So, let us see how the simple function technique
is used to prove this result.
So, let us start. So, we want to show that
for every g belonging to L 1 of X, S, nu integral
of g d nu can be represented as integral f
g d mu. So, where recall, we defined nu of
E to be equal to integral of f d mu over E.
So, this is the, this is the property we want
to prove. So, this is the property star we
want to prove for every function g.
So, as we said let us first. So, check this
property. So, step 1, let us take g is g is
a L 1 function. Let us say g is a function
which is indicator function of E. Let us take
g as the indicator function of E, E belonging
to S. In that case, the integral g d mu, the
left hand side is nothing, but nu of E right
because g is this is the indicator function
so, this integral of nu of E which by definition
is equal to integral chi E of f d mu. And
So, chi is g. So, this is equal to integral
g f d mu.
So, what it says, it say that the required
property star holds, when g is the indicator
function. And now let us take a nonnegative
simple function. So, that is step 2: Let us
take g is sigma a i chi of E i, i equal to
sum 1 to n; where E i is belong to S. And
our claim is that this property holds for
this g also. So, we are saying that next step
is to verify that the required property.
So, integral of g d nu by definition is equal
to integral of sigma a i indicator function
of E i d nu and what is that; by inherit property
of the integral it is sigma i equal to 1 to
n of a i, that is scalar times nu of E i,
because integral of the indicator function
is the measure. So, that is equal to sigma
i equal to 1 to n, a i and nu of a i by definition
is integral chi E i of f d mu right that is
the definition of nu of E i is this.
So, which I can write it again as sigma i
equal to 1 to n of a i we can take it out.
So, let us. So, this is and again by the linearity
property of that is integral a i chi E i times
f d mu, but this is nothing, but my function
g. So, this is integral of g f d mu. So, what
we are saying, that if g is summation a i
chi E i, then using linearity property this
is same as integral goes in. So, that is a
i integral of the indicator function of E
i that is nu of E i. And nu of E i by definition
is integral over E i of f d mu. And again
using linearity property of the integral I
can shift it outside. So, it is integral of
summation a i chi E i times f d mu which is
g. So, it says. So, the required property
holds. So, star holds, for nonnegative simple
functions g. So, that is what I said simple
function technique. And now let us try to
prove it, that this property also holds, when
g is a nonnegative measurable function. So,
let us look at now g.
So, let g on X to be measurable. Then we know
by the property of measurable functions implies;
there exist a sequence S n of nonnegative
there is sequence S n of nonnegative simple
measurable functions. Functions, such that
S n increases to the function g, so then by
Lebesgue by Modern Convergence theorem 
integral of g d nu with respect to nu must
be equal to limit, n going to infinity integral
s n d nu, but for nonnegative simple functions
just now we proved this the star holds; that
means, this can be written as limit. So, by
step 2: I can write this as integral of s
n times f d mu. So, with the integration of
a nonnegative simple measurable function with
respect to nu, can be converted into the nonnegative
simple measurable function multiplied by f
d mu. And now at this stage, we observe that
if s n is increasing to g, then s n time’s
f will be increasing to g time’s f.
And all are nonnegative simple measurable
all are nonnegative measurable functions.
So, once again Monotone Convergence theorem
is applicable, and this limit is nothing,
but integral of g f d mu that is; so, once
again we have used. So, this step this step
was by our step 2 that the property holds
for non negative simple measurable functions.
Integral with respect to nu is integral with
respect to mu of the product function. And
now once again we are applying Monotone Convergence
theorem. So, first integral of g is equal
to limit of integral s n s d nu by Monotone
Convergence theorem, and now by our earlier
step this is equal to integral of s n f d
mu again by Monotone Convergence theorem it
goes back so; that means, this implies that
star holds for nonnegative measurable functions.
