Lecture 10: Simple Functions

Key Takeaways

all right so let's uh continue with our discussion of measurable functions so last time we introduced the notion of measurable functions so if i have a measurable set e from or f from e to the extended real numbers is measurable or is a measurable function if for all alpha and r the inverse image of the half infinite open interval alpha to infinity is measurable right um and then we prove that uh you know if we have a function which is measurable then not only does is the pre-image of these intervals measurable but the pre-image of any borrel set any member of the burel sigma algebra which includes open sets closed sets and so on the pre-image of those sets is also measurable um and we proved that being measurable is closed undertaking soups imps limb soups lim imps limits in particular and changing the function a little bit on a or changing a function on the set of measure zero also preserves uh measurability and also being measurable is closed under the algebraic operations of taking uh linear combinations and products okay now this definition and properties we've worked out was for extended real valued functions you know quite often we'll be dealing with or in general functions that take values in the complex numbers so let me define um what it means for a complex value function to be measurable it's not too crazy let e be a subset of r be measurable we say a function f from e now to the complex numbers is measurable if the two functions given by the real part of f which now goes from e to r and the imaginary part of f which now is a function from e to r are measurable okay so function into the complex numbers you can always write as f equals real part of f plus the imaginary part of f times i so we just to say that f is measurable just require that it's real and imaginary parts are measurable okay um then you can verify the following simple theorem maybe i'll put it on an assignment or not i haven't decided yet or just parts of it if f g were measurable and i didn't say so but if we're talking about measurable functions the domain e the domain is always assumed to be a measurable subset of r so if these are measurable functions and alpha is a complex number then the functions alpha times f f plus g f times g and then we can do a few other things to complex numbers that or change it the complex conjugate of f and modulus of f are measurable functions okay um okay so and then we have the following theorem if fn from e to c is measurable for all in and these functions converge point wise to function f then the limiting function is measurable okay and you know this follows again so both of these theorems follow immediately from what we know about measurable extended real value functions so for example this one here follows from the fact that limit n goes to infinity of fn of x equals f of x if and only if limit as n goes to infinity of the real part of f x f n of x equals the real part of f of x and the limit as n goes to infinity of the imaginary part of f n of x equals the imaginary part of f x okay and if we're assuming fn is measurable for all n then the real part and imaginary parts of fn are measurable for all n and therefore they're point wise limits which is the real part of f and imaginary part of f are measurable by the theorem we proved about extended real valued measurable functions and so we conclude that the real part of f and the imaginary part are measurable and therefore f is measurable okay so you don't have to work very hard you can just use what you know of what we proved from from the previous lecture about extended real valued functions which are measurable [Applause] all right so now as far as measurable functions go we've shown that that if i have a continuous function then that is so this is from the previous time from previous lecture continuous functions are measurable and if i have a measurable subset of e then the indicator function of that set is measurable and we know that linear combinations of measurable functions are measurable so if i take linear combinations of indicator functions with say complex coefficients then that will remain measurable as well and those are func and those functions are kind of the simplest type in that they only take finitely many values okay they're so simple that we uh glorify them by giving them that name and we'll show that they're kind of uh you know that every measurable function is in a sense approximately a simple function so we have the following definition if e is measurable the measurable function fee from e to c is simple or we call it a simple function so we'll say is a simple function if the range of e phi of e is equal to finitely many values okay so a measurable function is a simple function if its range is finite so let me make a general remark about simple functions and also when i write that ph of e is equal to a1 up to a n you know this is a set um and when i write a 1 a 2 up to a n i am you know sort of implicitly writing here i'm not saying the set i'm saying each of these elements are different from each other okay so you know a simple function that just takes the value one i would not write its range as one one one one okay so although that's a kind of simple and silly uh remark to make i just wanted to make it now so v is a simple function then we can write it in sort of a canonical way with so if p is a simple function with phi of e equals a 1 up to a n then for all i the set a sub i which is equal to the inverse image of so here phi is going from e to c the inverse image of the single set singleton which is