Lebesgue Integral Example

alright thanks for watching and as promised today I will give you a specific example of a function for which we can calculate the lebesgue integral just to illustrate how the process of lebesgue integration works this is what's called the contour ternary function or the devil staircase function and you'll see that has nice property it's very non-intuitive but this is not why we're here today we're talking about the interval test case so if you know contr stuff okay it has to do with splitting up the interval 0 comma 1 into 3 pieces so let's define FS wallet on the counter instead if you don't know what the Cantor set is don't worry about it give you an explicit definition as follows so f of X equals to the following if you take the interval 0 comma 1 and chop it into 3 pieces you get 0 1/3 2/3 one on the middle piece define F to be 1/2 1/2 on the first count of 7 so on 1/3 2/3 and then continue on the first on the interval 0 comma 1 3rd divided up into 3 pieces so we get 192 9 and here on the middle piece they find F to be 1/4 ah one night two nights and well this is already taken but on this interval you can use the same speed so seven nine eight nine and here because we want the function to be non decreasing define it to be three quarters here three quarters oh seven nine eight ninths and then you can just continue for this interval split it up into two pieces and define your function to be one eighth and then here also split it up into two pieces define your function to be three eighths also while this is taken here split it up into three pieces they find it to be five eighths and finally here split it up into three pieces and define it to be seven eighths so it's one eighth on if you want the interval one over twenty-seven to over twenty seven then three eight second next interval five eggs and then seven eighths and so if you continue this process you get this weird thing you know this what's called the devil staircase which is a function that just say the devil is happy you know it makes you go up stairs that are like infinitely high right you go infinitely many steps and in general on the interval 1 over 3 to the N 2 over 3 to the N that becomes a 1 over 2 to the etc etc sweet and you get the next one is three over two to the end that the top up to two to the n minus one over two to the N so I know - 2 T and - wanna be on the very last piece so that's a technical definition but you know we get this very artistic picture and the question is once you have this graph what is the integral so question for today what is integral 0 to 1 of f of X DX other words what is function 3 colors look at the area below this graph what is this area okay and well good luck doing that with the Riemann integral I'm not even sure if it's integral boy in the riemann sense but using loopback integrals is actually much much easier to do because remember how you build up your back integrals you just start with two simple functions which are like indicator functions and you just take a limit of indicator functions and this is perfect because we actually have even though this graph is very weird it's actually an infinite sum of indicator functions that's why it's very prone to the back integration before I forget this function is also known for something else because notice that every step the derivative is zero so this is a function for which the river is 0 almost everywhere but you see this function is definitely not constant so that's what an analysis we like this function simple functions and equals to the limit as n goes to infinity of where exactly what I said you build the functions in steps namely F of 1 you start by saying it 1/2 times the indicator function of 1/3 2/3 in other words just as I said this is the interval 0 comma 1 and you take the first third the middle third and you define the value here to be 1/2 then in fact it was the first step in our construction and this if you like this is your F of 1 so f 1 and then we can just build up you know our function here F 1 so it's still 1/2 that's an indicator function of 1/3 2/3 but now you add this one quarter 3/4 right so 1/4 times the indicator function of one night two nights plus 3/4 times the indicator function of seven ninths eight nights so it looks again something that you have the interval 0 comma 1 you take the first third 1/3 you defined it to be one half here and then on the middle third here you let it be one quarter and then here you let it be three quarters so that's your f2 and you'll see if you continue this process you actually find that you know you do obtain The Devil's Staircase box here staircase formation why is that nice remember one of the steps of lebesgue integral equation says that the integral of F is really just a limit of the integrals of this FN and you know or you can justify this with its dominated convergence theorem because those functions R is less than 1 as it's on the interval 0 comma 1 ok so what I said my definition of the lebesgue integral is by definition the limit of the integral okay but now let's just calculate the integrals but what is the integral of F 1 remember it's an integral 1 half and the indicator function of 1/3 2/3 and that's 1/3 times 2/3 minus 1/3 sorry 1/2 times 2/3 minus 1/3 which is 1/2 times 1/3 and I don't simplify it like that because we want to find a pattern so f2 would be in this case integral again 1/2 times indicator function of 1/3 2/3 plus 1/4 times the integral indicator function 9 to 9 plus 3/4 times an indicator function of 7 9 8 9 and that equals to 1/2 times 1/3 plus 1/4 times 1/9 plus 3/4 times 1 minus because for the first interval the length is 1/3 the second one the length is 1/9 but in fact just for the next step let's group the intervals by the order and what I mean is here things is in terms of 3 