And now let us come to the last part namely:
So, let us final step 3: is let g belong to
L 1 X, S, nu. G be a integral function. Now
then what is g equal to, g plus minus g minus,
where g plus is a nonnegative measurable function
g minus is a non negative measurable function
and by step 2. So, by step 2: we know that
integral g plus d mu is equal to integral
of g plus sorry d nu. So, let me write it
again integral g plus d nu is equal to integral
g plus of f d mu and integral of g minus d
nu is equal to integral g minus f d mu. So,
that is by step 2. And now because g is L
one, so, that implies g is equal to g plus
g minus. So, g plus is in L 1 of nu and g
minus also belongs to L 1 of nu right.
A function g is integrable, if and only if;
its positive part and negative parts are integrable
so; that means, this quantities right they
are all finite, these are all finite quantities
so; that means, what and f is nonnegative.
So, that implies integral of g f d mu is equal
to integral of g plus fd mu minus integral
g minus f d mu. By definition of the positive
part and the negative part of the function
g f f is nonnegative. So, the positive part
of the function g f is same as g plus times
f and the negative part is nothing, but g
minus f. And now both of these are finite
quantities so; that means, implies that g
f is L 1 and by these two this is same as
integral of g d nu. So for a so the step 3:
for a g which is integrable, we have deduced
that this property is true. So, this is step
3. So, this is what I called the simple function
technique. So, let me go back and show you
once again, what we have done? So, we wanted
to show that, so this is property star, we
wanted to show for every function g which
is L 1.
First step is, so this is my step 1: that
look at the functions g which are indicator
functions. So, I want to verify this for the
indicator function, g to be the indicator
function. So when g is the indicator function,
so this left hand side is integral of over
E of the constant function 1. So, this is
equal to integral d nu is nu of E, which by
definition is integral f over E, which I can
write as integral f E; so, that is true. So,
step 1 is to verify the required thing holds
for characteristic function. And step 2: by
using the property that the integral is linear,
we show it is true for every non negative
simple functions.
So, take g a nonnegative simple measurable
function and apply. So, g is equal to integral
of nonnegative a i indicator function a i
and interchange and so, it required property
holds. So, the step 2 as that the required
property holds for nonnegative simple for
nonnegative simple measurable functions and
then using an application of Monotone Convergence
theorem so that is step 3. That if g is a
nonnegative measurable function then we know
it is a limit of nonnegative simple measurable
functions increasing limit. So, an application
of Monotone Convergence theorem together with
the earlier step gives us that integral of
g d nu is equal to integral of g f d mu.
So, that is the next step to show that it
holds for nonnegative measurable functions.
And once that is done, the final step that
it holds for all integrable functions is via;
splitting the function g into the positive
part minus the negative part. And g integrable
means both are integrable and for each one
of them, the required claim star holds. So,
by putting them together we get that the required
claim holds property star holds for all functions
g. So, which are L 1. So, this is what I normally
call as the simple function technique. So,
while proving results, about integrable functions
one uses quite often the simple function technique.
And while proving some properties about subsets
of sets recall; we had the sigma algebra Monotone
class theorem technique. So, for proving properties
about sets, one uses the Monotone class sigma
algebra, Monotone class technique and for
proving results about integrals one normally
uses what is called the simple function technique.
So, with this we have defined and proved general
properties about integral of functions on
sigma finite measure spaces. Now we will try
we will specialize this property this construction
when X is real line. So we want to specialize
this thing, for the real line. So, let us
see what we get.
So, while you looking at the special case,
when X is real line. The sigma algebra is
L that of Lebesgue measurable sets and the
measure mu will be the lambda the Lebesgue
measure. So, we will be working with the measure
space X S mu which is same as real line Lebesgue
measurable sets and Lebesgue measure. So,
the spaces of all integrable functions on
this measure space are L and lambda is called
the space of all Lebesgue integrable functions
and is also denoted by L 1 of R or L 1 of
lambda. So, this is the space of all Lebesgue
integrable functions.
So, we want to study, this space of Lebesgue
integrable functions in some more detail.