a closed uh well this is going to be so the inverse image of this guy here is measurable because it's you know it's equal to the intersection of two measurable sets it's equal to the intersection of when the real part equals the real part of phi equals the real part of a sub i intersect the set of all x's where the imaginary part of phi equals the imaginary part of a sub i okay so that's why this set is measurable is uh measurable and we have a few properties of this guy if i take two different uh elements in the range again this is sort of why i made that comment that when i write the range this way i'm not i'm listing the distinct elements of the range or the image of e for all i equals not equal to j these two sets are disjoint and if i take the union of i equals 1 to n of the ai's this is equal to the total set e okay because this is just the inverse image of this set which is just e since uh u v equals that okay and finally for all x and e we can write p of x as sum from i equals 1 to n of a i chi of a sub i of x okay so for a simple function i can write it in a canonical way where it's just a uh linear combination of indicator functions where the sets that those indicator functions are non-zero on are disjoint from each other okay and their union uh gives me uh e the domain okay so these three are kind of the simple but important properties of how to represent a simple function so it's not difficult to verify again just from the definition of a simple function that you know scalar multiples linear combinations and let's say in products of simple functions are again simple functions okay okay um so i've said that these functions are so simple that that's what we call them uh and that they are somehow um universal that in a sense any measurable function is almost a simple function so what in what sense do i mean that so that's the content of the following theorem which is that uh if f from e to and let's first uh we're going to do this for extended real valued functions and then the proof will essentially carry over to the complex value so i'm just going to do it for the extended real valued non-negative functions and again i'll indicate what the difference is once we go to complex valued measurable functions so if e is uh from so now this is extended real value but it's non-negative okay so if this is a measurable function then there exists a sequence of simple functions fee sub n such that three things hold um we can do this let's write down let's go over here three things hold roll x and e or let me hold off on stating that f dominates these simple functions and these simple functions are point wise increasing for all x and e 0 is less than or equal to phi 1 of v 0 of x is less than or equal to v1 of x is less than or equal to v2 of x and so on and they all sit below f of x okay for all x and e these fees are converging to f of x so a and b says that there exists a sim uh a sequence of measurable of simple functions that increase to f of x okay the last part is that if this function if this measurable function is bounded or wherever it's bounded this convergence is not just point wise but uniform for all b bigger than or equal to zero the sequence c v n converges to f uniformly on the set where f is bounded okay so again the take home is that for every non-negative extended real value measurable function we can find a sequence of simple functions that well approximates f okay and in what sense does this will approximate f they increase to f and if f is bounded then that convergence is in fact uniform or you know more precisely wherever f is bounded convergence is uniform okay so this is in what sense every measurable function is um you know almost a simple function all right so let me let's get started with the proof and first i'm going to draw a few pictures so that you can kind of get the idea of how we're going to do this or how we're going to construct the sequence of simple functions um and then you know turn that into math which might be a little bit jarring if i went that route first and then drew pictures so uh and this picture i'm going to draw kind of looks like what i was talking about where we split the domain the range up rather than the domain when we were motivating why we would you know even introduce the concept of a measurable function so let's say we have um our function f now how i'm going to build these fees are that what i'm going to do is so here i'm going to draw v0 now there's a yellow shot what i do is this will indicate kind of the power of two how high up i'm going and also the resolution how much how much am i dividing how high up i'm going uh into smaller parts so fee 0 you should think that my height will be well my height is going to be 1. and so what i do is how i define this this simple function phi 0 is i look at where f is above the final value 1 and my simple function will be that final value 1 there okay and let me move this over a little bit and that just leaves you know where f is less than one and there i set the value of my simple function to be uh zero and zero simply because that's the the lower uh bound or that's the the smallest value of this interval zero one okay and that's how i um define v0 it goes up to height one and i only split each um you know each part into one so there's only one part here now maybe some of that didn't make any sense to you that's okay we're going to drop e1 and then i'm going to stop there because this is going to grow exponentially which means i'll probably draw an exponentially worse picture each time so now we're going to draw feed two and what i do is