the outer measure of an interval is its length

[[concept]]

Topics

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Theorem

If $I$ is an interval, then $m^{\ast }(I)=\ell (I)$

Proof

Suppose $I=[a,b]$. Then $I\subset (a-ϵ,b+ϵ)$ for all $ϵ>0$. But then we have $m^{\ast }(I)\le \ell (a-ϵ,b+ϵ)=b-a+2ϵ$ Thus we have $m^{\ast }(I)\le b-a$.

Now we show $b-a\le m^{\ast }(I)$. Let $\{I_n{\}}_n$ be a collection of open intervals such that $[a,b]\subset {\bigcup }_n{I}_n$.

Now, since $[a,b]$ is compact (by the Heine-Borel theorem), there exists a finite subcover $\{J_k{\}}_{k=0}^m\subset \{I_n{\}}_n$ such that $[a,b]\subset {\bigcup }_{k=0}^m{J}_k$

Since $a\in {\bigcup }_{k=0}^m{J}_k$, there exists $k_1$ such that $a\in J_{k_1}$. Rearranging intervals, I let $k_1=1$ and $a\in J_1=(a_1,b_1)$. If $b_1<b$, then $b_1\in [a,b]$. So ther eis some $k_2$ such that $b_1\in J_{k_2}$. By rearranging again, I assume $k_2=2$. So $b_1\in J_2=(a_2,b_2)$.

Continuing in the same manner, we conclude that there exists a $K,1\le K\le m$ such that $b<b_K$. And for all $k\in \{1,2,\dots ,K-1\}$ we have $b_k\le b$ and $a_{k+1}\le b_k<b_{k+1}$.

But then we have

\sum_{n} \ell(I_{n}) & \geq \sum_{k=1}^m \ell(J_{k}) \\ & \geq \sum_{k=1}^K \ell(J_{k}) \\ &= (b_{K}-a_{K}) + (b_{K-1} - a_{K-1}) + \dots + (b_{1}-a_{1}) \\ &= b_{K} + (b_{K-1} - a_{K}) + (b_{K-2}-a_{K-1}) + \dots + (b_{1}-a_{2}) - a_{1} \\ & \geq b_{K} - a_{1} \\ &\geq b-a \\ &= \ell(I) \end{align}$$ Thus we have $m^*(I) \geq \ell(I) = b-a$ Thus we conclude $m^*(I) = \ell(I)$ $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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