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04:44
I see! but I think -
If $E \subset F$ then $m(E) \leq m(F)$
Let $U$ be the open set containing $E^c$ then $E^c \subset U$ then $m(E^c) \leq m(U)$
or $m(U) \geq m(E^c
)$
ohhh
We have $m(E^c) = 0$
so $m(U) \geq 0$
so were we checking that this $m(U)$ is necessarily infinite or not?
@MichaelGreinecker
Also, I was thinking of a previous line -
@MichaelGreinecker Here $E =\cup_{\alpha \in \Bbb{R}-\Bbb{Q}} S_{\alpha}$ then $E^c = \cap_{\alpha \in \Bbb{Q}} S_{\alpha}$
05:16
@BAYMAX $E^C=\bigcup_{\alpha\in\mathbb{Q}}S_\alpha$.
By de Morgan's law it is the intersection of the complements of spheres with irrational radius.
So it is exactly the set of points whose distance from the origin is not an irrational number. But then the distance must be a rational number $q$ and such a point is then in $S_q$.
hm
$E =A \cup B$
then $E^c = A^c \cap B^c $ ?
Am I confusing somewhere?
You switched the index set in the process
The complement of $\bigcup_{\alpha\in\mathbb{R}-\mathbb{Q}}S_\alpha$ is
$\bigcap_{\alpha\in\mathbb{R}-\mathbb{Q}}S_\alpha^C$, same index set
but this set equals $\bigcup_{\alpha\in\mathbb{Q}}S_\alpha$, different index set
@BAYMAX
Oh i see!!!
cannot $m(U)$ be infinite?
as $m(U) \geq 0$
can we provide some upper bound on $m(U)$
?
05:33
If $U$ is an open set that includes $E^C$, then $U$ can certainly have infinite measure. For example, $U=\mathbb{R}^2$ would work. The point is that even though $E^C$ is a dense subset of $\mathbb{R}^2$, it need not have large measure.
A common confusion of beginners is that open dense sets must be the whole space or almost the whole space.
@BAYMAX
@MichaelGreinecker Oh I see this is a nice point!!
Any reference regardiing the connection b/w denseness and measure?
any particular theorem ?
@BAYMAX Under Lebesgue measure, every nonempty open set has positive measure. But open dense sets can have arbitrarily small measure, as in the example I gave. A great thin book that relates measure and concepts such as being open dense is "Measure and Category" by John Oxtoby.
Thank you so much@MichaelGreinecker!!!
@BAYMAX You are welcome! I have to leve now, but I'll check in later in case you still have questions. Cu
C u :)
good day1

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