there's one feet or feet one sorry one again should indicate the power of two that i'm both going up and resolving uh the axis so now i go up to let's make this a so again this will not be quite to scale but hopefully that's okay um all right so let's let's make this a little closer to scale okay so one now i have two parts two to the one so this is um parameterizing how the tallest i'm going in and cutting up the the range of my function f and then i'm going to now resolve each part in half okay 2 to the minus 1. so now i take this is now 3 halves and this is now one-half two to the minus one okay so why am i writing two two to the minus one again because this one here corresponds to and that one here corresponds to i'm cutting the whole uh increments into halves if i go on to feed two i'm going to be going up to two to the two which is now four and from four to three three to two two to one i'm going to be cutting those into fourths okay and again what i do is i look at i cut now i look at where the function is in these widths so if it's above my highest bound then my simple function will be uh 2 on that set of x's where it's at its highest where f goes past the largest number which i'm resolving the axis in okay and then now here for example uh so you should try and draw your own picture and not just go off of mine because again mine is looking kind of rough already but now i look at the function the the set of x's where the function is between two and three halves right so that's going to be um you know this piece and this piece and then i set my simple function to be equal to the value the lower bound on this interval that i've cut the range into okay and then i do that from three halves to one and for example that's happening here on this x and i set it equal to i set my simple function equal to the again the lower bound on this uh interval that i've cut the range up into which is one and so and so on okay and now uh the function between one and a half it's this is kind of only the last piece everywhere else is filled in and there's no f in between here so i don't assign a a value there okay all right now that was me talking my way through how this function how the sequence of functions look now i'll just write down what these functions how these functions are defined so for so for n equals 0 1 2 and so on for k between 2 to the 2n minus 1. i define sets e k n this is equal to the set of all x's and e such that f x is between k times 2 to the minus n is less than or equal to k plus 1 2 to the minus n okay which is just me explicitly writing out that this is the inverse image of k times 2 to the minus n k plus 1 2 to the minus n closed okay and this is a an interval and since f is measurable assumed to be measurable this is a measurable the inverse image of that interval is a measurable set okay and then i define fn to be the inverse image of when f uh exceeds my top value of how i'm cutting up the the range two to the n okay again which is measurable and finally i will take um my simple function v n to be sum from k equals 0 to 2 to the 2 n minus 1. k times 2 to the minus n which is again this is the lower this would be this would correspond to the lower part of my e k n that i'm looking at so for this example uh if this e k n is the two three halves and three halves is the lower part times uh chi e k n plus 2 to the n times where f exceeds 2 to the n okay so i encourage you to maybe write out fee one okay what that actually is i mean you know there's uh in fact here i'll do that you know i drew the picture that goes over here let me write out what v1 actually looks like v1 is equal to 0 times the indicator function where f is between 0 and 1 half plus a half times the indicator function where f is between one half and one plus three halves times the indicator function of when f is between uh no that's not three halves that should be one half uh no no what am i doing one times the indicator function where f is between one and three halves three halves times chi f inverse of where f is between now three halves and two so 2 is you know how high up i break up the range and then plus this last part this fn part which is 2 times chi where f the indicator function of where f is bigger than 2 to the n so bigger than 2. okay so um you know all of these sets are disjoint the e k ns and fns for k different from k prime and fixed n is are disjoint and uh and they are disjoint from this set so you know this is a simple function the finitely many values it takes is 2 to the n along with k times 2 to the minus n or at least the finitely many values it can take is a subset of that so this is a simple function for each n and by design it's always sitting below f right so let me in fact bring it up here well let me make the statement and then i'll say so by definition for all x and e b in is non-negative and always sits below f x right so how do we see that um you can see it from the general formula but i'll just indicate why just for say fee one let's say so x has to be in one of these sets let's say it's here in uh where f is between one half and one then phi one of x is equal to one half which is less than f of x because f takes on the value between one half and one right okay in fact here i'll give a brief argument here if x is in e k n then by definition this means that k 2 to the minus n is less than f x is less than or equal to k plus 1 2 to the minus n which implies that v n of x which is by definition k times 2 to the minus n just again by how we've defined the feast of ends where did we define the fisa bins there is less than f of x okay