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If $E=\cup I_n$ is a countable union of pairwise disjoint intervals ,prove that $m^*(E)=\sum _{n=1}^\infty l(I_n)$.

Since $E=\cup I_n\implies m^*(E)=m^*(\cup I_n)\le \sum m^*(I_n)=\sum l(I_n).$

But I am not getting how the equality comes here.I am getting only inequality here.

Please help.

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    Is there something you can use about the inequality being an equality when the intervals are disjoint? It would help if you'd define your terms and what assumptions you can make too... for instance, it's clear to me that $m^*$ is a measure or something like one, but not exactly which of the many related notions. – 2017-01-02
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    ^m* is the outer measure. – 2017-01-02
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    ahh ok, then you should go at it from the concrete definition of outer measure rather than from the subadditivity. – 2017-01-02
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    You can show that intervals are $m^*$ measurable, so $m^*$ follows a measure's additivity on intervals. – 2017-01-02

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Without loss of generality, we assume that all the intervals are open intervals. (Because the endpoints of countable intervals are Null set). Then write $I_n=(a_n,b_n)$, for any $\varepsilon>0$, we have \begin{align*} m^*(E)&=m^*\left( \bigcup_{n} I_n \right)\\ & = m^{*} \left( \bigcup_n (a_n,b_n) \right)\\ &\geq m^{*}\left( \bigcup_{n=1}^{\infty}\left( a_n+\frac{1}{2^{n}}\varepsilon, b_n-\frac{1}{2^{n}}\varepsilon \right) \right)\\ & = \sum_{n=1}^{\infty} m^{*}\left( \left( a_{n}+\frac{1}{2^{n}}\varepsilon, b_{n}-\frac{1}{2^n}\varepsilon\right) \right)\\ & = \sum_{n=1}^{\infty}\left[l(I_n)-\frac{1}{2^{n-1}}\varepsilon\right]\\ & = \sum_{n}l(I_n)-2\varepsilon. \end{align*} Letting $\varepsilon$ go to zero gives $m^{*}(E)\geq \sum_n l(I_n)$. Notice that the third equality holds because all these intervals have positive distance away from each other.

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    If you change the $=$ sign in your second line to $\geq$ then the "WLOG assumption" that each $I_n$ is open (which is ok ) becomes unnecessary. Just a suggestion for "polishing" the proof. – 2017-01-02