightarrow \mathbb{R}, \ \mu(A) = \sum_{i=1}^m g(b_i) -g(a_i) $ and $ A=\bigcup_{i=1}^m (a_i,b_i]$. The intervals are disjoint. g is a left-continous and monotonically increasing function. $\mu , \mu^*$ are measures.

Now I can apply Carathéodory's theorem on $\mu$ to get $\mu^*$, which is defined on $B(\mathbb{R}) := \sigma(F^1)$ $$ \mu^*(A)= \inf\left\{\sum_{i=1}^{\infty} \mu\left((a_i,b_i]\right) : A \subset \bigcup_{i=1}^m (a_i,b_i] \right\} $$

Can anybody help me calculating $$ \mu^*((a,b)), \mu^*((a,b)], \mu^*([a,b])? $$

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  • $\begingroup$ can you explain the notations used in your question? What tare $\mu$ and $\mu*$? WHy are you considering $g$ and what is $f$? $\endgroup$ – Mike V.D.C. May 6 at 18:42
  • $\begingroup$ Sorry I will edit that $\endgroup$ – Steven33 May 6 at 18:47
  • $\begingroup$ If you are attempting to typeset $\mu^*$, use \mu^*. $\endgroup$ – angryavian May 6 at 18:50
  • $\begingroup$ Thanks for your advise:) $\endgroup$ – Steven33 May 6 at 18:53
  • $\begingroup$ I mixed up f and g. f is wrong. Can you understand everything now? $\endgroup$ – Steven33 May 6 at 18:54

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