okay so that's for x and e k n and if x is in f sub n then that means f of x is greater than 2 to the n uh which is always bigger than or equal to phi n of x well i mean this is actually equal to phi n of x all right so we always have the fiends are non-negative and they always sit below f of x so now let's prove that they are in fact increasing so part a okay so now we're proving part a that the fiends increase to f so and again to to since the for fixed in e k n and f n and the e k ends and f n uh form a disjoint union of e i just need to check that what i want holds uh on each of the ekns and fn so suppose x is in ekn then f of x is less than or equal to k plus 1 times 2 to the minus n and it's bigger than k 2 to the minus n um and which by a silly trick of just multiplying and dividing by 2 tells me that f is between 2 times k times 2 to the minus n minus 1 is less than f of x is less than or equal to 2 k plus 2 times 2 to the minus n minus 1 which implies that x is in the union of e 2 k n plus 1 right 2 k n plus 1 union 2 k plus 2 and plus 1. if x is in e2 k n plus 1 then i get that phi n of x is equal to by definition k times 2 to the minus n which equals 2 k 2 to the minus n minus 1 which because x is in e 2 k n plus 1 this is equal to the n plus 1 of x and if x is in e 2k plus 2 n plus 1 um no i shouldn't this shouldn't be a 2 this should be a 1 i'm sorry because this goes from 2k up to 2k plus 1 and then from 2k plus 1 up to 2k plus 2. okay so uh it's either in e2 k or 2k plus 1 and if e is in a 2k plus one then vn of x is still i mean x is still in uh e k n so phi n of x is equal to still k times two to the minus n which is equal to two k times two to the minus n minus one which is less than two k plus one times two to the minus n minus one which is by definition since x is in e2 k plus 1 n plus 1 n plus 1 of x okay and similarly if x is in fn then n of x is less than or equal to the n plus one of x okay so we verified for all x uh since e is equal to this union over k equals 0 2 to the 2n minus 1 e k n union f n this implies that for all x and e v n of x is less than or equal to v n plus one of x okay all right and this proves a now how to prove b and c we'll prove these things will follow from part a and a simple estimate that we're going to prove so b and c will follow immediately from the following claim in part a which uh the claim is that um for all x in the set y and e that f of y is less than or equal to 2 to the n this part we already know that f of x minus v n of x is non-negative but in fact this is bounded by two to the minus n okay right so then b and c from a and this claim okay why does b follow from from this claim well you know um wherever x is so let's i don't want to have to also explain what happens if f is you know equal to infinity that also follows um basically from the definitions not necessarily from this estimate but the more important part is when f is finite uh so let's assume f is just finite uh you know for every x so then every x in e is eventually in one of these sets right so um there exists let x be fixed then for n sufficiently large x is in one of these sets okay since f of x is finite um and therefore for all ends of it for capital n uh for all uh capital n sufficiently large this minus v n of x is less than or equal to 2 to the minus n but then f of x minus v to the m is also less than or equal to two to the minus n for every m bigger than or equal to n right because the phi ends are increasing yeah if v n is this close to f then phi m is also that close to f if m is bigger than or equal to n right again because they're increasing all right so that proves um that's why point wise convergence b follows from from this estimate as far as part c that also follows from this estimate since um you know if i have a fixed b then i can choose a natural number just depending on what b is so that that set depending on b in the original statement i'm pointing at it but i don't think you can see it from the camera that set which depends on b is contained in one of these sets and therefore for all x in the set where f is bounded by b this holds uniformly in x and that's where the uniform convergence comes from um but the whole point is that this is the estimate that that gives us b and c once we've proved a all right so let's prove this claim it's not hard it's basically because we are cutting up the range not only to height 2 to the n [Applause] but with resolution 2 to the minus n at each stage at each end right so to prove the claim we have that the set of of y's and e such that f of x is less than or equal to 2 to the n this is equal to the union k equals 0 2 to the 2 n minus 1 of the e k ends okay so if i want to check that bound i just have to check it for each if x is in one of these okay so suppose x is in e to the k n then i mean it really is just following from the fact that we're cutting up um the range uh with resolution two to the minus n right so i'm just going to draw a small little picture here we have um the axis here and here's k plus 1 times 2 to the minus n here's k times 2 to the minus n if x is in there um that means the there's a we're looking at the portion of f that sits between k 2 to the minus n and k plus 1 2 to the minus n then k times 2 to the minus n is less than f of x is less than or equal to k plus 1 2 to the minus n and remember the simple function on this piece evaluated in here so here's a x gives the value at the lower bound right so and therefore we get that f n of x minus phi n of x this is equal to not fn sorry f of x minus k 2 to the minus 1. now x again is an e