 So we are still in the middle of trying to prove nice properties for almost minimal sets, and I still don't have any boundary available because things are simpler there. And here I'm going to talk a little bit about rectifiability. So almost minimal sets are rectifiable, OK? And the proof, so it's amusing because with Steven we had a very complicated proof of rectifiability. Angren had the proof of rectifiability, but we didn't read. And we didn't understand why you could prove rectifiability without proving uniform rectifiability. And apparently, yes, you can, and it's not so hard. So let me try to talk a little bit about that. Also, since I was asked to do more pictures of a feather flaming projection, so I'll try to do at least one more picture of a feather flaming projection. So I'm in a cube, and let me talk about the basic step, which is I'm in a phase of dimension, let's say, d plus 1, and d will be 1 because it's simpler to draw pictures like this. I have a set of dimension 1 in there, and I want to project it in here. But the set, for a strange reason, let's imagine that the set is not rectifiable. So I was hesitating a little bit about, for me, all my sets are alphas regular from now on. And if you're alphas regular and you're not unrectifiable, I think it still means that you look like a counter set. At least you should not be connected because connected sets we find out are rectifiable. So it looks a little bit like this. So that's my set E. So of course, this set is not going to be a nice one of our almost minimal sets, but it's like this. And I still want to do some feather flaming projection on it. Now there is, so I'm going to use a variant of the basic of each feather projection theorem, which you might have seen coming. But I hesitated, but I will do just this picture. So imagine now that I'm projecting this set on a line, a random line. The basic of each feather projection theorem is that for almost every direction of a line that I look at this, when I project this totally unrectifiable part, let me say irregular, the projection of this set has zero measure. That's the basic of each feather projection theorem. Let me recall, so it was supposed to be before, and some people said, for me, rectifiable means that you are up to a set of zero measure covered by a countable number of, let's say, C1 graphs, something like this. Every set of finite measure has a decomposition into a rectifiable piece and the rest, OK? The rest does not meet, let's say, C1 curve too loud. It is too loud. OK, I'll try to be, I think, it's better here, right? OK, so any set has a decomposition into a rectifiable piece, which is rectifiable and a totally unrectifiable piece, which I also call irregular, which essentially never meets a set which is rectifiable, OK? And here I tried to draw the irregular part, all right. So let me return slowly. Projection theorem says that almost all projections vanish, OK? What we need is exactly the same thing, except that it's supposed to be almost all radial projections. And it turns out that by using Foubigny and writing down correctly, and this I owe to Vincent Favrier, but probably some people did this before. It's probable that Almgren knew how to do this before. But anyway, the same story happens with radial projections, OK? So I usually like to take one half of this one here. And what I'm saying is that if I pick a random psi here, almost every psi, and I project this set, this is likely annoying, right? Is there a way? Do you know if there is a way? No. OK, all right. OK, so if you take almost any center here, which I usually call psi, then the projection on the boundary will have zero measure. So it's going to be something like this, but really, really small, OK? Vanishing measure. OK, our sets are not always going to be purely unrectifiable. Usually they have a rectifiable piece and a totally unrectifiable piece. So in the next proof, there will be the irregular part that I have here, plus a tiny piece, rectifiable part. And this one will project somewhere here. But if the measure here is small, the measure will stay small when I project, OK? So that was one step. Of course, I'm supposed to iterate this, because maybe I'm not in the dimension one. When I iterate this, so here I will be able to get rid of the totally unrectifiable part inside the face. What is on the boundary stays on the boundary. So maybe there was an irregular part in the boundary too. But then next time I will project on faces of dimensions one less. And at the interior of the face of dimension one less, I will be able to kill again, and so on and so forth, OK? Right. So what I started doing here is preparing the proof of rectifiable piece. So let me now try to set myself correctly. I don't have a thing anymore, but I think essentially I'm trying to describe the story here, OK? So you want to prove that the set, your almost minimal set, is rectifiable. If it is not rectifiable, it has a totally unrectifiable piece, irregular. And by