2
$\begingroup$

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 $E$.


enter image description here

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Type $G_\delta$ means countable intersection of open sets. $\endgroup$ – Matt A Pelto Jul 12 '18 at 22:42
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.