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Why does the author let $V_m = \bigcap_{k=m}^\infty H_k$? I think $V_m = H_m$ so there is no need to define $V_m$. Can we just use $H_m$ and let $H=\bigcup H_m$?


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The point of taking the intersections $V_m = \bigcap_{k=m}^\infty H_k$ is so that the $V_m$ are increasing. Note that without any further assumptions, the $H_m$ are not necessarily increasing, even though they have the same outer measures as the $E_m$. Hence we do not necessarily have $H_m = V_m$. So we replace $H_m$ by the increasing sequence $V_m$ to apply Theorem 3.26.


Just as a simple example to see that the $H_m$ are not necessarily increasing, suppose that we are working in $\Bbb R$ and let $E_m = B_{1-1/m}(0)$, the ball of radius $1-1/m$ centered at $0$. Take $H_m = E_m\cup\{m\}$. Then clearly $H_m$ is measurable, and $|H_m| = |E_m|_e$, but $H_m\not\subset H_{m+1}$, so the sequence $H_m$ is not increasing. However, $V_m = E_m$ is increasing.

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    $\begingroup$ I'm not OP but I appreciated the example. Thanks :) $\endgroup$ – Calvin Khor Jul 11 '18 at 22:51

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