general density results, in fact, you can even select points where the density of what is not irregular. So in other words, the density of rectifiable piece is zero. Almost every point in the irregular set has density zero for the regular part, OK? So at this point, I will select an initial cube, and I'll draw it like this, OK? And in this cube is exactly what I was saying, which is there may be a large irregular piece. It doesn't look large, but irregular sets never look large when you look at them because of a basic of each projection theorem. And a rectifiable piece, which is small, as small as you want, because of a density story, OK? So you do the federal flaming projection, OK? And you will end up, when you do the projection, into a set of measures zero, which comes from the irregular part, plus a set of small measures, which comes from a rectifiable part, OK? And then again, we are exactly as in the same situation as for lower alpha's regularity. We have a projection which never contains a full face. So if it never contains a full face, you can do one more step of a projection, project on d minus 1 dimensional skeleton. And for all that matters, this set you make it disappear, OK? Right? And I decided that I wanted to save a little bit of time, so I will talk more and show slides less. So the same sort of proof that we had before works again, which is you have this set on a cube. So in principle, you do that. There is a more complicated picture because maybe you start it with a cube of dimension three. Then this is one of the inside cubes, and it was connected to another one here and so on and so forth. On the boundary, on a thin annulus, you have to do something slightly different, so you cannot kill the set completely. But if the annulus is thin enough, it doesn't count in the estimate so much. And at the end, you show that in this case here, if a set was almost entirely composed of unrectifiable pieces, you kill the set inside. Maybe you pay for it a little bit on the annulus, but you still get the contradiction because you were able to erase a big part of the set and not pay too much for it. One thing that I meant to say before I started and I forgot is now we know that the set is all false regular. So for instance, we know that the total measure of a set here inside is reasonably large. So if I can erase whatever is inside, I really erased a lot. I know this and this makes the estimate easier. The second thing that I should have said also is that now I know the set is all false regular. So in fact, it's not as dispersed as you would think. So for instance, in the middle here of my cube, I know that there will be a cube or a ball in the complement of a set because of a set, a false regular set cannot be sort of dense, which makes things easier because it means that the leap sheets mapping that I'm going to talk. So my psi, instead of taking it here, I will take it somewhere in the middle here. And then the mappings that I will get are even leap sheets with uniform estimates. So estimates are easier once you know that you're a false regular. So that was rectifiability. OK. And again, I don't want to insist so much. So this was the estimates that I decided not to show you. Now, there is the last thing that I want to do about, there is the last thing that I want to do again about using federal flaming projections, which is the property of projections. And if it looks a lot like one of the slides of Antonio, sorry, but this is what happens. And it's exactly for the same sort of reasons that we have to do that anyway. OK. So the theorem is the following. You take an almost minimal set. And you suppose that there is a plane which is fairly close to the set or that the set is fairly close to a plane. I just draw the picture. The plane is here. The set is your almost minimal set here so maybe I like to draw a ball just to say I'm working in here. And again, this is my assumption. The main assumption is that the set is almost minimal and stays close to the plane. That's my assumption, which is in principle this one here. OK. Then there is the projection, the orthogonal projection on the plane. And I'm saying that the orthogonal projections of a set in the plane essentially contains all it could contain. OK. And this will be proved by contradiction. And again, it's going to be a picture that you've seen before. So I was stupid enough to make a plane which is not horizontal, but that's still going to be OK. I want to prove that the projection is essentially on two. And I proceed by contradiction. So I suppose not. OK. Supposing not meaning that there is, for instance, this point here where the orthogonal project complement. And if a set is one-dimensional, I mean R3 maybe is going to be just hyperplane. But don't worry too much about this. OK. Anyway, I will do the proof in co-dimension one because it's easier, or most of the proof. So I suppose again that this thing here does not meet the set. OK. And the set is contained in a very thin