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Welcome to lecture number 14 on measure and integration. Today, we will start looking
at functions on measurable spaces; these are called measurable functions.
To start with, we assume that we have a measurable space X, S. X is a set and S is a sigma algebra
of subsets of the set X. We have a function f defined on X taking extended real values;
R star denotes the set extended real line; that is, the set of all real numbers together
with plus infinity and minus infinity and with possible operations that we had defined
earlier. We will be looking at functions which are extended real values defined on the set
X. To start with, we want to prove the following;
namely, for this function f, the following statements are equivalent: inverse image of
the open interval c, closed at infinity - if you take the inverse image of any such interval,
then that belongs to the sigma algebra S. We will show that this is equivalent to saying
that the inverse image of the closed interval c to infinity belongs to S for every c, the
real number. This is equivalent to saying that the inverse image of the interval minus
infinity to c, minus infinity closed and c open, also belongs to the sigma algebra S.
Then, we will show that this is also equivalent to saying that the inverse images of all the
intervals of the type minus infinity to c, c closed, belongs to S for every c belonging
to R. We will show that these four are equivalent to each other and also these are all equivalent
to the following; namely, the points f inverse of plus infinity and the set f inverse of
minus infinity along with f inverse of every set E, E a Borel set in R, belong to S.
We will show that for a function f defined on X taking extended real numbers as the values,
these five conditions are equivalent. The methodology is going to be this: we will prove
(i) is equivalent to (ii); (ii) is equivalent to (iii); (iii) is equivalent to (iv); and
any one of them is equivalent to (v).
Let us start proving these properties. First property: we are given that f inverse of c
to plus infinity belongs to S for every c belonging to R. We want to prove the same
property for f inverse of .... Keep in mind what is f inverse; f inverse of c to plus
infinity is the set of all points x belonging to X such that f of x belongs to c to plus
infinity. This is a set of all points x in the domain which are mapped into the interval
c to infinity. f inverse does not mean that the function is invertible or anything; this
is a symbol used ; it is a pull-back of the points which go into this c to plus infinity.
We want to look at f inverse of closed interval c to plus infinity; we want to show that this
belongs to S. To show that, let us observe the simple set-theoretic equality; namely,
the closed interval c to plus infinity can be written as intersection of look at the
open interval c to c minus 1 by n to plus infinity and look at the intersection of all
these intervals. Keep in mind here is c and here is c minus 1 over n .
If you take this open interval, this open interval c minus 1 by n to plus infinity includes
this closed interval c to infinity for every n; the intersection also we had included.
Actually, this is equal because any point which is slightly bigger than c can be excluded
by taking n as sufficiently large. So, c minus 1 over n converges to c; that is the basic
idea. This is a simple identity about intervals which should be easy to prove.
This implies that the f inverse of c to plus infinity is equal to f inverse of intersection
n equal to 1 to infinity of c minus 1 over n to plus infinity. Here is another simple
observation that the inverse images of intersections are same as intersection of the inverse images.
This is equal to f inverse of c minus 1 over n to plus infinity closed. We are given that
whenever the interval is of the type c to plus infinity, c open, the inverse image is
in S; so, each one of these sets belongs to S; S is a sigma algebra; so, intersection
belongs to S; so, this belongs to S . Basically, what we have done is the closed
interval c to plus infinity is written as an intersection of open intervals c minus
1 over n to plus infinity. Observing that the inverse images of intersections are intersections
of inverse images, we get that f inverse of the closed interval c to plus infinity belongs
S. We have proved (i) implies (ii).
Let us show that (ii) also implies (i). What is the statement (ii)? Statement (ii) says
f inverse of the closed interval c to plus infinity belongs to S for every c belonging
to R; that is the statement (ii). We want to now prove the same thing for open intervals.
The idea is the open interval c to plus infinity can be expressed as union of the closed intervals
c plus 1 by n to plus infinity, n equal to 1 to infinity. That is quite easy to verify;
the interval c plus 1 by n to infinity is inside the interval c to plus infinity; so,
this union is inside it; converse is easy to check, because c plus 1 over n goes to
c; this is actually equal to c to infinity. Once again, observe that the inverse image
of c to plus infinity is equal to f inverse of the union n equal to 1 to infinity c plus
1 over n to plus infinity. Once again, a simple observation is that the inverse images of
the union is union of the inverse images; that gives us that this is f inverse of c
plus 1 over n to plus infinity. We are given that each one of them belongs to S. This is
a union of sets in S; S is a sigma algebra; it implies that this set also, f inverse of
c to plus infinity, belongs to S . Hence, we have shown that (ii) implies (i).
So, (i) implies (ii) and (ii) implies (i). Thus, we have shown that the statement (i)
implies statement (ii) and the statement (ii) implies (i). If you see the proofs carefully,
in both of them we have just tried to represent a closed interval as an intersection of open
intervals; also, in the (ii) implies (i) we have tried to use the fact that you can represent
an open interval as a union of closed intervals. Similar facts are used in proving the remaining
statements; let us just prove the remaining statement, namely, (ii) implies (iii).
