Correction at 9:40. If we integrate g(x), then we can interchange the sum with the integral do to the non-negativity of each term in the series. This gives us SUM 2^n Measure(E_n).
@NewCalculus
4 ай бұрын
Lebesgue measure is a load of incoherent rubbish.
@davidsnyder6898
4 ай бұрын
Analysis : A course on applications of the Triangle Inequality lol, still one of my favorites.
@Freddy-ss7mb
5 ай бұрын
Your videos inspire me so much to pursue a PhD, thanks for existing!!
@jamesromano3288
5 ай бұрын
Yes....and you will be teaching inner city high school kids how to count to 5,,,,,,
@shivanshuojha1307
4 ай бұрын
@@jamesromano3288maybe in mathematics, but definitely not in engineering.
@abdulsalamyussif6391
5 ай бұрын
I would be very grateful if you could do REAL ANALYSYS videos for beginners. I like your presentation and most probably you'll teach us to understand
@user-qb8fp8oj1p
5 ай бұрын
Good job ❤Professional 👍Outstanding 💪😍Terrific 😘🤩🌹
@larryyonce
5 ай бұрын
Nice. You are good at explaining proofs. 👍
@koled224
5 ай бұрын
Been working through the more interesting problems in Evans Pde and it's interesting to see the different ways these same results are used
@OOobstkuchenOO
5 ай бұрын
The second problem is pretty nice. I think you could make it a bit prettier by having the sum on the right start at - infinity instead of zero, then you can drop the requirement that your measure is finite. (and for finite spaces the conditions are equivalent)
@retr0foxx_osu
4 ай бұрын
very cool, i only managed to somewhat understand the solution to the first problem. i am self studying math and is still a beginner in proof based math and when i watch your videos, it always reignites my motivation to continue self studying real analysis (i am using stephen abbott's understanding analysis book) so yeah thanks for the video !!
@boyankugiyski
5 ай бұрын
9:40 - this is not due to the definition of the Lebesgue integral. It is, in fact, true that you can switch this infinite sum with the integral because all of your functions are non-negative, i.e. you can apply the monotone convergence theorem and you're done. But I stress again - you cannot justify the interchange of integral and infinite sum by just saying that it follows from the definition of the Lebesgue integral.
@Heuyy123
5 ай бұрын
Exactly what I thought
@Heuyy123
5 ай бұрын
Koled. Here ist what boyan meant: the characteristic functions (and the multiplication with powers of 2) in the definition of g are non-negative and thats why the sequence of partial sums is increasing. So you can apply the monotone convergence to the series (i.e. limit of sequence of partial sums) I completely agree with Boyan: Saying that this follows from definition ist rather questionable, especially from a grad student in analysis
@boyankugiyski
5 ай бұрын
@@Heuyy123Exactly.
@koled224
5 ай бұрын
Huey123 OK, I deleted it. Been a while since I took real analysis lol. Would you not also be able to apply the dominated convergence theorem (albeit with extra steps)?
@lugia8888
4 ай бұрын
@@Heuyy123 its alright we know what he meant
@KurtGodel-po3zl
4 ай бұрын
Very interesting video. Thanks! Btw, what pen are you using? Would you recommend it?
@ProfessorNoobster
5 ай бұрын
The saga continues!
@mini059
5 ай бұрын
Goood job 💙but please do linear algebra next video 🙏
@rotem1201
4 ай бұрын
Just a random thought, but have you ever considered talking about some non-math classes you took in college or grad school? It would be cool to see what other subjects you're into outside the number world. No pressure though, just curious! As always love the content
@Math.A-level-student
5 ай бұрын
nice video
@mathematicsforfunforever
4 ай бұрын
question for the first problem: you get that |x - 1/2| is less than delta, so you can make |f(x_n) - f(1/2)| arbitrary small. Would a convergence argument be equivalent, where since f is continuous as a limit of a uniform convergent sequence of continuous functions, we get f(x_n) = f/(1/2) when n goes to infinity, therefore if we chose N large enough we could make the distance |f(x_n) - f(1/2)| arbitrarily small? I get that you would show the same thing, but in previous solutions for problems like this that I have seen, they argue that the distance is arbitrarily small by making x and x_n close, and using continuity, instead of finding a large N. Hope my question wasn't too confusing or stupid, but what I am asking is, does it make any difference as to which argument I go for in that case?
@mathematicsforfunforever
4 ай бұрын
also, in the second problem. Would it be possible to use Markov's inequality so solve it somehow? Just a thought.
@anshsehgal7043
5 ай бұрын
Wassup bro
@Griezmanndagoat
5 ай бұрын
1:57 this problem is trivial using the Uniform limit theorem.
@strikeemblem2886
5 ай бұрын
How so? That theorem only concludes that f is continuous. I do not see how the answer can be simplier than the one presented.
@baobin82
5 ай бұрын
@@strikeemblem2886if it converges uniformly isnt the limit function always continuous ?
@strikeemblem2886
5 ай бұрын
@@baobin82 Yes, but that is not what you are asked to prove.
@xhumanoid5116
5 ай бұрын
Nice video. I only understood the part where you said 0+0+0+0...=00000 lol... Ah, I wish I had a hundred thousand years so I could understand math more.
@Heuyy123
5 ай бұрын
19:30 Dominated convergence theorem would work if g_n convergences pointwise a.e. to zero. Here you merely have convergence in measure which doesnt imply pointwise convergence a.e. so the proof doesnt work
@aimsmathmatrix
5 ай бұрын
But that's exactly the assumption made? If g_n convergences to zero, then we have convergence in measure, not the other way around.
@Heuyy123
5 ай бұрын
No the assumption is convergence in measure, which is denoted by the m above the arrow.
@Heuyy123
5 ай бұрын
He also says in the video that the assumption ist concergence in measure
@PhDVlog777
5 ай бұрын
You can use DCT with convergence in measure. Just do a quick google search :)
@aimsmathmatrix
5 ай бұрын
@@PhDVlog777 oh yeah, my bad; was super tired, head was elsewhere. Love the content! Also, quick question. When did you start with material covering measure theory in your academic career? Curious as someone from another country / continent entirely. It's my second year (third semester) as a maths student and we started our degree with real analysis in the first semester, up until now, currently doing measure theory in the third. I just passed that exam as well! Your content has been nice and relatively soothing to watch!
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