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Measure of open intervals and half-open half-closed intervals


Solution to Measure, Integration & Real Analysis by Axler Section 2A Exercise 2A6

Prove that if $a,b\in\mathbf R$ and $a<b$, then $$|(a,b)|=|(a,b]|=|[a,b)|=b-a.$$


Solution 1: Since $(a,b)\subset [a,b]$, by 2.5 and 2.14, we have $$|(a,b)|\leqslant |[a,b]|=b-a.$$Hence to show $|(a,b)|=b-a$, it suffices to show that $|(a,b)|\geqslant b-a$.

Let $\varepsilon>0$ be any positive number which is sufficiently small, for example provided $$2\varepsilon < b-a.$$ Note that $[a+\varepsilon,b-\varepsilon]\subset (a,b)$. Therefore we obtain from 2.5 and 2.14 that $$|(a,b)|\geqslant |[a+\varepsilon,b-\varepsilon]|=b-a-2\varepsilon.$$Because $\varepsilon$ is arbitrary, we conclude that $|(a,b)|\geqslant b-a$. So we have $|(a,b)|=b-a$.

To show $|(a,b]|=b-a$. Notice that $(a,b)\subset (a,b]\subset [a,b]$. It follows from 2.5, 2.14, and $|(a,b)|=b-a$ that $$b-a=|(a,b)|\leqslant |(a,b]|\leqslant |[a,b]|=b-a.$$Therefore $|(a,b]|=b-a$. The part $|[a,,b)|=b-a$ is similar.


Solution 2: By 2.4, we know that $$|\{a,b\}|=|\{a\}|=|\{b\}|=0.$$Hence the part $|(a,b)|=b-a$ follows from Exercise 2A1 by setting $A$ to be $(a,b)$ and $B=\{a,b\}$. The other two can be obtained in a similar way.

Of course, it is also possible to use Exercise 2A3.

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

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