Welcome to lecture 21 on measure and integration.
In the previous lecture, we had
started looking at the properties of sequences
of integrable functions and we started
proving an important theorem called Lebesgue’s
dominated convergence theorem.
Let us continue looking at that; after that
we will start looking at the special case
of
integration on the real line and that will
give us a notion of Lebesgue integral.
.
Let us recall what we had started proving,
namely, dominated convergence theorem,
which says, that if f n is a sequence of measurable
functions such that there exists a
function g belonging to L 1, say, that all
the f ns are dominated by this integrable
function g, ‘almost everywhere’ x for
all n and if f n converges to f, then the
limit
function is integrable and integral of f is
nothing but the limit of integrals of f ns.
So, the theorem basically says, that if f
n is a sequence of measurable functions, all
of
them dominated by a single integrable function
g, then all the f ns become integrable, of
.ourse. And if f ns converge to f then f is
integrable and integral of f is nothing but
the
limit of integrals of f ns.
We had proved this theorem in this particular
case when instead of this ‘almost
everywhere’, that mod f ns are dominated
by g everywhere, and f ns converge to f
everywhere. So, to extend this case to ‘almost
everywhere’, we have to do only a minor
modification.
.
Let us define the set N to be the set of all
x belonging to X, where mod f n x is not
dominated by g. So, g of x or union the set
of all those points x belonging to X, say
that,
f n x does not converge to the function f
of x.
So, on N compliment we have f n x is less
than g x and f n x converges to f of x. For
every x belonging to N compliment and mu of
the set N is equal to 0, because we are
saying that this mod f n x is less than g
x ‘almost everywhere’ and f n x converges
to f of
x ‘almost everywhere’. So the set where
it does not hold, that is, the set N and that
set
has - N - has got measure 0. Now, let us consider
the sequence indicator function of N
compliment times f n.
.
This is a sequence of functions, which are
dominated by for n bigger than or equal to
1;
they satisfy the property, namely, this, the
indicator function of N compliment f n mod
of
that is less than g for all x everywhere and
the functions converge to the indicator
function of N compliment f.
.
By our earlier case, what we get is the following:
namely, that indicator function of N
compliment times f is L 1 is integrable and
integral of f n limit n going to infinity,
indicator function of N compliment times f
n, the integral of that converges to the
.ntegral of indicator function of N compliment
times f. So that is by the earlier case when
everything is true for all points.
That means this is the same. But note that
mu of N is equal to 0, that implies that the
integral of f over N d mu is equal to 0. We
already know that on N compliment f is
integrable. So, that together with this fact
implies integral of mod f d mu is finite,
implying that f belongs to L 1. And this equation,
which said that integral of f n over N
compliment converges to integral of f over
N compliment and mu of N being 0 together,
gives us the condition that integral of f
n d mu converges to integral f d mu, because
integral f n d mu is the same as integral
of f n over n plus integral over N compliment
and integral over N compliment converges to
integral over N compliment of f and on n
both are 0. This gives us the required result.
.
That is how from ‘almost everywhere’,
conditions are deduced from the fact that
something holds everywhere. This dominated
convergence theorem holds for whenever
the sequence f n is dominated by g and f n
converges to f ‘almost everywhere’, then
f
limit function is integrable and integral
f converges to integral of f n.
As I said, this is one of the important theorems,
which helps us to interchange the limit
and the integral side. Let us look at some
minor modifications of this theorem. One more
thing - we can even deduce that integral of
mod f n minus f d mu also converges to 0.
.
To deduce that part, we just have to observe
that mod f n minus f is less than or equal
to
twice of g and mod f n minus f goes to 0.
So, again, an application of dominated
convergence theorem - which we proved just
now - implies that integral of mod f n
minus f d mu goes to 0. That is another modification,
another consequence of the
dominated convergence theorem.
.
Let us prove what I call as the series version
of this theorem; namely, that if f n is a
sequence of functions which are integrable
and integrals of f n summation 1 to infinity,
.o sum of all the integrals of mod f ns are
finite. Then, the conclusion is that the series
f
n x converges ‘almost everywhere’ and
if you denote the limit, the sum giving f
of x,
then the function is integrable and integral
f is equal to summation of integral f ns.
So essentially, this theorem says that if
the summations of mod f ns are finite then
this,
the series f n x is convergent ‘almost everywhere’
and integral of f is equal to integral of
summation of integral f ns, that is again
interchange of limit essentially.
.
Let us see how from dominated convergence
theorem we can get this. To show that this
series is convergent ‘almost everywhere’
we will actually show that it is absolutely
convergent. For that, let us define g n of
x to be equal to summation mod f k of x k
going
from 1 to n, the partial sums of the absolute
values 1 to n.