Let us prove (ii) implies (iii). The statement (ii) is regarding closed intervals. We are
given that f inverse of this belongs to S for every c belonging to R; that is the statement
(ii) which is given. We want to show that f inverse of minus infinity to c open belongs
to S for every c belonging to R. If you look carefully, this interval and this interval
are related with each other. They are complements of each other.
The given statement implies that because this belongs to S, X minus f inverse of c to plus
infinity also belongs to S. Here is a small observation: the complement of the inverse
image is nothing but the inverse image of the complement. This set is equal to f inverse
of R star minus c to plus infinity. That is equal to f inverse of minus infinity to c
open, because here c is closed; so, this also belongs to S, because S is a sigma algebra;
if a set belongs, its complement belongs; so, this belongs to S. All these statements
are reversible; if this belongs, then its complement belongs; these are all if and only
if statements. So, (ii) implies (iii) is obvious by taking complements.
Let us prove (iii) implies (iv). The statement (iii) implies what is given to us is f inverse
of minus infinity to c open belongs to S for every c belonging to R. Here, we want to conclude
this closed interval.... Note that the closed interval to c is equal to... We want to include
the point c inside; it is nothing but look at minus infinity to c plus 1 over n - the
open interval. Here is c and here is c plus 1 over n . This interval - the closed interval
- is already inside c plus 1 over n for every n.
If I take the intersection of all this, that will give us the closed interval minus infinity
to c; it is similar to the earlier argument. This implies that f inverse of minus infinity
to c is equal to f inverse of the intersection; that is, intersection of the inverse images
- f inverse of minus infinity to c plus 1 over n open; each one of them belongs to n
and so this belongs to S. So (iii) implies (iv). If f inverse of minus infinity to c
open belongs, then f inverse of minus infinity to c closed also belongs. So, (iii) implies
(iv).
Let us prove the converse statement namely (iv) implies (i) . We are given f inverse
of minus infinity to c closed belongs to S for every c belonging to R. We want to look
at f inverse of the open interval c. Once again, it is a similar situation; that is,
minus infinity to c; it is here . We want to look at the open interval. Let us look
at the union of intervals minus infinity to c minus 1 over n, n equal to 1 to infinity.
Here is c minus 1 over n . These are all inside it - closed interval; the unions will give
us this open interval. Once again, taking the inverse images minus infinity to c is
equal to f inverse of the union n equal to 1 to infinity; f inverse of the union is union
of the inverse images; that gives us minus infinity to c minus 1 over n. Each one of
them is given to be inside S; that implies that this belongs to S; so, (iv) implies (iii)
is also true. What we have shown till now is that the first
four statements are equivalent to each other. The first statement was about intervals of
the type c to infinity, open; the next one was c closed; next was minus infinity to c.
Inverse images of all these types of intervals are inside S; all these statements are equivalent
to each other. Now, let us prove that this implies that f inverse of plus infinity and
f inverse of minus infinity and f inverse of every Borel set is inside S.
Let us assume any one of (i) to (iv) and hence all because they are equivalent. So, we know
that f inverse of an interval belongs to S for every interval I which looks like c to
plus infinity or looks like closed c to plus infinity or it looks like minus infinity to
c open or minus infinity to c closed. For all these four types of intervals, any one
of the first four statements implies they belong to S.
Now, look at any other interval. Supposing I is an open interval a to b. We can write
this open interval a to b as minus infinity to b open interval intersection with the open
interval a to plus infinity. We know that inverse image of this interval belongs to
the sigma algebra and inverse image of this belongs to the sigma algebra . That will give
us that the inverse image of a, b is equal to f inverse of minus infinity to b intersection
f inverse of a to plus infinity. Both belong to the sigma algebra and so this will belong
to the sigma algebra S. What I am trying to say is that any one of
the statements (i) to (iv) implies that inverse image of every open interval also belongs.
Similarly, we can take actually a closed interval also; for example, a closed interval a, b
can be written as minus infinity to b intersection a to plus infinity. A similar argument will
imply that inverse of this interval also belongs to S.
If we assume any one of the statements (i) to (iv), that implies that f inverse of every
interval belongs to S for every interval. Recall that any open set in R is a countable
union of open intervals; say, the open set is U; then, U can be written as union of Ijs,
j equal to 1 to infinity, Ijs open. Actually, you can write it as a disjoint union of open
intervals also - countable disjoint union of open intervals.
So, f inverse of U will be equal to f inverse of disjoint union of Ijs which is the same
as union of f inverse of Ijs, j equal to 1 to infinity; actually, disjoint is not needed,
but anyway, that is okay. f inverse of every open interval belongs to S; so, this belongs
to S . If we assume any one of the four conditions, then that implies that f inverse of every
open set is in the sigma algebra. Here is the sigma algebra technique: consider the
class A of all subsets in BR such that f inverse of E belongs to S. Just now we showed that
the open sets are inside A; it is easy to check that A is a sigma algebra. Let us check that A is
a sigma algebra.