k n so f of x is between these two numbers so this is less than or equal to k plus 1 times 2 to the minus n minus k 2 to the minus n and this equals 2 to the minus n all right okay so whenever we have an x in one of these sets it has to be within 2 to the minus n the simple function evaluated at that x has to be within 2 to the minus n of f again this is just by construction by how we've cut up the uh y-axis if you like if the range okay we're cutting it up into res with resolution uh two to the minus n all right so that proves the claim which as i said along with part a proves b and c so um okay so that proves that you know every measurable function is almost a at least non-negative extended real valued measurable function is a limit of a sequence of simple functions now this theorem carries over without difficulty to complex valued functions after i just introduce kind of a breakup of a function in general so if e is a function from minus infinity to infinity so an extended real valued function now we define its positive and negative part f plus s f plus of x to equal the max of f of x 0 so this is the positive part of f f minus of x is equal to the min of the max i'm sorry minus f of x 0 this is the negative part of f all right so what about these positive and negative parts then f is equal to f plus minus f minus okay you can just check that i mean take any x if f is positive or non-negative on that then uh i get f of x out if it is negative then i will get minus f of x which is the absolute value times minus gives me back f of x okay and the absolute value of f is equal to f plus plus f minus okay so let me make a comment that you know if i this is just a definition for you know an arbitrary function from e to the extended real numbers if f is measurable then each of these functions is measurable right because this is if you like the supremum of the sequence of functions given by f and then 0 afterwards and this is given by you know the supremum of functions given by minus f and 0 afterwards all right so um if f is measurable it's positive and negative parts are also measurable functions okay and they're also non-negative okay so maybe that wasn't uh clear or at least it should be clear this is the max always involving zero so it's always non-negative okay so now the construction we did a minute ago essentially carries over to the case of complex valued measurable functions so let e be measurable and f from e to c be measurable then there exists sequence of simple functions v in such that kind of analogs of those three properties hold a is for all x and e these are increasing in in modulus an absolute value 0 is bigger than or equal to of course phi 0 of x is less than or equal to the absolute value of v1 of x less than or equal to the i say absolute value in modulus same thing is less than or equal to the absolute value of f of x [Applause] um these fee ends are converging to f point wise and finally convergence is uniform on sets where f is bounded so for all b bigger than or equal to zero all right say a minute ago oh vn converges to f uniformly on the set x in e such that the absolute value of f of x or the modulus of f of x is less than or equal to b okay so this theorem follows kind of immediately from the previous theorem because now what i do is i just take f i split it into its real and imaginary parts and then i split the imaginary the real and imaginary parts into both positive and negative parts so um i will leave it to you to to actually fill in the details but uh you apply the previous theorem to the real part of f plus or minus which are now non-negative measurable functions and the positive and negative parts of the imaginary part of f again which are non-negative measurable functions and then you just take linear combinations of these um simple functions that you know add up to f okay you know you will take the sequence of simple functions corresponding to the real part of f and subtract the sequence of simple functions that you got for the minus for the negative part of f real part of f and you will take that and add i times the positive part or the sequence of simple functions converging to the positive part of the imaginary part of f minus the sequence of functions simple functions converging to the negative part of the imaginary part of f okay so that's what i mean by apply previous theorem to the positive and negative parts of the real and imaginary parts of f okay so what's the significance of this theorem not only um let me just say if you know also showing that somehow measurable functions are well approximated or almost simple functions this also gives us a way of maybe defining the integral at least of non-negative functions that way we don't have to deal with possible deals with the in you know subtracting infinity from infinity but uh by simply defining uh its integral to be the limit of you know these integrals of these simple functions and for a simple function we would you know presumably know how to define an integral it would just be the numbers time the measure of the sets that uh appear in the indicator functions for these simple functions you know and like i said last lecture kind of if you wanted to define the lebeg integral that way you would run up against well you know does this number depend on the sequence of simple simple functions you chose to approximate f but we're not going to define the labeg integral in that way we're going to define it a little bit differently which is what we're going to move on to now which is the labeg integral of a non-negative function and then we will uh