Multidimensional Integration 14 | Proof of the Regularity of the Lebesgue Measure [dark version]

[00:00:00]Hello and welcome back to multi-dimensional integration the video course where we talk about the integration with respect to the lebec measure in Rn and indeed in today's part 14 we will show that this lebec measure is a regular measure this regularity of the measure is something that can really help for showing other facts in the integration theory however before we go into the technical proof I first want to thank all the nice people who support the channel on steady here on YouTube or via other means. And please don't forget with the link in the description, you can download a lot of additional material for the videos. And now without further ado, let's immediately recall the proposition about the regularity of the lebec measure from the last video.

[00:00:45]Indeed, there we have two statements that should hold for a lebec measurable set A. The first one is what we call the outer regularity of the lebec measure and it just tells us that the measure of a can be approximated by open sets from the outside. And the second one is the inner regularity which is similar but now we want to approximate from the interior with compact sets K. And there you should see that the picture here is quite simple because the set U should always extend our given set A and the compact set should completely lie inside A. And now since a measure is monotonic, we immediately know that the measure of U has to be larger or equal than the measure of A. And on the other hand, the measure of K is smaller or equal than the measure of A. So in conclusion, for each statement here, one inequality is immediately given. So this inequality is given for every open set U with this property and therefore it also holds for the infyum. And on the other hand we have the other inequality for the second statement. So the measure of K is always smaller or equal than the measure of A.

[00:01:54]So it also holds for the supreum.

[00:01:57]Therefore if we start with the proof we actually only have to show one inequality each. And that's exactly what we do. Now the whole video will just be about showing these two statements. And in order to show this approximation, it's good to have a fixed distance that we can call epsilon. So this epsilon is completely arbitrary and it can be as small as we want. Okay. Then let's start with the first case where we already know that we come from the outside with a larger set. And now by the definition of the libec measure, you also know that we can cover the whole set A with rectangles. Of course, these are n-dimensional generalized rectangles and we can call them RJ. So we want to have here is that the countable union over these rectangles is a supererset of A.

[00:02:48]However, most importantly by the definition of the Lebec measure, we know that we can choose these rectangles in such a way that we are as close as we want to the measure of A. So more concretely if we sum up the leback measures of the given rectangles we are closer to the measure of a than our given epsilon. In other words if we add epsilon to the measure of a we are larger or equal than the sum of the rectangles. So if you don't remember this property please check out part 13 again. Indeed there you can see that in general this rectangle is not an open set. But this does not matter so much because a rectangle can always be embedded into an open set. Indeed, we just have to make the volume of the rectangle a little bit bigger and then you can see the whole rectangle lies inside an open set. And now let's simply say that this new open set is called B with index J. So it's a supererset of the rectangle and also open. And now most importantly the volume we have to add around the rectangle can be as small as we want again and of course the difference in volume here could be given by our epsilon again however now I also want that this depends on the chosen index j let's say we have 2 ^ j in the denominator as well so this just means the larger this index j is the smaller the difference in the volume should be and we want to have set because now I want to define a new open set by the union of these open sets and this will be our set U that covers our set A. So this is quite clear because we cover the rectangles with these open sets we also cover the set A. And now you see we can just use this inequality for every index J to calculate the measure of U. It's definitely smaller than summing up all the measures of the BJS. Therefore, we simply get two infinite sums out. And the first one is smaller or equal than the measure of a plus epsilon. And the second one by construction is just an infinite sum over the powers of two. So it's just a geometric series which we can calculate by the common formula and we get out exactly one. Hence our inequality here just has a plus 2 epsilon in the end. So we see for a given set a we always find a covering open set u that satisfies this inequality. And since the infyum here can only be smaller or equal than the left hand side. It also satisfies the same inequality. And this is the crucial thing. It satisfies this inequality no matter how small our epsilon is. And that's the important part here because epsilon is arbitrary. The difference between the two things here can be as small as we want it. So the only possibility is that the difference is less or equal than zero which means we have this inequality without the epsilon. And there you see this was exactly the missing inequality for point A. Therefore I would say let's immediately go to part B where we have to talk about compact sets. And since in our space Rn compact sets are always bounded, it would be best to first consider the case that A is a bounded set as well. So let's write that our first case is that A is bounded, which also implies that the Lebec measure of A is a finite number. And moreover, it also implies that the closure of A is definitely a compact set. However, we don't need to work with the closure because we could take any larger set C that is also compact and contains A.

