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Problemathic
United Kingdom
Приєднався 17 вер 2020
Abstract concepts made clear!
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In this youtube channel we go through advanced math areas on detail so that everyone can become an expert starting from the ground up!
If you like our videos please consider donating on KoFi: ko-fi.com/problemathic
In this youtube channel we go through advanced math areas on detail so that everyone can become an expert starting from the ground up!
Product measures | Motivation and Properties
In this video we will motivate the definition of product measures and product sigma-algebras and set the topics that we'll need to study .
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Timestamps in this video:
00:00 Introduction.
03:45 What we'll study.
05:20 Properties of product measure.
#maths #measuretheory #analysis
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Timestamps in this video:
00:00 Introduction.
03:45 What we'll study.
05:20 Properties of product measure.
#maths #measuretheory #analysis
Переглядів: 375
Відео
Lusin's Theorem | What does it mean? | Proof
Переглядів 5067 місяців тому
In this video we will prove Lusin's Theorem. Which states that measurable functions are continuous on very large sets. ☕ Make a small donation on Ko-fi: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram, join us! problemathic...
Folland Exercise 2.4.42 | Measure Theory
Переглядів 1827 місяців тому
Solution to Exercise 42 in section 2.4 of Folland's Real Analysis. We prove that for the Counting Measure, convergence in measure and uniform convergence are equivalent. ☕️ Buy us a Coffee: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram...
Folland Exercise 2.4.33 | Measure Theory
Переглядів 1988 місяців тому
Solution to Exercise 33 in section 2.4 of Folland's Real Analysis. We prove a stronger version of Fatou's Lemma for convergence in measure. ☕️ Buy us a Coffee: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram, join us! probl...
Is Egorov's Theorem true for infinite measures? | Checking the hypothesis | Examples
Переглядів 3248 місяців тому
In this video we show that Egorov's Theorem (or Egoroff's theorem) is not valid when the measure is infinite. We show this with two examples. ☕ Make a small donation on Ko-fi: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram, join us! ins...
Egorov's Theorem | Almost everywhere and uniform convergence | Proof
Переглядів 1,1 тис.8 місяців тому
In this video we learn and prove Egorov's Theorem (or Egoroff), that states that for finite measure spaces, convergence almost everywhere and uniform convergence are very related. ☕ Make a small donation on Ko-fi: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40...
Does Cauchy in measure imply convergence? | Convergence in Measure and L1
Переглядів 3039 місяців тому
In this video we learn two results about convergence. We prove that if a sequence is Cauchy in measure, then it converges in measure to a function and that if a sequence converges in L1 then there exists a subsequence that converges in measure. ☕ Make a small donation on Ko-fi: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great ma...
More types of convergence | Convergence in measure
Переглядів 4119 місяців тому
In this video we learn about convergence in measure and what it means for a sequence to be Cauchy in measure. ☕ Make a small donation on Ko-fi: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram, join us! problemathicvideos 💬...
What does it mean to converge? | L1 and pointwise a.e. convergence
Переглядів 4039 місяців тому
In this video we learn about modes of convergence. Uniform, pointwise, pointwise almost everywhere, and L1 convergence. ☕ Make a small donation on Ko-fi: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram, join us! problemathi...
Folland Exercise 2.3.27 | Measure Theory
Переглядів 28510 місяців тому
Solution to Exercise 27 in section 2.3 of Folland's Real Analysis. We prove several properties of the function f(x) = ae^(-nax) - be^(-nbx) ☕️ Buy us a Coffee: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram, join us! probl...
Folland Exercise 2.3.26 | Measure Theory
Переглядів 25910 місяців тому
Solution to Exercise 26 in section 2.3 of Folland's Real Analysis. We prove continuity of a special function. ☕️ Buy us a Coffee: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram, join us! problemathicvideos 💬 Join our Disc...
Folland Exercise 2.3.19 | Measure Theory
Переглядів 27810 місяців тому
Solution to Exercise 19 in section 2.3 of Folland's Real Analysis. About uniform convergence and convergence in L1. An application of the Dominated Convergence Theorem. ☕️ Buy us a Coffee: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram,...
The most important convergence theorem | Dominated Convergence
Переглядів 40710 місяців тому
In this video we prove The Dominated Convergence Theorem. Arguably, the most important convergence theorem in Measure Theory! ☕ Make a small donation on Ko-fi: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram, join us! probl...