so let's group all the things with 3 here the level if you want is 9 so let's factor out 9 from everything so it's still 1/2 times 1/3 plus in this case here the functions is 1/4 so 1/4 times 1/9 so this is the value because we have value 1 over 4 3 over 4 and this is the length of the interval 1 9 and then we just have 1 plus 3 evaluate this but if you follow this pattern and I think by the definition then use the definition of f of F we can show the following integral FN is again 1/2 times 1/3 times 1 and then plus 1/4 times 1/9 times 1 plus 3 and then just follow this pattern then the next step is the intervals is 1 over 3 to the N or it is 3 to 1 over 27 the value of n becomes 1 over 8 no 3 over eights 508 7 over 8 so it's really 1 over 2 to the third power times 1 plus 3 plus 5 plus 7 and then if you look at the pattern the last thing then becomes 1 over 2 to the N times 1 over 3 to the N times 1 plus 3 plus 5 plus 7 up to the last term if you remember the definition it's 2 to the N minus 1 over 3 and now let's try to simplify this gibberish so this is our answer in terms of n let's try to simplify that what happens if you sum up 1 plus 3 plus 5 plus 7 plus dot dot dot plus 2 L minus 1 it really becomes I know I guess 2 times 1 minus 1 plus 2 times 2 minus 1 plus dot dot plus 2 times L minus 1 and that's if you write it in Sigma notation it's something 1 up to L of 2 L minus 1 sorry sum from 1 to L of 2 I minus 1 and that becomes 2 times sum from I to L of I minus so this is once sum of I from 1 to L and then there's a nice fact that you should know if this becomes 2 times L times L plus one over two and that's just because L and so u of l squared plus L minus L and that's ever like the magazine elsewhere Time magazine okay therefore all this junk here can actually be simplified much much more nicely because then we have the following integral FN then becomes we still have 1/2 times 1/3 times 1 plus 1/4 times 1/9 times so 1 plus 3 here l equals to 2 l equals to 2 so we have 2 squared and then if you have here L equals to 3 no no L equals to 4 here so we have 2 4 squared plus if you want 1 over 8 times 1 over 27 times 4 squared but just to make stuff easier later on for let's write it as 2 to the 2 so if you wanted to to the 4 freely and then you just continue and you left with 1 over 2 to the N 1 over 3 to the N and well what does that correspond to that's really 2 times 2 to the N minus 1 minus 1 so L is 2 to the N minus 1 so what this becomes is 2 to the N minus 1 squared yes which here becomes 2 to the 2 and minus 2 let me erase that gibberish ah to make this an even more suggestive let's do that it is 1 over 2 to the 1 1 over 3 to the 1 and then this you can write 1 is 2 to the 2 times 1 minus 2 ok good us 1 over 2 squared 1 over 3 squared 2 times because we want to write this in this form right 2 times 2 minus 2 plus 1 over 2 power 1 over 3 cube 2 times 2 times 3 minus 2 and then hopefully now you see the pattern so it's 1 over 2 to the N 1 over 3 to the N 2 to the 2 n minus 2 so now we can write the integral as a song in Sigma notation the song from K equals to 1 ok to n of 1 over 2 to the K 1 over 3 to the K 2 to the 2k minus 2 now we can simplify this a little bit and by the way that's why I didn't write 1 over 6 to the K so k equals to 1 to N of 1 over 3 to the K 2 to the 2k minus 2 minus K and then equals the sum from k also 1 to N of 1 over 3 to the K 2 to the K minus 2 let's hope there is no decay here ok and then this 2 to the minus 2 we can pull it outside so it's 1/4 some k equals to 1 to n of 2 to the K over 3 to the K that's two-thirds to the K and remember what is this gibberish it's the integral of F an integral as an yes I guess from 0 to 1 equals to 1/4 sum from k equals to 1 to N of 2/3 okay all right and now what you know will actually reach your final answer what is the integral from 0 to 1 of F well it's by what I said if the limit as n goes to infinity of this junk into your from 0 to 1 of f n of xdx which becomes in this case simply the sum from k to 1 to infinity of 2 thirds K but this just becomes a geometric series so almost because this is 1/4 times 2/3 plus 2/3 squared and therefore just factor out at 2/3 so 1/4 times 2/3 times 1 over 1 minus 2/3 and that's because if you want 1/6 if you want 1/6 times 1 over 1/3 and that becomes 1/2 tada so again challenge sure you try to do it through the riemann integral which is I think a bit harder but using them the bag integral it's so elegant enough that we can just say that the integral of F is just the limit of the simple functions one little thing though there isn't oh my god way of doing this and then we quickly present this to you it turns out if you look at the definition of f you get that F of 1 minus x is in fact equal to 1 minus f of X so in case you're wondering what kinds of functions satisfy this well then now integrate integral from 0 to 1 of F of 1 minus X DX becomes integral of 1 over the interval which is 1 minus integral f of X DX but now by a change of variables if you're like you the 1 minus X you can show that it's the same as integral from 0 to 1 of f of X DX and so this is 1 minus integral from 0 to 1 f of X DX and so you literally get an equation for this unknown and if you solve for it you've got the interval from 0 to 1 f of X DX equals to 1/2 oh it's nice but it doesn't really straight to the begging to grow much you know this is nicer I think all right so if you like this you know Rebecca integral why gosh and if you want to explore Mormo please make sure to subscribe to my channel thank you very much

Channel: Dr Peyam

Views: 32,001

Rating: 4.9081516 out of 5

Keywords: lebesgue integral, lebesgue, simple, step, math, limit, sum of odd integers, symmetry, analysis

Id: kb0MZr0D-zc

Length: 22min 52sec (1372 seconds)

Published: Mon Feb 12 2018