define the le vague integral or la greg labeg integrable functions these will be complex value functions now for which we can define an integral four and that is the the full theory of and that's the you know end of the um end of the game for as far as defining lebeg integral and then we'll prove some um convergence theorems along the way which make the le big integral stronger than riemann integrals so now we're moving on to the le vague integral of a non-negative function okay and why start with a non-negative function again because you know i just pulled this trick on you a minute ago that if somehow we know how to do stuff for uh non-negative measurable functions then by playing this game where we take the real and imaginary parts and split it up into positive and negative parts hopefully we can do something for general functions so that's why we start out with um introducing the or defining the labeg integral of a non-negative function so definition if e is a measurable subset of r we define l plus of e this is the set of all extended real valued functions non-negative real value functions that are measurable okay and now of what the goal is is to define the integral of a function like that's in uh class and it may be an infinite number it may not be and in order to do that we're first going to define how to integrate the simplest type of functions uh well simple functions so let v be a simple function let's write v in this kind of canonical way so phi is equal to a sum of aj chi aj where for all i not equal to j well first off for all j aj is a subset of e for all i not equal to j a i intersect aj is empty so these are disjoint and their union gives me set e okay the labag integral of the simple function fee is the number which i said is kind of the simplest way or what you would expect to define how to define the integral of a simple function right so we know how to measure sets and remember we initially built up measure so that the integral which would be a theory of area underneath the curve should be the integral of the indicator function should be the area underneath the curve of one area being the measure of the set where that indicator function is one and therefore by um you know linearity this will be how we define the labeg integral of a simple function so the labeg integral of phi is uh the following number e phi this is how it's defined sum from j equals 1 to n aj measure of aj okay and you know this number could be infinite all right and you know instead of writing just the integral over e of b i might add a dx in there i'm just warning you ahead of time okay all right so this is the labeg integral of a simple function again we split it up into this kind of canonical way where it's just um the indicator function of disjoint sets which whose union gives the the set of the domain and where these uh numbers out in front these coefficients give you um you know the the numbers that go uh of the elements of the range so for example again i mean this is let's say a is from or let's say e is the interval a b you can check that well i mean it's just from the definition that if my simple function is in fact you know these sets are just intervals so if this is what my simple function looks like you know it takes finitely many values this one this one this one this one this one and uh on these sets that whose disjoint union forms a b then you know as i've uh defined and since as i've defined the integral and since the measure of an interval equals the length of the interval uh again this is you should count this as some more comments this integral equals you know the sum of uh so this would be aj then how i've defined the lebeg integral it spits out the area underneath you know this simple function that is uh taking these values on these intervals okay i hope that's clear it's towards the end of the day so maybe my explanations are getting a bit wonky but i hope it's clear okay and then we're going to use this definition of simple of how to integrate simple functions we're going to extend it to general uh elements of l plus of e the the non-negative measurable functions but first uh let's prove a few uh properties of how we've defined the integral for simple functions let's take two simple functions first is that if c is bigger than or equal to zero then the integral of c times phi over e this is equal to c times the integral of e v okay two the integral the lebeg integral over e of phi plus uh psi is equal to the big integral of phi plus the lebeg integral of psi and third property is if phi is less than or equal to psi on e so i'm writing that shorthand by that so what i'm saying in that statement is for all x and e v of x is less than or equal to psi of x then what you expect is uh the integral of phi is less than or equal to the integral of psi okay and let me include in fact one more very simple property if f is so let's add a little punctuation if f is a measurable subset of e then phi which is a simple function on e is also a simple function on f it takes only finitely many values on f as well and therefore it has an integral over f and my claim is is that that is equal to the integral over uh e of phi times the indicator function of f okay and that's less than or equal to the integral over e okay [Applause] now i will leave problem four to you i might even put on the assignment but um it will follow once you've seen how we prove one two and three okay you'll be like okay so i know how to do this so number one is pretty easy um simply because multiplying by a non-negative constant just carries through and just changes the constant