So, let us first agree to call integral f
d lambda to be the Lebesgue integral of the
function f. So whenever, f is integrable or
nonnegative integrable f d lambda will be
called the Lebesgue integral of f. Sometimes
we have to look at functions which are defined
on subsets of E. So, for any subset E, which
is Lebesgue measurable, L 1 of E will denote
the space of all integrable functions on the
measure space E, so underlying set is E, L
intersection E is the collection of all Lebesgue
measurable sets inside E and lambda is the
Lebesgue measure restricted to subsets of
L intersection the sigma algebra L intersection
E.
A of particular interest, for the time being
is going to be, the set when E is a closed
bounded interval a, b. So, we will start looking
at the space L 1 of a, b that is the space
of all Lebesgue integrable functions defined
on the interval close bonded interval a, b
and we also have the space r a, b namely the
space of all Riemann integrable functions
on a b. So, we want to compare these two spaces
on one hand we have got the space of Lebesgue
integrable functions on a, b. On the other
hand, we have got the space of a Riemann integrable
functions on a, b, and we want to see the
relation or establish a relationship between
the two and that was one of the starting points
for our discussion of this subject namely;
the space of Riemann integrable functions
had some difficulties, some problems, some
drawbacks, and for which we wanted to extend
the notion to a larger class and this is the
larger class L 1 of a, b.
So, what we are going to show is R a, b the
space of all Riemann integrable functions
is a subset of L 1 of a, b and the notion
of Riemann integral is same as the notion
of Lebesgue integral for Riemann integrable
functions. So, that is called the relation
between the Riemann integral and the Lebesgue
integral.
So, to be more specific we want to prove the
following theorem, namely; if f is defined
on a close bounded interval a, b is Riemann
integrable function, then f is also Lebesgue
integrable and the Lebesgue integrable is
same as the Riemann integral of the function
f. So, this is what we wanted want to prove.
So, let us start looking at, how do we prove
this? So, the proof of the theorem
So, we are given that the function f belongs
to R a, b, it is a Riemann integrable function;
that means, so, let us recall how is the Riemann
integral of a function define. It is defined
via; limits of upper sums and lower sums of
partitions. So, implies there exist a sequence
P n of refinement partitions with norm of
P n going to zero as n goes to infinity, partitions
n going to infinity with the upper sums of
P ns with respect to f, limit of that is same
as integral the Riemann integral of f is same
as the limit of the lower sums L P n of f.
So, that is the meaning of saying that a function
f is Riemann integrable.
So, we can find, Riemann integrable implies
there exists a sequence of partitions P n,
which are refinement partitions. Refinement
means P n plus 1 is obtained from P n by adding
one more point. And norm of these partitions
the maximum length of the sub intervals goes
to 0. And integrability means that the upper
sums and the lower sums, both converge to
the same value and that is the Riemann integral
of the function f. So, this is this is the
property of saying that f is Riemann integrable.
So, now from here, let us look at what is
U P n, f upper sum. So, let us write down
the partition P n as something. So, let us
say P n looks like a. So, interval is a to
b, so a the point x naught less than x 1 less
than x n which is equal to b. So, let us say
that is the partition P n. So, in the picture
it will look like, here is a, here is b. So,
this is the x 0, this is x n, and here is
x 1, x 2 and so on. So, to construct the upper
sums, what one does? To construct the upper
sums, one looks at the maximum value of the
function in this interval and the minimum
values in this interval.
So, let us write let us let us write M k to
be the maximum value of the function, in the
interval x say x i minus x k. So, let us x
k minus 1 to x k. I am just trying to be trying
to make the intervals disjoint. Maximum in
this interval of maximum in this interval
maximum of maximum of f of x maximum in of
f of x. And similarly M k, let us write it
is a minimum in the interval x k minus 1 to
x k of f of x. Only at the end points you
have to make it close, but that is not going
to matter much. So, that then we define, what
is U P n, f? that is, essentially looks like
summation of the maximum value into the indicator
function of that sub interval. And the lower
sum with respect to P n, f looks like summation
small m k, the minimum value of the function
in that sub interval x k minus 1 and x k.