Let us observe this sequence g n. Note g n
is a sequence of nonnegative measurable
functions and g ns are increasing to some
function, that is, they are going to increase
to k
equal to 1 to infinity mod of f k of x, and
let us call that as g x, and they are increasing
to
the function g of x. That implies, by monotone
convergence theorem, we have integral of
g n d mu must converge to integral of g d
mu.
.
But integral of g n d - that is the same as
saying integral g d mu is equal to limit n
going
to infinity of integral g n d mu. But what
is integral of g n? It is the sum of absolute
values of f k 1 to n. So, by linearity property,
this is nothing but limit of n going to
infinity of summation 1 to n k equal to 1
to n of integral mod f k d mu. And this limit
is
nothing but 1 to infinity and that is given
to be finite. Let us write that.
.
This is equal to summation k equal to 1 to
infinity of integral mod f k d mu, which is
given to be finite. Hence, what we get is
- thus g is integrable - g is an integrable
.unction. Saying that g is integrable implies
- recall that if a function is integrable
and g
is a nonnegative function - g of x is finite
‘almost everywhere’, that is, nonnegative
function which is an integrable function must
be finite ‘almost everywhere’.
.
We get that g is finite ‘almost everywhere’.
And what is the function g? So, g is nothing
but the limit of the absolute values of f
k x. That means, that proves the series; hence,
sigma k equal to 1 to infinity mod f k of
x is finite ‘almost everywhere’ x. Once
the
series is absolutely convergent, it is also
convergent, that implies that sigma k equal
to 1
to infinity, f k x is finite for ‘almost
everywhere’ x.
Let us denote this limit by f of x; this is
f of x. As observed earlier, note f of x,
we can
also write f of x as the limit n going to
infinity of summation k equal to 1 to n f
k x. And
if these functions are called something, say,
phi n, then note that mod phi n - what is
mod
phi n? It is the absolute value of 1 to n
f k x. Absolute value of that and that is
less than
or equal to summation 1 to n mod of f k 1
to n and that is nothing but our g n which
is
less than or equal to g.
All these partial sums, which we have called
phi n(s) are all dominated by g and phi n(s)
converge to f by dominated convergence theorem.
.
.
What we have got is: all the phi n(s) are
less than or equal to g and phi n(s) converge
to f
‘almost everywhere’. That implies by dominated
convergence theorem, that integral of
phi n(s) d mu must converge to integral of
f d mu.
.
This is nothing but - this phi n - this is
what we called phi n, that is, summation 1
to n. So
this is nothing but summation of 1 to n of
integral k equal to 1 to n of integral f k
d mu
must converge to integral f d mu and that
is same as saying that integral f d mu is
equal
to summation k, equal to 1 to infinity integral
f k d mu.
.
That proves the theorem, namely, if f k is
a sequence of functions which are integrable
and the sum of the integrals is finite, then
the series f n x n 1 to infinity itself is
.onvergent ‘almost everywhere’ and the
limit function is integrable and integral
of the
limit function is equal to summation of integrals
of f ns.
.
This we will refer to as the series version
of dominated convergence theorem. There is
another interpretation of the dominated convergence
theorem, when the underlying
measure space is a finite measure space, then
one has that if X S mu is a finite measure
space and f n is a sequence of measurable
functions, such that all of them are dominated
by a single constant M ‘almost everywhere’
and f n x converges to f of x then integral
f
ns converge to integral f.
This is a particular case of dominated convergence
theorem when the underlying
measure space is a finite measure space and
the only thing to observe here is that because
let us see how does this follow from our is
a dominated convergence theorem.
.
We are given that mod f n x is less than or
equal to M for ‘almost everywhere’ x.
Now look at this constant function M. Look
at the function g of x which is equal to M,
for every x belonging to X. The constant function
is measurable, note that g is a
nonnegative measurable function because it
is a constant function. Note, that its integral
g d mu is equal to integral the constant function
M d mu and that is equal to M times the
measure of the whole space x which is finite.
What we are saying is that on finite
measure spaces a constant function is always
integrable. This implies g is L 1.
So, f n x bounded by M- that is a constant
function, that is an integrable function.
Once
we have that and f n x converges to f of x
‘almost everywhere’. So, now, dominated
convergence theorem is applicable and that
implies integral f d mu is equal to integral
f n
d mu limit n going to infinity.
.
The main thing is on finite measure spaces,
a constant function becomes integrable
because of this reason. This is what is called
bounded convergence theorem and it is
quite useful when underlying measure space
is a finite measure space.