but not the sets so phi times c times v is equal to c times a sub j therefore the integral of c times phi over e this is equal this is by definition equal to j sum from j equals 1 to n of c times a sub j measure of a sub j and this is equal to c times sum from j equals 1 to n of a sub j m of a sub j and this equals c times the integral of phi over e all right so to prove two we write fee as again as in this kind of canonical form sum over a sub j's times the indicator function of a sub j is where again these sets are disjoint and their union gives me e and then also i do the same thing for psi now k equals 1 to m maybe take on different number of values b k chi b k where again the b k's are disjoint and their union gives me e okay then so since you know since the union of the ajs gives me e and the union of the bks give me e this implies that if i want to look at one of these ajs it's equal to a union of a certain type is equal to the union k equals 1 to m of aj intersect b k because this is just going to be equal to aj intersect the union of the b k but the union of the b case gives me e okay and similarly for the b case because [Music] the union of aj's gives me e okay and again these are and these unions are disjoint you know the ajs are disjoint from each other the bks are disjoint from each other so for all j and k aj intersect b k is going to be disjoint from another a j prime intersect uh b k prime okay when j or k do not equal each other all right okay so since um these are disjoint union unions we have from the additivity property of the bag measure we get that the integral over phi e plus the integral over e of psi this is equal to by definition j equals 1 to n of aj measure of aj plus k equals 1 to m b k measure of b k now aj is written as this disjoint union and therefore the measure of aj is equal to the sum of the measure of these of a j intersect b k plus and then the same thing here the measure of so bk is equal to this disjoint union so its measure is the sum of the measure so this is this sum now includes j so let me just rewrite this as sum over j k a j plus b k measure of a j d k all right but the point is that the sum of these two simple functions you can check this is equal to the sum over j k of a j plus b k times the indicator function of aj intersect bk right so when x so at those x's when phi equals aj and at those x's where psi equals k then i'm in this set and the two sides agree okay this implies that the met that the integral over e of 5 plus psi is equal to the measure of aj intersect b k which as we just saw is equal to the sum of the integrals okay okay so that was 2 3 is not too difficult either so again let's assume that phi and psi are written in that way in this canonical way then for all x and e p of x is less than or equal to psi x this is equivalent to um a j is less than or equal to b k whenever aj and bk is not empty okay thus again we're going to use the additivity of lebeg measure and the fact that the unions of these bks give me e then if i look at the integral of e of phi this is sum j equals 1 to n aj measure of aj this is equal to the sum j k a j times the measure of a j b k because the aj is a union over aj intersect b k's and those are that's a disjoint union and now whenever this is um whenever this is non-zero that means that aj intersect b k is uh um non is not empty and therefore that aj appearing there is going to be less than or equal to b k and whenever this is 0 well whatever's there is also always less than or equal to b k times the measure of aj intersect b k so this is less than or equal to j k b k measure of aj intersect b k okay all right so whenever this is so maybe that previous uh explanation was not good whenever this is non-empty um i will always have aj less than or equal to bk so i should have just said that i don't know what i was going on about measure zero stuff but ignore that whenever this is non-empty like i just said a minute ago aj is less than or equal to b k all right and now we just reverse coarse and this is equal to k equals 1 to m of bk measure of b k because union over j's of these sets give me b k and this is equal to the integral of psi v and again i will leave 4 to u as a very simple exercise all right okay um i'm about out of time so what we've done is we've defined the labeg integral of a simple function and you know as that picture shows i hope or at least convinces you that you know if the simple function takes on that of a step function meaning it's the aj's are just intervals then the labeg integral of that step function will in fact be the area underneath fee so it's a you know again it's a this is you can think of that as two ways as the le beg integral is giving a theory of the area underneath the curve you may also think of that as being the first indication that when i have a riemann integrable function it will be also the bag integrable because if i have a step function like phi is in the picture then that is a riemann integral function and the riemann integral is the area underneath the curve which also agrees with um you know the the definition of the lebeg integral so um that like i said should maybe indicate that um that the lebago integral reduces to the riemann integral whenever we're uh integrating a riemann integrable function and so next time we will define the integral of a non-negative measurable function using uh how we've defined the integral of simple functions and prove some uh basic properties including two big two of the main convergence theorems um that go along with this theory of integration so we'll stop there you