open intervals with upper bound infinity are measurable

[[concept]]

Topics

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Proposition

For all $a\in \mathbb{R}$, the interval $(a,\infty )$ is measurable.

Proof

Suppose $A\subset \mathbb{R}$. Let $A_1=A\cap (a,\infty )$ and $A_2=A\cap (-\infty ,a]$. Now we want to show that $m^{\ast }(A_1)+m^{\ast }(A_2)\le m^{\ast }(A)$.

If $m^{\ast }(A)$ is infinite we are done. So suppose that $m^{\ast }(A)<\infty$. Now, let $\{I_n\}$ be a collection of intervals such that ${\sum }_n\ell (I_n)\le m^{\ast }(A)+\epsilon$ And define $J_n=I_n\cap (a,\infty )K_n=I_n\cap (-\infty ,a]$ Then for each $n$, each of $J_n,K_n$ are either an interval are empty and

• $A_1\subset {\bigcup }_n{J}_n$
• $A_2\subset {\bigcup }_n{K}_n$
• $\ell (I_n)=\ell (J_n)+\ell (K_n)$ Thus we have

m^*(A_{1}) + m^*(A_{2}) &\leq \sum_{n} m^*(J_{n}) + m^*(K_{n}) \\ &= \sum_{n} \ell(J_{n}) + \ell(K_{n}) \\ &= \sum_{n} \ell(I_{n}) \\ &\leq m^*(A) + \varepsilon \end{align}$$ Then take $\varepsilon \to 0$ and we have the desired result. $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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