.
Let us look at what we have proved till now.
We have looked at the space of integrable
functions, proved linearity property and an
important theorem called dominated
convergence theorem.
.f you recall, for nonnegative measurable
functions we had two theorems. One was
monotone convergence theorem; namely, that
was a theorem when f n is a sequence of
nonnegative measurable functions increasing
to a function f. Then, integral of f is equal
to limit of integral. That means interchange
of limit and integration is possible by
monotone convergence theorem whenever the
sequence f n is monotonically increasing
and a sequence of nonnegative measurable functions.
The second theorem, which involved sequences
of measurable functions was again for
nonnegative measurable functions and that
was called Fatou’s lemma.
There we do not emphasize, we do not require
that the sequence f n be nonnegative and
measurable. We only want the sequence f n
to be a sequence of nonnegative measurable
functions- they need not be increasing. For
such a sequence we had that integral of the
limit inferior of the sequence f n is less
than or equal to limit inferior of the integrals
of f
n.
That was Fatou’s lemma. Now we have the
third theorem- dominated convergence
theorem, which again helps you to interchange
the notion of integral and the limiting
operation under the condition that all the
f ns are dominated by a single integrable
function.
So, these are the three important theorems,
which help us to interchange limit and the
integral signs.
Let us at this stage emphasize one more point
about this technique of integration. So
basically, for integral, we started with simple
functions and then we go about
nonnegative functions and then we defined
it for integrable functions.
This process of step-by-step defining the
integral can be useful in proving many results
and I call it - the simple function technique.
This is a technique, which is used very often
to prove some results about integrable functions
and nonnegative measurable functions.
What is the technique? Let me outline that
and then I will give an illustration of this.
.
Suppose you want to show that a certain property
- let us call that property, star - holds
for all integrable functions.
To prove that the property holds for all integrable
functions, the technique is as follows;
basically, show that this property, star,
holds for all nonnegative simple measurable
functions.
If you want to show that a property holds
for all integrable functions, first show that
it
holds for the class of nonnegative simple
measurable functions. Next, show that star
holds for nonnegative measurable integrable
functions by using the fact that nonnegative
measurable functions are limits of increasing
limit of simple measurable functions.
There one normally uses the monotone convergence
theorem. Using monotone
convergence theorem, one extends the property
star from simple measurable functions to
nonnegative measurable functions or nonnegative
integrable functions.
.
Then, keeping in mind that for a function
f it can be split into positive part and negative
part: f can be written as f plus minus f minus
and if a property holds for nonnegative
functions - about integrals - f for f plus
that will hold for f minus hold, then conclude
from there that it will hold for f also.
So, this is what I call as the simple function
technique to prove results about integrable
functions.
.
.o give an illustration of this, let us look
at the following result. Let us take a measurable
space X S mu, which is sigma finite measure
space and let us look at a function f which
is integrable on this measure space and is
nonnegative. So we have got a sigma finite
measure space and f is a nonnegative integrable
function on this measure space.
Let us define nu of E for every set in the
sigma algebra. Let us define nu of E to be
integral of f d mu over the set E - integral
of f over the set E - is denoted by nu of
E for
every set E in the sigma algebra S.
We had already shown that this nu, the set
function nu, is in fact a finite measure on
S.
We have already proved this. But what we want
to prove now is that, if g is any
integrable function on the measure space X
S nu - this nu is the new measure. If g is
integrable on X with respect to nu, then the
product function f into g is integrable with
respect to mu and this relation holds integral
f d mu - so integral of f with respect to
nu is
equal to integral of f into g with respect
to mu.
What we have done is, by fixing a function
f, which is nonnegative, we had defined a
new measure on the measurable space by nu
of E to be equal to integral of E f d mu.
And
we are saying that if we want to integrate
a function with respect to a function g, with
respect to this new measure, then it is the
same as integrating the function - the product
function f g - with respect to the old measure
mu.
So let us see how the simple function technique
is used to prove this result.
.
Let us start. 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.
Recall, we defined nu of E to be equal to
integral of f d mu over E. This is the property,
star, we want to prove for every function
g.
As we said, let us first check this property.
Step 1 - let us take g is a L 1 function.
Let us
say g is a function which is an 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 because g - this is the
indicator function - so this is integral of
nu of
E which by definition is equal to integral
chi E of f d mu.
.
So chi E is g, this is equal to integral g
f d mu. What does it say? It says that the
required
property, star, holds, when g is the indicator
function. Now let us take a nonnegative
simple function - that is step 2, let us take
g sigma a i chi of E i, i equal to sum 1 to
n,
where ‘E i’s belong to S.