stripe here. And for some strange reasons, since I made a fuss of saying that I need a card shock, here is I'm drawing in advance a wall. So if I'm in dimension 3, the wall is just a thin thing around, but precisely, but then the set would be two-dimensional. OK. And how do I use this line? Let's stay in the plane, right? How do I lose this line that does not meet the set? It's easy because now I can project points exactly parallel to this direction, essentially irradially after you project or something like this. And you will get a new set. So the image of a set may be not the whole green thing here, but it will be contained in the little green walls. OK. Remember, we're trying to look for a contradiction. Very close to the contradiction. The set was at first regular, so there was a lot of mass here. I mean some amount of mass. Then you project it, and you project it on something like this here, which has a small length here, epsilon times r, an additional small length here, epsilon times r. OK. So you were able to manage to find a set which is a deformation of your initial set and which has much less mass contradiction. OK. Let me say two words. I also should revive this if I don't want to type my password all the time. OK. So suppose now we were in higher co-dimensions. If you are in higher co-dimensions, you do this projection, and you find yourself in the wall. But the wall has an infinite host of measure of dimension d. That's not good. It's not so bad either. Near the wall, you can do an extra feather flaming projection and, in fact, project on something which is essentially a grid around here. And the estimate, it works. So it's not infinitely harder in higher dimensions. It's just that you have one more step where you have to take a grid, you want to replace this wall by a grid of a right dimension. OK. That's the, OK. And the reason why I mentioned this, there are two reasons. The first one is that if you know that the set is uniformly overactive variable from this, you did use some extra property, like big pieces of elliptic graphs. But the main point is not this one. It's because it's useful for some other estimates that we'll show up later. OK. So that was this. We only do this in dimension n minus 1. So I'm sorry, this is the proof. But you will not even have time to read it. And I think that's the end of a list of regularity properties that I promised to you. And it doesn't go very far, because alphas regular is just pleasant, but it's not really a regularity property. Rectifiability is not very impressive. So we would expect better. OK. And this story about not having holes is not shocking either. I mean, it's good to know. OK. The last thing that I should have said also is that, of course, since the set is rectifiable, there are lots of situations where this happened. Because the set has, it's also alphas regular, so it has tangents almost everywhere. And near a tangent, you have a situation like this. So essentially it means that when you, I mean near a tangent, when you have a tangent plane and the ball is small enough, you have a situation like this with no holes in the projection. OK. Limits. So limits, again, you're not supposed to be very impressed by this theorem that's up there. You would be much more impressed if I told you it was not true, because then you would understand that we're in trouble some way. So here, this is the logical thing that is supposed to happen. You have a sequence of almost minimal sets. It converges to a limit. I will give you the definition in one second, but it's the normal one. OK. And you want to say two things. The first one is that host of measure along the sequence doesn't increase suddenly when you're taking the limit, lower semi-continuity. And the second one is that the limit is also an almost minimal set. And that's, I mean, it's the second one that really is interesting to you most of the time. And of course, the second one, there's no chance of proving that the limit is almost minimal, or let's say, think even minimal, if its mass is too large. So in the proof of part two, part one will be very important, of course. But unfortunately, as I'm speaking, and I hope it's going to get better, but OK, we'll see. Part one is easy to prove, and part two takes a lot of energy for some reasons that are technical. And maybe that will disappear within two months, depending on whether some person here that I don't see is working hard enough. OK. The tummy pretends not to be. OK. Right. So this is supposed to be the goal for a good part of this hour, except that again, I'll talk about part one and not part two. And I hope the thing is sort of correct. So I should say one or two things about it. The result is a little stronger than what I said here first, because it also works for quasi-minimal sets instead of just almost minimal, so a little bit more possibilities. And the second one is that I said for a limit of almost minimal sets, it also works for minimizing, I mean, let's say some minimizing