In the second paragraph, the author defines $H_k = \bigcap_{m=k}^\infty U_m$, and in the third paragraph, he writes $H=H_1 \cap Z^c$. Are these two $H_1$ the same? Why do we get $E\subset H_1$? Note that it's necessary to define $H_k = \bigcap_{m=k}^\infty U_m$ since we need to guarantee that $H_k \nearrow H$. $U_m$ may not $\nearrow \bigcup U_m$.

Theorem $3.28$: $E$ is measurable if and only if $E=H\cap Z^c$, where $H$ is of type $G_\delta$ and the measure of $Z$ is zero.

Type $G_\delta$ means countable intersection of open set. $|E|_e$ means the outer measure of a set $E$, and $|E|$ means the measure of a set $E$.

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    $\begingroup$ In standard usage, $\cap$ is used in things like $A\cap B$ and $A_1\cap \cdots\cap A_n,$ and $\bigcap$ is used in things like $\bigcap_{m=k}^\infty A_m.$ I edited the question accordingly. $\endgroup$ – Michael Hardy Jul 12 '18 at 20:11
  • $\begingroup$ @MichaelHardy Thank you. I appreciate it. $\endgroup$ – user398843 Jul 12 '18 at 20:12
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    $\begingroup$ Type $G_\delta$ means countable intersection of open sets. $\endgroup$ – Matt A Pelto Jul 12 '18 at 22:42

Are these two $H_1$ the same?

No, otherwise $E$ would have finite measure. Basically, $H$ (this is the same set $H$ defined in the preceding paragraph) is measurable since $H$ is of type $G_{\delta \sigma}$. And so this Theorem $3.28$ says that $H = H_0 - Z$, where $H_0$ is of type $G_{\delta}$ and $|Z|=0$.

Note: I decided to exchange this somewhat abusive $H_1$ for $H_0$. Since $H_k=\bigcap_{m=k}^\infty U_m$ $(k=1,2,\ldots)$, and $U_m$ is of type $G_\delta$ for each $m=1,2,\ldots$, we do know $H_k$ is of type $G_\delta$. Hence "$H$ is of type $G_{\delta \sigma}$."

Why do we get $E\subset H_0$?

Since $E_k \subset H_k$ for $k=1,2,\ldots$, $E=\bigcup_{k=1}^\infty E_k$, and $H=\bigcup_{k=1}^\infty H_k$, it follows that $E \subset H$. Since $H=H_0-Z$, we know that $E \subset H_0 = \left(H_0-Z\right) \cup \left(H_0 \cap Z\right)$. Notice how if we had $H_0=H_k$ for one of the sets $H_k$ defined in the second paragraph, then we would have $|E|_e<+\infty$ since $|H_k|<+\infty$ which would contradict the hypothesis of the case being considered in the last 2 paragraphs of this proof.


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