.
Our claim is that this property holds for
this g also. We are saying that the next step
is to
verify the required property. Integral of
g d nu, by definition, is equal to integral
of sigma
a i indicator function of E i d nu. What is
that? By inheritive property of the integral,
it is
.igma 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. 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 - that is the definition of mu of E i.
Which I
can again write as sigma i equal to 1 to n,
you can take this a i out and again by the
linearity property that is integral a i chi
E i times f d mu. But, this is nothing but
my
function g, this is integral of g f d mu.
What we are saying is that if g is summation
a i
chi E i, then using linearity property this
is the same as integral - goes in - so that
is a i
integral of the indicator function of E i,
that is nu of E i - 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. The required property
holds, so, star, holds for nonnegative simple
functions g. That is what I said - a simple
function technique. Now, let us try to prove
that this property also holds when g is a
nonnegative measurable function.
.
Now, let us look at g. Let g on x be measurable.
Then we know by the property of
measurable functions it implies that there
exists a sequence s 
n of nonnegative simple
measurable 
functions such that s n increases to the function
g.
Then by Lebesgue’s, by monotone 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 - we just 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. 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.
.
At this stage we observe that if s n is increasing
to g then s n times f will be increasing to
g times f. All are nonnegative simple measurable
- all are nonnegative measurable
functions. Once again, monotone convergence
theorem is applicable and this limit is
nothing but integral of g f d mu. Once again
we have used this step was by our step 2 that
the property holds for nonnegative simple
functions integral with respect to nu is integral
with respect to mu of product function. Now,
once again we are applying monotone
convergence theorem.
First, integral of g is equal to limit of
integral s ns d nu by monotone convergence
theorem. Now, by our earlier step, this is
equal to integral of s n f d mu. Again, by
monotone convergence theorem it goes back.
That means - this implies - that star holds
for nonnegative measurable functions.
.
Now, let us come to the last part, namely,
final step 3. Let g belong to L 1 X S nu - g
be a
integrable function.
Then what is g equal to? It is, g plus minus
g minus - where g plus is a nonnegative
measurable function, g minus is a nonnegative
measurable function. By step 2 we know
that integral g plus d mu is equal to integral
of g plus - sorry - d nu - 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, that is
by step 2. Now, because g is L 1, 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.
A function g is integrable if, and only if,
its positive part and negative parts are
integrable; that means these quantities - they
are all finite.
These are all finite quantities. That means
what? And f is nonnegative - that implies
integral of g f d mu is equal to integral
of g plus f d 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 - 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 - both
of these are finite quantities. That implies
that g f is L 1 and by these two, this is
the same as integral of g d nu.
.or step 3 - for a g which is integrable - we
have deduced that this property is true. This
is step 3. This is what I call the simple
function technique. Let me go back and show
you
once again what we have done.
.
We wanted to show that – this is property
star - we wanted to show for every function
g,
which is L 1. This is my step 1, that - look
at the functions g which are indicator
functions - I want to verify this for the
indicator function g to be the indicator function.
When g is the indicator function, this left
hand side is integral of over E of the constant
function 1. This is equal to integral d nu
is [int/integral] nu of E, which by definition
is
integral f over E, which I can write as integral
f E. So, that is true.
.
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 that it is true for every nonnegative
simple functions. Take g - a nonnegative simple
measurable function and apply. So, g is
equal to integral of nonnegative a i indicator
function E i and interchange and show that
required property holds. Step 2 was that the
required property holds for nonnegative
simple measurable functions.
.
.hen, using an application of monotone convergence
theorem - that is step 3, that, if g is
a nonnegative measurable function, then we
know that 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.
.
That is the next step - to show that it holds
for nonnegative measurable functions. 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,
which are L 1. This is what I normally call
as the simple function technique.
While proving results about integrable functions,
one quite often uses the simple function
technique and while proving some properties
about subsets of sets, recall, we had the
sigma algebra monotone class theorem technique.
For proving properties about sets, one uses
monotone class - sigma algebra monotone
class - technique and for proving results
about integrals one normally uses what is
called
the simple function technique.
.ith 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.
.
We want to specialize this thing for the real
line - let us see what we get. You will be
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 the same as real line
Lebesgue measurable sets and Lebesgue measure.
The space of all integrable functions on this
measure space - R 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.
This is the space of all Lebesgue integrable
functions. We want to study this space of
Lebesgue integrable functions in some more
detail.
.
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
integral, 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 the
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. Of particular interest for
the time being, is going to be the set: when
E is
a close bounded interval a b.
