A Problem on Outer Measure

April 2025
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There is a problem in Real Analysis by Halsey L. Royden, Patrick M. Fitzpatrick.

Suppose that outer-measure is defined by covering sets by countable collections of closed, bounded intervals rather than coverings by open, bounded intervals. Show that the outer- measure remains unchanged.

I give my answer here.

Proof.

Let $A$ be a set of real numbers. $m_1=\inf \left\{\sum_{k=1}^{\infty}l(I_k)\mid A\subseteq \bigcup_{k=1}^{\infty}I_k\right\}$ , $m_2$ defined similarly with intervals to be closed.

Now consider $I_i=[a,b]=a\cup (a,b)\cup b$ , we form the collection $B=\cup_i a_i\cup_i b_i$ . This set has measure zero, as they are countable. Thus, $m_2(B)=m_{1}(B)=0$ . We know that $A$ is covered by $\{I_{i}/(a\cup b)\}$ . Thus $m_2(A)\leq m_1(A)+m_1(\cup_i a_i\cup_i b_i)=m_1(A)$ .

On the other side, we could do similar operation. Consider $I_k=(a_k-\epsilon/2^{k+1},b_k+\epsilon/2^{k+1})$ . Then $\sum_{k=1}^{\infty}l(I_k)=\sum_{k=1}^{\infty]l([a-k,b_k])+\epsilon$ . Taking $\epsilon$ to $0$ . We get $m_1(A)\leq m_2(A)$ , giving the desired result.

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