Integrating any function and defining L1 spaces | Properties
Переглядів 33010 місяців тому
In this video we define integration of Real valued functions and L1 spaces using the equivalence relation of almost everywhere equal functions ☕ Make a small donation on Ko-fi: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop 💥 Over 40k followers on Instagram, join us! in...
Munkres Exercises 4.34.3 | Topology
Переглядів 26410 місяців тому
We solve Exercise 3 in Chapter 4, Section 34 of Munkres' Book. A compact Hausdorff space is metrisable if and only if it is second countable. An application of Urysohn Metrisation Theorem. ☕ Make a small donation on Ko-fi: ko-fi.com/problemathic 💸 Support us on Patreon, get great perks and items! www.patreon.com/problemathic 👕 Buy great math t-shirts! www.redbubble.com/people/problemathic/shop ...
Folland Exercise 2.2.13 | Measure Theory
Переглядів 43410 місяців тому
Folland Exercise 2.2.13 | Measure Theory
The most important Theorem in Topology | Urysohn Metrisation theorem
Переглядів 38211 місяців тому
The most important Theorem in Topology | Urysohn Metrisation theorem
A very important convergence theorem! Fatou's Lemma | Proof
Переглядів 1,2 тис.11 місяців тому
A very important convergence theorem! Fatou's Lemma | Proof
The Monotone Convergence Theorem | Consequences | Measure Theory
Переглядів 34411 місяців тому
The Monotone Convergence Theorem | Consequences | Measure Theory
The most important lemma in Topology | Urysohn Lemma | Part 2
Переглядів 38511 місяців тому
The most important lemma in Topology | Urysohn Lemma | Part 2
Folland Exercise 2.2.12 | Measure Theory
Переглядів 44611 місяців тому
Folland Exercise 2.2.12 | Measure Theory
The most important lemma in Topology | Urysohn Lemma | Part 1
Переглядів 1,2 тис.Рік тому
The most important lemma in Topology | Urysohn Lemma | Part 1
The Monotone Convergence theorem | Proof | Measure Theory
Переглядів 797Рік тому
The Monotone Convergence theorem | Proof | Measure Theory
Every Metrisable space is normal | Proof | Topology
Переглядів 244Рік тому
Every Metrisable space is normal | Proof | Topology
Lebesgue Integration | Non-negative functions | Measure Theory
Переглядів 650Рік тому
Lebesgue Integration | Non-negative functions | Measure Theory
Regular + 2° Countable? | On the way to Uryshon's Metrization Theorem | Topology
Переглядів 133Рік тому
Regular 2° Countable? | On the way to Uryshon's Metrization Theorem | Topology
Approximating Measurable Functions| Simple Functions | Measure Theory
Переглядів 960Рік тому
Approximating Measurable Functions| Simple Functions | Measure Theory
Is the Sorgenfray Plane Normal? | Regular and Normal spaces | Topology
Переглядів 518Рік тому
Is the Sorgenfray Plane Normal? | Regular and Normal spaces | Topology
Why 5:48?
Fj = ∩ i=j to infinity (Ej), you said Fj ⊆ Fj+1 means that it’s an increasing sequence, can you explain how? Because when i am trying to understand its decreasing sequence Fj+1 ⊆ Fj, thanks
@@VineetDairashri Asj increases, you’re intersecting less and less things, so Fj+1 being an intersection of less sets than Fj, contains Fj . It would be decreasing if the intersection was from n=1 to j, but in this case the intersection starts at j
@@Problemathic Oh, now i got it, thanks alot for this clarification! [Pov: i am just a beginner in learning advance maths:)]
@ no worries! I do sometimes make mistakes, so it’s always good to check! 🥰
Great video :))
Concerning the first exercise: Is Q convex in R? If not, then your counterexample doesn't work, right?
Hi! Thank you for your content, I love your videos they are very neat and organized. I was wondering what software do you use to write them on, if you don't mind.
@@micheldraybi3159 Thanks for the support! I use an app called “Write” on MacOS, and a graphic tablet
Thank you.This video is much needed for my exercise.
@@devjyoti5614 I’m very glad! 😌
This is very good lecture, thank you.
@@wargreymon2024 I’m glad you find it helpful! 🥰
I have a question: if fn converges to f pointwisely, if f'n converges to f' pointwisely and if f'n and f' are continuous, then, can we conclude that f'n converges to f' uniformly?