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 bounded
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 end we have got the space of
Lebesgue integrable functions on a b, on the
other hand we have got the space of
Riemann integrable functions on a b; we want
to see the relation or establish a
relationship between the two. That was one
of the starting points for our discussion
of
the subject, namely, the space of Riemann
integrable functions had some difficulties,
.ome problems, some drawbacks, for which we
want you to extend the notion to a larger
class - this is the larger class L 1 of a
b.
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 the same as the notion of
Lebesgue integral for Riemann integrable functions.
.
That is called the relation between the Riemann
integral and the Lebesgue integral. 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 integral is the same as the
Riemann integrable of the function f; this
is
what we wanted to prove.
.
So, let us start looking at how we prove this.
The proof of the theorem - we are given that
the function f belongs to R a b. It is a Riemann
integral function. Let us recall how the
Riemann integral of a function is defined
- it is defined via limits of upper sums and
lower sums of partitions.
It implies that, there exists a sequence P
n of refinement partitions with norm of P
n
going to 0 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 the Riemann integral of f, is the same
as the
limit of the lower sums L P n of f.
That is the meaning of saying that a function
f is Riemann integrable. We can find that
Riemann integrable implies, that 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 subintervals - goes
to 0. And, integrablity 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.
This is the property of saying that f is Riemann
integrable. Now, from here, let us look at
what is U P n f upper sum. Let us write down
the partition P n as something. 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.
.et us say that is the partition P n. In the
picture it will look like - here is ‘a’
here is ‘b’
this is x 0, this is x n, and here is x 1,
x 2 and so on.
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.
.
Let us write M k to be the maximum value of
the function in the interval x, say x i minus.
Let us write x k minus 1 to x k.
I am just 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. Similarly, M k, let
us
write - it is the minimum in the interval
x k minus 1 to x k of f of x. Only at the
end
points do you have to make it closed, but
that is not going to matter much.
Then, we define what is U P n f. That essentially
looks like, summation of the maximum
value into the indicator function of that
subinterval. The lower sum with respect to
P n f
looks like summation small m k - the minimum
value of the function in that subinterval,
x k minus 1 and x k.
Let us do one thing. 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.
.hese 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
sum P n f is equal to - the integral of
- Riemann integral of this function psi k
of x d x. These functions phi k and psi k,
which
are linear combinations of indicator functions
are in fact nonnegative measurable
functions on the measure space a b - the interval
a b. That is the observation that we
should note.
.
Let us note down, phi k and psi k are - measurable
functions, non negative, sorry, simple measurable
functions. Say, 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. 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. That
is, the phi k was maximum, so this should
be bigger than or equal to like this - because
phi k is taken as the supremum. This is the
upper sum P k of f and that is bigger than
or
equal to the - upper sum, sorry - lower sum
with respect to the P k of f and in the limit
both of them are converging.
Here is the second observation: that the upper
sum with respect to the partition of f is
the
same as - so what was it? - That was equal
to sigma M k into the length of the interval
x
k minus x k minus 1, that is the upper sum
- that is also the Riemann integral. In fact,
this
is also equal to length - so this is the length
- so you can write this as a length.
. k times the 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 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.
.
.imilarly, the lower sum P k f is equal to
integral of psi k d lambda. Now, essentially,
the
idea is to put them together, because phi
k and psi k - they are between these two.
.
Let us look at integral of - look at the sequence.
Consider the sequence - psi k minus phi k
minus psi k. Recall phi k is bigger than f
x less than psi k, so phi k minus psi k is
nonnegative for every k. 
Saying that the upper sums and lower sums
converge to the
same value is saying that the integral of
phi 
k minus psi k d lambda, that goes to 0. So,
that goes to 0 because the phi k d lambda
is the upper sum, this is the lower sum and
that
goes to 0.
So that implies that limit so that means this
implies that the limiting function f is trapped
in between. That means limit phi k x is equal
to limit psi k x ‘almost everywhere’.
Why
is that? That we can deduce from the fact
that applying Fatou’s lemma. To deduce this
look at the limit inferior of phi k minus
psi k integral d lambda will be less than
or equal
to limit inferior of integral phi k minus
psi k and that is 0 - so this is 0 - so this
says that
integral of a nonnegative function is 0, so
the function must be 0 ‘almost everywhere’
and that is the same as saying this must be
0 ‘almost everywhere’.
And f is trapped in-between. 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 - that means that the
function
f is measurable.
So we will continue the proof of this tomorrow,
in the next lecture. 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 the
same as the Lebesgue integral.
We will continue the proof in the next lecture.
Thank you.