sequences, which means that you can prove that limits of minimizing sequences are minimal sets. OK, so that's OK. That's one of the ways to try to find minimal sets. OK. And it sort of fits with what Antonio was saying. In this case, his proofs were easier than the one here. And again, I hope to revert that at some point of time, but that's not clear. OK. Right. So what are the ingredients? There is one ingredient which is important, and it is a little surprising. It is the rectifiability of any limit of those guys. OK. I think I forgot also one or two comments. No, OK. So I'll put the definition of limits here and discuss something else so that if you're in principle, you're not supposed to be surprised here. So one thing that I put in the slide and I should have said is the following. Again, my main estimate is lower semi-continuity for host of measure. Of course, if you take just any sequence of sets, this is not true. For instance, these sets, dotted lines, you can easily make them converge to a line or a line segment. And surprisingly, the line segment has something like 10 times more mass than the dotted lines. If you did the dotted lines, OK. So this is what you have to be afraid of or fight at least. It's not going to be so hard, but anyway. But at first, I mean, dotted lines we've seen in the previous case that they cannot be almost minimal, so there is no contradiction in math. It's just we have to be careful about that. The same way, I was about to say that the main ingredient was the rectifiability of limits. And you know very well that if you take a sequence of rectifiable sets, for instance, try to take your favorite description of a 1 fourth counter set using little squares. You'd get a sequence of little squares that are all rectifiable, bounded mass and so on and so forth, so there's nothing fishy there. The limit is totally unrectifiable, but the sets were rectifiable. So I mean, there is no way, in principle, you're not supposed to hope that if the sets are rectifiable, then the limit is rectifiable. Yet this is exactly what I'm saying here. And again, it's because it's special sets. So by now, you probably had time if you were worried to read what is the limit of sets. Anyway, OK, so I have my, it's good in any way. I have my normalized Hostert distance up there, which says when you want to see whether a set is close to another set in a ball, you're normalized by dividing by the radius of a set. And then you take the maximum distance from a point of the set to the other one and of a point of the other one to the set. And you get a Hostert distance slightly different from the usual one localized between two sets. And I'm saying that a sequence of set tends to a limit when the regular, when the normalized Hostert distances in any balls turn to zero, OK? Not surprisingly. We don't have so much choice in this definition because we want to be able to extract subsequences that converge. So in fact, this is the, I mean, this is essentially the only notion where we can do that. You could try to define more refined topologies. It would be good because then you would be able to prove better estimates, but it would be bad because again, I mean, given the sequence of set, I want to be able to extract a subsequence that converges. And then I have no choice. I have to take this definition, OK? Or an equivalent one because there is lots of ways to make them equivalent, OK? But you're not shocked. This is probably the idea you had of converging to limits. And it's better the limit is defined up to density or something like this. So we can always take it closed and that's what we'll do, OK? Rectify the limits, OK? So I made sure not to erase my picture here because now you imagine that in fact, this is not the set that I started with for a single guy. I write it as the limit set of a sequence of almost minimal set, OK? The thing that I call E infinity inverse light, OK? And again, I start the argument exactly as I did before. Suppose the limit is not rectifiable and I can find a place where the rectifiable piece is very small and the totally unrectifiable piece is most of it. Limit of alphos regular sets with uniform alphos regular constants. It's also alphos regular. The limit is alphos regular. Everything that I said here is fine. The set is alphos regular so I can suddenly find a bone away from my limiting set which is far from the limiting set, OK? In this picture, I can project the totally unrectifiable piece to a set of measures zero for the limit. The rectifiable piece is small so I can project it on a small piece. And the main point is that I can iterate the construction and find another mapping which is still lip sheets because I picked points far away from the thing which essentially takes my limiting set and sends it to a grid of dimension d minus one. OK? Right? So I have a limit set. I have a lip sheet mapping that sends the set onto a grid of dimension d minus one. And I'm essentially telling you now that, you know, I