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Show that $\mu: \widetilde{I} \cap(0,1] \to [0,\infty] $ be defined by $\mu(a,b]:=\bigg\{b-a$ if a $\ne 0$, $0\lt a \lt b \le 1$ $\bigg\}$ and $+\infty$ otherwise, is finitely additive, where $\widetilde{I} \cap(0,1]$ is the class of all left open right closed intervals in $(0,1]$ . Is $\mu$ countably additive also??

I could show that $\mu$ is not countably additive by taking $I_{n}=\left(\frac{1}{n+1},\frac{1}{n}\right]$. Then $\bigcup_{n=1}^{\infty}\left(\frac{1}{n+1},\frac{1}{n}\right]=(0,1]$. But $\mu\left(\frac{1}{n+1},\frac{1}{n}\right]=\frac{1}{n}-\frac{1}{n+1}$. Hence $$\sum_{k=1}^{n}\left(\frac{1}{k+1},\frac{1}{k}\right]=1-\frac{1}{n+1}$$. Thus $$\lim_{n \to \infty}\sum_{k=1}^{n}\left(\frac{1}{k+1},\frac{1}{k}\right]=\lim_{n \to \infty}\left(1-\frac{1}{n+1}\right)=1$$ where as $\mu (0,1]=+\infty$.

But I am not sure how to prove the finite additiveness.

tattwamasi amrutam
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  • The notation $\tilde{I}\cap(0,1]$ is poor - I would recommend just defining $\tilde{I}$ as the set you describe. I think also you should probably define $\mu$ on the generated field or $\sigma$-field. – Jason Feb 13 '15 at 10:51

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wWe have $$\bigcup_{k=1}^{n} \left(\frac{1}{k+1},\frac{1}{k}\right]=\left(\frac{1}{2},1\right]\bigcup \left(\frac{1}{3},\frac{1}{2}\right]\bigcup\ ...\ \bigcup \left(\frac{1}{n+1},\frac{1}{n}\right]=\left(\frac{1}{n+1},1\right],$$ where the sets are disjoint.

You've showed that $$\mu\left(\bigcup_{k=1}^n\left(\frac{1}{k+1},\frac{1}{k}\right]\right)=\sum_{k=1}^{n}\mu\left(\frac{1}{k+1},\frac{1}{k}\right]=\mu\left(\frac{1}{n+1},1\right]=1-\frac{1}{n+1}.$$

By these two expressions above finite additivity is shown for the ring of sets generated by the intervals whose lover limit is not zero. Since, however, $\mu$ is not $\sigma$-additive it cannot be extended to the Lebesgue measurable sets of $(0,1]$. So finite additivity cannot hold for "arbitrary" sets.

However if $\epsilon>0$ then $\mu$ can be extended to the Lebesgue measureable sets of $(\epsilon,1]$. As a result finite additivity is shown for the Lebesgue subsets of $(\epsilon,1]$.

zoli
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