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

Question

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.

Answer 1

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.