had a sequence of set that tends to this thing. The fact that it tends, it means that in this cube, the distance between my sets, so I would also have a set, let's say, e k which I, you know, I can try to draw. But the way I will draw it is by, oh, here we go, is by repeating the same story here. OK? It's close to the other one. My mapping is defined near the set here because this is the way it works. So I'm essentially saying I just can follow the construction that I had for the limiting set and project the other set up to a grid of dimension d minus one. So for k large enough, my set e k, I also can project it in on something small and make it disappear. And then I continue the proof as before. OK? So I get the same contradiction as before using the limit mapping that is corresponding to the limit set. OK? So this is one of the cases where rectifiability for the limit is just as simple as rectifiability of all the sets. Plus a little bit of being careful about having uniform estimates. But that's not so bad. OK. Lower semi-continuity of hd. And again, I will sort of move my hands around. But the proof is not more complicated than what I'm saying. OK? So with this lower semi-continuity of hd, it means you have this sequence of almost minimal sets. They converge to a set infinity. And you want to show that in any open set, the measure of infinity is less than the measure of the limits. OK? Let me try to at least draw a picture, so that I don't have to. OK. So here in the proof that I was talking about, there was an open set v. And the proof that I should have at last at least read the statement. And then I have this set e infinity, which is the limit of our sets e. And maybe there were sets e k. So I don't know if you have already here. But we're close by. And the thing that I want to say does not happen is something like this, right? That the measure of a limiting set is much larger than the measure of a set e k. OK? How do I prove this? Well, there is the usual way you start all these arguments by covering lemma. So we'll use the vitally covering lemma to essentially cover almost all of the limit set by little balls. And the little balls will be disjoint. They will cover almost everything. And they will have some properties. And the properties will be properties that, for almost every point, small balls satisfy. OK? So this is just a little bit of techniques. So here I will cover my infinity by tiny little balls with some properties. I'll draw one of those tiny little balls here. OK? And in this ball, I want two things. One of the main things is the set. At almost every point, there is a tangent plane. So I use this to say that for small balls, the set is very close to a plane. So this is the picture that I'm supposed to have here. And if needed, and I think it's needed, I also require the host of measure of the limiting set here, which is rectifiable. So in particular, it has all the properties of densities. So if I want, I think it should be here of the host of measure of the limiting set in the little balls that I cover is close to what it should be, which is omega d times the radius to the power d. OK? I get a covering by balls like this. And now what I have to do, so I have a covering. In principle, it's an infinite covering. But if I want to cover, let's say, 99.9% of a set, I just need a finite number of balls. They're still disjoint, so I take this final number of balls. And I will just prove that in each ball, there is enough, I can compare the mass of the limits to the mass of e k. All right? So I draw e k. k is large enough. And again, don't worry about the large enough because I have a finite number of balls to take care of. And I have to tell you why the measure of this set here is less than the measure of a red set there. And that's easy. The measure of this set here is not so large compared to the measure of its projection on the plane. And this set e k here, look at the way I can point to my pictures. OK, so this red set here, it satisfies exactly the assumptions of the previous lemma, which implies that the projection is essentially on 2. OK? So if the projection is on 2, the measure of a set e k is larger than 1 minus epsilon times the measure of the disk. And the measure of the disk is what I have up there. OK? Nothing more. OK? And I'm supposed to, yeah, it's even written here. I'm supposed to say that this proof is, so, it's amusing to me how simple this proof is. Because again, previously with Stephen, we had more complicated proofs using something which is called the Dalmazo, Morel, and Solimini concentration theorem. We've, I mean, it looked more complicated. And then what happened is that Fung had to prove the same thing for elliptic integrants. And then there was less choice. So this proof works for elliptic integrant. And the way we rediscovered or discovered that proof, in this case mostly him, was having to do with integrants. Right? And OK. So that's, again, that is supposed to be the main part of the limit.