With continuity from above, we require one of the sets to have finite measure. I'm wondering, if all sets have infinite measure and the measure is Lebesgue measure, does then the intersection also have infinite measure?
I know for e.g. counting measure on $\mathbb N$, this may not be the case.
My sad story got so sad that now I find it funny. There is a chess tournament happening in my institute today and tomorrow. I didn't know about it earlier. When I finally knew about it, the 3rd round had already started.
I registered, got paired in 4th round. My opponent didn't show up. In the next round my next opponent didn't show up either XD
@psie maybe I should get specific; we have a set $E$ and the sets in question are the union of the dyadic cubes that intersect $E$. These dyadic cubes have vertices in the lattice $(2^{-k}\mathbb Z)^n$ and side length $2^{-k}$. So that union decreases as $k$ increases.
@psie basically I'm wondering if $\overline{A}(E,k)$ denotes the union I was talking about here, does $\bigcap_{k=1}^\infty \overline{A}(E,k)$ have infinite measure?
basically I was wondering if continuity from below still holds if $E$ is unbounded, meaning all $\overline{A}(E, k)$ have infinite measure. But I guess your line example is a counterexample then.
@Jakobian you simply haven't shown that $m(\bigcap_{k=1}^\infty \overline{A}(E,k))=0$. Note, it's not possible to apply continuity from above here because all $\overline{A}(E,k)$ have infinite measure.
Whatever, I think we are misunderstanding each other. I believe one has to declare $m(\bigcap_{k=1}^\infty \overline{A}(E,k))=\infty$ whenever $E$ is unbounded. I don't see how else one can "show" this.
It has to be a sort of convention. I don't see how to show it simply because continuity from above does not apply.
@Jakobian could you explain once more how $ \overline{A}(E,k)$, being a possible infinite union of closed sets, is closed? That goes against the foundations in topology, doesn't it?
If $\mathcal{A}$ is a family of sets which is locally finite, then its hereditarily closure preserving in the sense that $\overline{\bigcup \mathcal{B}} = \bigcup \{\overline{B} : B\in\mathcal{B}\}$ where $\mathcal{B}\subseteq\mathcal{A}$
@psie just prove it
If $\mathcal{A}$ is locally finite then so is $\mathcal{B}\subseteq \mathcal{A}$
This is probably one of these topics with more than 2000 years of discussion behind it
To me it always seems that logic is actually not central here What is a "possible world"? - that is the question And I would say a fairly pointless one
rather than tell everyone what a possible world is, i'm going to show you. please take something from that package of pills that i mailed out last week
@VladimirLysikov yes. it seems like philosophers just define the set of possible worlds to be whatever set of worlds they want
if it's just upto imagination, I can conceive of worlds where there's no logic. e.g. take a world where there only exists a point. now, logic has no meaning there
but our world has the structure of mathematically rich physics models, in which there's determinism and differential equations. All of this structure is defined using logic
Consider the collection $\mathcal Q_k$ of dyadic cubes of sides $2^{-k}$ with vertices in the lattice $(2^{-k}\mathbb Z)^n$. Let $U$ be open and define $\underline{A}(U,k)$ to be the union of the cubes that are contained in $U$. Then $\underline{A}(U,k)$ is increasing with $k$. Define $\underline{A}(U)=\bigcup_1^\infty \underline{A}(U,k)$.
Now, I'm reading a lemma of the fact that $U\subset \mathbb R^n$ open is a countable union of cubes with disjoint interior. Write $$\underline{A}(U)=\underline{A}(U,0)\cup\bigcup_1^\infty [\ \underline{A}(U,k)\setminus \underline{A}(U,k-1)\ ].$$
Then $\underline{A}(U,0)$ is a (countable) union of cubes in $\mathcal Q_0$, and for $k\geq 1$, the closure of $\underline{A}(U,k)\setminus \underline{A}(U,k-1)$ is a (countable) union of cubes in $\mathcal Q_k$. These cubes all have disjoint interiors and the result follows.
Question: Why does the author mention the closure of $\underline{A}(U,k)\setminus \underline{A}(U,k-1)$? What relevance does the closure have here?
I have drawn the rectangle with vertices at $(0,0),(1,0),(0,2),(1,2)$. For $k=0$, we can fit two dyadic cubes into that rectangles, for $k=1$, well, we just bisect the cubes we already have and get $8$ cubes and so on. But clearly $\underline{A}(U,k)\setminus \underline{A}(U,k-1)$ will be the empty set for $k=1,2$. I guess the statement about closure is not false, but I still do not comprehend why the author used that word.
In fact, probably $\underline{A}(U,k)\setminus \underline{A}(U,k-1)$ equals its closure?
