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While browsing through some old questions and problems with regard to measure theory I found the following one, for which I first thought I might have a solution :

Let $\mu$ be a finite measure on the measurable space $([0,1], \mathfrak{B}([0,1]))$ with $\mu([0,1]) \neq 0$. Show that there is a sequence $(I_n)_{n\in\mathbb{N}}$ of closed intervals with $$[0,1]\supseteq I_1 \supseteq \cdots \supseteq I_n \supseteq I_{n+1} \supseteq \cdots $$

such that every $I_n$ has length $L(I_n) = 2^{-n}$ and $\mu(I_n) \neq 0 $.

Since it must hold that $$L(I_n) = b_n - a_n = 2^{-n},$$ I first thought that the sequence might be $$I_n = \left[1 - \frac{1}{2^{n+1}}, 1 + \frac{1}{2^{n+1}}\right]$$ but I quickly noticed that this is just wrong because of the required inclusions above. Hence, what is the solution here?

1 Answers 1

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We could start be letting $I_1' = [0,\frac{1}{2}]$ or $I_1'' = [\frac{1}{2}, 1]$. As $\mu([0,1]) \neq 0$, we must have that either $\mu(I_1')$ or $\mu(I_1'')$ is of positive measure. Suppose $\mu(I_1') > 0$. Let $I_1 = I_1'$. Then we construct construct $I_2$ in a similar way: let $I_2' = [0,\frac{1}{4}]$ or $I_2'' = [\frac{1}{4}, \frac{1}{2}]$. Since $\mu(I_1) > 0$, we must have that either $I_2'$ or $I_2''$ is of positive measure etc... At each step, we break down $I_n$ into two pieces $I_{n+1}'$ and $I_{n+1}''$, one of which must have a positive measure such that we select $I_{n+1}$ inductively. By construction, $L(I_n) = 2^{-n}$ and $I_{n+1} \subset I_{n}$ for all $n$