@@toopytoopy8547 Hi. I think not, for example fn(x)=(x^2)/n + x converges point wise to f(x)=x, both are continuous. But it doesn’t converge unif to f
@@Problemathic thank you. what if we add the assumption that f' is bounded? in this case, we have fn converges uniformly to f , right?
I have a question: if fn converges to f pointwisely, f'n converges to f' pointwisely and if f'n and f' are continuous, then can we conclude that the convergence of f'n to f' is uniform? Thank you in advance.
Hello thank u for your videos..can u provide us with a video about radin measure
Do you know why inner regular is not also defined as open set too? Or not defined as closed set?
For what it's worth I would appreciate more Munkres videos when you've got the time to make them.
@@williamwright1440 Thanks! Hopefully we will be able to continue with this reproduction list later 🥰 it’s good to know that it would be useful to people!
Constadinos Caratheodorys , a Greek Mathematician who helped Albert Einstein.
Thanks for the video
@@RavshanbekKodirov-w2c I’m glad it’s helpful!
can't thank you enough for producing these videos - saved me a LOT of pain this semester! (i don't think i like point-set topology very much 🤣) i think there's a small mistake at 6:30 - i think you mean x_0 \in X and (a,b) i f(x_0), and that you're using a characterization of continuity (Thm 18.1 (4)) currently it's written as x_0 \in [0,1] and (a,b) i x_0, and you state you're using the definition of continuity
@@counting1234 I’m glad they’re helpful! Thanks for spotting the error! 🥰
thanks for your video. It was helpful
@@mdakashmolla5253 thank you! I’m so glad 😌
What a great explanation, thanks !
@@Useremu I’m glad it’s helpful! 🥰
Looking forward to the next videos
6:12 I am really confused why X tilda = Union of Intersection of that set. We also have "for all k", so shouldn't it be equal to Intersection of Union of Intersection of the set?
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Thank you very much for your work here. You would deserve much more views and like! (Maybe only a little more volume at your agreeable voice 😉)
@@yopenzo thank you! Your support means a lot to me. Newer videos have better audio quality!
Thank you!
@@ivanrioscervino4369 I’m glad it’s helpful 😊
I think the definition of N_r is wrong. It should be the union of two sets which you defined.Not the intersection.
up!
thank you!
hmm, not sure about your proof for R^omega in the box topology - I think we do have to show that B is open. Also, I'm only seeing a theorem in Munkres that X is connected if and only if the only subsets of X that are both open and closed in X are the empty set and X itself, but I think in your video you present it without closed being a necessary condition.
Prove B is open is similar to A. Also, B is the complement of A, so A is closed if B is open
i have a question. G is not a sequence, so what does it mean to say μ(G) --> M?
That’s a good point, what I meant was that the measure of the union of Gn from n=1 up to N tends to M when N tends to inf
im wondering if you could perhaps make a video on how to identify the triangle inequality in weird situations and apply it? like an "advanced triangle inequality" video. i struggle with recognizing how to use it when it's not directly identifiable, like in p. 122 or p.125 in this chapter of metric spaces in munkres. thank you! (:
Nice ASMR
And also a good exercise and explanation.
@@TheNumberCruncher0913 thanks 😂
great video, once again!! thank you!! it takes me such a long time to read munkres and i feel concerned when i don't understand a small detail; you help me identify and comprehend the main storyline.
@@counting1234 thank you! It’s a lot if effort making this videos, it’s great news knowing that they’re actually helping people! 🥰
@@Problemathic Thank you so much. these videos are really helping me out.
@@Nate3145-zt8rh I’m so glad to hear that. Thanks for the kind words 🥰
i cannot thank you enough for your videos. my professor is extremely unclear, and i find munkres difficult to parse (and am frequently wanting more concrete examples, counterexamples, or images - as an example, i was not sure about what kind of objects his notation represented in his definitions in this section). your videos are a lifesaver in my topology course this fall!!
@@counting1234 Thank you for the kind words, I’m so glad you’re finding it helpful
I don't think the proof for part 1 is correct, you used the same x in two different variables.
Please explain exercises 3.26.1 please help tomorrow is seminar😢
Hola! Pregunta: en el minuto 05:00 ¿no es lo mismo decir que el complemento de tal conjunto B es abierto y eso probaría que B es cerrado? Digo, porque la elección de x0 es arbitraria y siempre vas a poder hallar un abierto Wx que es disjunto de B. Gracias!