[00:06:38]This doesn't make a big difference, but it makes the picture easier to grasp because then we can just look at a difference between C and A. So, we have a set that does not contain any points of A anymore. And now it might be helpful to abbreviate this new set by a tilda simply because for this new set we can just apply part a and we approximate it by an open set. So we know that there exists an open set that is really close to a tilda measured with the lbec measure. And here please don't forget in the picture this open set also has this part inside. So we approximate from both sides. And let's call this new open set U. So we get the existence of such a set U that is a supererset of a tilda and also an open set and satisfies the following inequality. We want to be closer than a difference of epsilon to the volume of our a tilda. So if we add epsilon on the right hand side we can choose u such that we have a smaller or equal sign. And this is guaranteed because we have our infinum inequality from part a. And with that we have everything we want and we can finally define our compact set that approximates A from the interior. And the picture already tells you that this should be this whole region here. And you see we simply get that by defining C without U.

[00:08:05]Or other words this one is the compact set C intersected with a closed set. And obviously the closed set is just the complement of the open set U. Therefore, the resulting set that we can call K is definitely a compact set as well. And indeed, as the picture suggests, this completely lies inside our original set A. So, this is quite easy to show. And then in the next step, we can just calculate the Lebec measure of K or more precisely what we want to have is the difference between the Lebec measure of A and the Lebec measure of K. And first of all you should see that our compact set C is split up into two disjoint sets namely A and A tilda. Therefore we can simply express the lebec measure of A as this difference here. And there we can bring our open set U in because we have this inequality. So we subtract something that is bigger than lambda of U minus epsilon. Therefore we get this inequality for our LEC measure of A. And because of minus minus we have a plus epsilon.

[00:09:13]And then in the next step we can use that the union of k with u definitely covers our c. So we get an inequality again and we can replace the lebec measure of c by the addition of these two lebec measures. This means finally the lec measure of k comes into the game. And in fact you should see we are already done because the lebec measure of u cancels out. So the difference between the volumes of A and K is definitely smaller or equal than epsilon. And since the supreum can only get bigger, we also get the same inequality for the supreum. And as before, the only thing remaining on the right here is our epsilon. But there you already know this was arbitrarily given.

[00:09:58]So we can make it as small as we want.

[00:10:01]So it's the same reasoning as before.

[00:10:03]The inequality has to hold even without the epsilon. And that's it. This is the approximation with compact sets from inside. However, please don't forget the whole proof was only possible because our A was bounded.

[00:10:19]Hence, now in the second case, we can consider the case that A is an unbounded set. Indeed, this whole thing is not a big problem because we can just scoop it with bounded sets. This means we take our bounded set and make it bigger and bigger such that in the limit we get the whole set a back. So we have a whole chain of bounded sets where the union is a again. And the good thing is then we also know that the lebec measure of the set a j converges to the lebec measure of a. So this is what I meant when I said scooping we work with bounded sets and in the limit we get our set a back.

[00:10:59]So this is quite nice and now we can even distinguish two sub cases here.

[00:11:04]Either this measure is infinity or it's a finite number. Anyway, both cases are quite simple. Now this is because we can apply part one to each of the bounded sets AJ. This means we always find a compact set KJ that is included in AJ.

[00:11:24]And moreover, we know from before that we can always satisfy this inequality when we add an epsilon on the right hand side. However, now since the left hand side is unbounded when j goes to infinity, the right hand side goes to infinity as well. And that's all we need because we simply want that the supreum given here is also equal to infinity.

[00:11:47]And this already completes our first subcase here. And now the second subcase just means that a gets smaller when it goes to infinity. So it's unbounded but still with finite measure. Therefore in this case the supreum is not allowed to go to infinity. However now the good thing is that this means that with our bounded sets we can get as close as we want to the measure of a. So let's say we find an index m where the difference is already smaller than epsilon. In other words, we have our standard inequality where we just have plus epsilon on the right hand side. However, for this am, we can also apply part one again and have the same inequality as before. So, you see in the end here, we have the same inequality again just with plus 2 epsilon. Therefore, with the same reasoning as always, the inequality also holds for the supreum. And then again, the last step is just that we can skip the epsilon altogether. And that's it.

[00:12:48]The inequality is proven also in this subcase. So this means in total the whole regularity of the lebec measure is proven. And that's all what I wanted to show you today. What we can do with this regularity I show you in another video.

[00:13:05]So I really hope I meet you there again and have a nice day. Bye-bye.

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