[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math

Search:


View post   

>> No.10027115 [View]
File: 6 KB, 334x188, fdw.gif [View same] [iqdb] [saucenao] [google]
10027115

>>10027007
>I don't understand how covers work in the definition
Just look at the picture as an easy example. The union of red lines/points is the set you are trying to find the Lebesgue outer measure from. First of all, the intervals that make up your open cover are, well, open. The sequence of intervals that are supposed to cover your set CAN contain the empty set (under most definitions at least), and even when it can't the result doesn't change, it's just that the building process is more annoying. We do know beforehand that the length of any open interval of the cover that has the form (a,b) is |b-a|.

Now, note that with this cover, we are overshooting. We are covering more than what is included in the original set, so of course the outer measure is going to be inaccurate by excess.

Ignore that set for a little while. Imagine we are covering a single point x; one of our covers may include inaccurately the interval (x-1/2,x+1/2). But then another of our covers can include (x-1/4,x+1/4). Maybe the next one (x-1/8,x+1/8). And you can probably see where we are going with this.

The reason we take the infimum of all the covers is precisely this, to make the excess as small as possible. So if you have an infinite amount of covers of said point, it would be the same as just having a sequence of intervals (x-(1/2)^n,x+(1/2)^n). The length of these intervals is |x+(1/2)^n-(x-(1/2)^n)|=|(1/2)^(n-1)|, and the infimum of all these lengths will be 0. Keep in mind that not all the covers of the point will have the form I mentioned earlier, but since the ones that do are actually included in our set of all possible covers, we can build this sequence of intervals.

The same process goes for all the subsets you see in the picture, no matter whether they are open/closed/neither, you can find countable open covers because of Lindelöf's Covering Theorem. You just want to make the covers as "narrow" as possible, while still strictly containing the set.

Navigation
View posts[+24][+48][+96]