Si! es básicamente lo mismo que hacemos en el video, solo que dicho con otras palabras 😊
I want to thank you about your great videos I found many things that are not found anywhere else only here ❤
Thanks! I'm so glad you find it helpful! 🥰
We can not take the case Ai's are countable except two uncountable sets A1 and A2 since the measure of the union will still 1 but the sum will be 2 so we must take one uncountable set not two .. ok why ? Is the reasons are the formula of the segma algebra & Ai's are disjoint ? I proved this using contapositive l said let Ai's are countable except two sets A1 and A2 where their completens are countable.. Since Ai's are disjoint then A1 is a subset of the completent of the union of Ai's (with A2) And if we take the complement of each side then we will replace each side I mean the union will be subset of the complement of A1 wich is a contradiction ( uncountable subset of countable is a contradiction)... the union is uncountable since it has A2 which is uncountabl and the complement of A1 is a countable from the formula of the segma algebra ..ok is it a logic reason or proof ? So there exist no case which say Ai's are countable except two ? If yes what about if we take all A are uncountable? .. we will have the same problem What about this case ?
Hi, exactly as you say in your comment and what I say in the video. There can only be at most one index j for which Aj is uncountable. So the only two possible cases are - All Aj are countable - All but one Aj are countable (and that other one is of course uncountable) So the case that you mention, with two uncountable sets is not a valid option, it never happens!
@Problemathic Thank you so much..lots of love ❤️❤️❤️
Hi why in proving countable union of measure le set is measurable you suppose the sets are disjoint?
she supposes the sets are disjoint when doing a countable union because its all thats needed, in particular since you can always make a countable sequence of sets a countable disjoint sequence of sets (just make every entry in the sequence subtract the union of every other member of the set)
nice
Thank you for the video. I had a question on the first part of the problem. I thought the wording countable implied that we also needed to prove that the basis was a countable one as well. How would you suggest we go about doing that part? Thank you!
Can you please make a video on extension?
@@HariwanMSalih What do you mean by extension?
I have just found it , I mean the caratheodory theorem ...thanks @@Problemathic
Well described , really that brought a clear idea about dictionary order topology. But I am struggling to understand why the dictionary order topology restricted to I=[0,1] is not same as the subspace topology on I×I obtained from the dictionary order topology on R×R...Can you please explain? I can understand that for inherited subspace topology there may be half-open intervals as it is just the intersection of an interval of R×R with the unit square... Is it for like the basis only consists of open intervals for this type of order topology ? Am I right?
@@debmalyaroy7639 Hi, thanks, I’m glad it’s useful! Regarding your question, this video might be helpful: ua-cam.com/video/t3_8cchermo/v-deo.htmlsi=eKucU1DPDUmOGby-
Urysohn’s lemma can also be used to prove the Portmanteau theorem.
@@爸爸到底-s9x Yes, it has lots of applications! It’s an amazing theorem 😍
Sound is very low...thanks
@@HariwanMSalih Yes, sorry about that, newer videos have better sound!
Finally a video where I get the intuition behind how the simple functions are defined. I was never able to understand the intuition till now. Thank you so much!
@@ayaneshlal7718 I’m so happy you find it helpful! Thanks for the kind words! 🥰
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in the first question's example...why is Y convex in X??
Very useful video and nicely put together. Gives the viewer a better intuition of the countability axioms by going through an example.
@@fadilawand4434 Thanks! I’m glad it’s helpful!
5:45 why it is inf Lim not the opposite Lim inf¿
@ 4:48 when we are considering other case, why are we doing "there exist for some..." when in first case we just start that E_i are countable for all i? Is it just there to eventually force the form (U E_i)^c is countable? Since, in the first case, we already forced (U E_i) is countable by starting all E_i are countable? Proving two case since for something to be in A: (something is subset of X AND something is countable) OR (smthing is subset of X AND smthing-complement is countable} where, in our case, that Something = (U E_i). Lastly, in your video, are we to assume fancy A is subset of Powerset of X? so Since we claimed E in A earlier. E is also subset of X. So union of subsets of X is still subset of X so (U E_i) is subset of X, satisfying first part of each 2 cases? I ask this because I think you only briefly mentioned in the last video fact A is subset of P(X)... sorry for my English.
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