So, let us let us do one thing, sorry this
is upper sums they are not these are not.
So, let us let me I am sorry this is not the
upper sum. Let us call this when in the interval
x k minus 1 to x k the value is capital M
k. Let us call that as, the function phi k
and when you are taking the minimum value
in that interval and summing up, let us call
that as psi k. So, these are functions, because
they are linear combinations of indicator
functions. And the upper sums and lower sums
are nothing, but the upper sum P n, f is nothing,
but Riemann integral a to b of phi k x d x
and the lower some P n, f is equal to the
integral of Riemann integral of this function
psi k of x d x. That this functions phi k
and psi k, which are linear combinations of
indicator functions are in fact, non negative
measurable functions. On the measure space
a, b the interval a, b. So, note: so that
is the observation that we should note and
then so let us note down.
Note: phi k and psi k are measurable functions
are measurable nonnegative sorry are simple
measurable functions. So that phi k is less
than or equal to at every point x is less
than or equal to f of x is less than or equal
to psi k of x. And as far as the integral
is concerned, the integral a to b of f x d
x is between the upper sum and the lower sum,
so that is the phi k was a maximum sorry this.
So, this should be bigger than or equal to
like this, because phi k is taken as the supremum,
so this is. So, this is upper sum P k of f
and that is bigger than or equal to the upper
sum sorry; bigger than or equal to the lower
sum with respect to P k of f. And in the limit
both of them are converging. So, here is the
second observation is that the upper sum with
respect to the partition of f is same as,
so what was it. So, that was equal to sigma
M k into the length of the interval x k minus
x k minus 1. So, that is the upper sum that
is also a Riemann integral.
In fact, this is also equal to so, the length.
So, this is the length of so you can write
this is the length. So, M k times length of
x k minus 1 and x k, which is same as the
Lebesgue integral of the function phi k d
lambda. So, this is, this is the important
observation that we should keep in mind. That
the building blocks for Riemann integral which
are these step functions are also Lebesgue
integrable and the Riemann integral of the
step functions phi k and psi k are same as
the Lebesgue integrals of phi k and psi k.
So, similarly the lower sum P k, f is equal
to integral of psi k d lambda. And now essentially
the idea is to put them together. So, because
phi k and psi k they are between these two.
So, let us look at integral of. So, look at
the sequence.
So, consider the sequence psi k minus phi
k minus psi k. Recall, phi k is bigger than
f x is less than psi k. So, phi k minus psi
k is nonnegative for every k and saying that
the upper sums and lower sums converge to
the same value is saying that the integral
of phi k minus psi k 
phi k minus psi k d lambda that goes to 0.
So, that goes to 0. So, because the phi k
d lambda is the upper sum, this is the lower
sum and that goes to 0.
So, that implies, so that implies that limit
so; that means, this implies that the limiting
function f is trapped in between so that means,
limit phi k x is equal to limit psi k x almost
everywhere, Why is that? So, that we can reduce
from the fact that, applying fatous lemma,
so to reduce this look at the limit inferior
of phi k minus psi k, integral d lambda be
less than or equal to limit inferior of integral
phi k minus psi k and that is 0. So, this
is 0 so; that means, and this is and so, this
says that integral of a nonnegative function
is 0. So, the function must be 0, almost everywhere
and that is same as saying this must be 0,
almost everywhere and f is trapped in between.
So, that implies that limit phi k x limit
of phi k x is equal to f of x is equal to
limit psi k x for almost everywhere x. So
that proves is equal to.
So, we are falling short of time so; that
means that the function f is measurable. So,
we will continue the proof of this tomorrow
in the next lecture. So, our aim is to prove
that the space of Riemann integrable functions
is inside the space of Lebesgue integrable
functions and the Riemann integral is same
as the Lebesgue integral. So, we will continue
the proof in the next lecture.
Thank you.