Why is A a sigma algebra? Clearly, the empty set and the whole space R belong to A, because
X and the empty set belong to S. Secondly, let us observe that if a set E belongs to
A, that means that f inverse of E belongs to S; that implies that f inverse of E complement
belongs to S, because S is a sigma algebra; that is same as f inverse of E complement
belongs to S. It implies E complement belongs to A; so,
A is closed under complements. Finally, if Ens belong to A, it implies that f inverse
of En belongs to S; it implies union of f inverse of Ens also belongs to S, because
S is a sigma algebra. Hence, that implies that union of inverse images is inverse image
of the union; so, union En also belongs to S; it implies union Ens belong to A. We have
verified that A is a sigma algebra This is a sigma algebra including open sets.
So, it must include the Borel sigma algebra inside, but it is already a subclass of Borel
sets; so, A is equal to the class of Borel sets. That means if we assume any one of those
first four conditions in the statements that we just now stated, then that implies the
statement that f inverse image of every Borel set is in the sigma algebra S. Let us verify
that the inverse images of the points plus infinity and minus infinity are also....
Note that plus infinity can be written as intersection of n to plus infinity, n equal
to 1 to infinity. f inverse of plus infinity is equal to intersection n equal to 1 to infinity
f inverse of n to plus infinity. f inverse of this is equal to f inverse of the right-hand
side . The right-hand side is the intersection; so, it is the intersection of the inverse
images. Each one is an interval; inverse image of each one of the intervals belongs to S;
so, intersection belongs to S; so, this belongs to S.
A similar argument for minus infinity will imply, because minus infinity can be written
as intersection of n equal to 1 to infinity of minus infinity to minus n. The inverse
image of this will be intersection of inverse images and will imply that f inverse of minus
infinity belongs to S. We have shown that if you assume any one of those four conditions
stated above, then that implies that the inverse image of the point plus infinity and inverse
image of every Borel set belong to the sigma algebra S.
The converse statement is obvious, because every interval is a Borel set. Saying that
statement (v) implies any one of the four statements above is obvious, because every
interval is a Borel set; that is a special case. We have proved this theorem; namely,
for a function f defined on a set X taking extended real-valued functions, all these
five conditions are equivalent to each other . If you assume any one of them, then other
will also hold.
A function which satisfies any one of these conditions is called a measurable function.
A measurable function on X taking extended real values is a function which satisfies
any one of those five conditions as stated here . These are going to be an important
class of functions for us to deal with. Let us look at some examples. The first example
is that of what is called the indicator function of a set. Let us look at what is called the
indicator function of a set X.
Let us take any set X and A is a subset of X. We define a function called the chi of
A; this is the Greek letter chi and lower suffix A. It is a function on X taking two
values 0 or 1. This is called the characteristic function or the indicator function. This function
takes a value; at a point x, the value is 0 if x does not belong to A; at the point
A, the value is 1 if x belongs to A. Here is the set X; here is the set A. On A,
it gives the value 1; outside A, it gives the value 0. It is a two-valued function;
the points where it takes the value 1 is exactly the points in the set A; so, this is called
the characteristic function or the indicator function of the set A. This is called the indicator
function of the set A.
X is a set; S is a sigma algebra; we have got the indicator function A of the set A
on X taking, of course, only two values. We can consider it as a function taking extended
real values. We want to know if it is measurable. Suppose the indicator function of A is measurable.
That implies that if I look at chiA inverse of the singleton point 1, that belongs to
S, but what is that value? What are the points where it takes the value 1? That is precisely
A; that is the set A; so, A belongs to S. If the indicator function is measurable, then
we get A belongs to S. Conversely, if A belongs to S, we claim that chi of A is measurable.
For that, look at chiA inverse of any interval I. What is that going to be? The inverse image
of an interval is going to be equal to the empty set if 0 or 1 does not belong to the
interval I, because then there is no point which goes to the interval; it is equal to
A if 0 does not belong to I and 1 belongs to I; similarly, it is A complement if 0 belongs
to I and 1 does not belong to I; it is equal to X if both 0 and 1 belong to I.
It is an empty set or it is a set A or A complement or X and all of these are elements of the
sigma algebra S. The inverse image of every interval is in S; hence, the indicator function
is a measurable function. What we have shown is that the indicator function is measurable
This indicator function is defined as 1 if x belongs to A and 0 if x does not belong
.
to A.
The characteristic function is measurable if and only if the set A belongs to S; this
is the simplest example of a measurable function. Let us consider a linear combination of the
indicator functions. Suppose s is a function defined on X such that s of X is equal to
ai times the indicator function of a set Ai at at x, i equal to 1 to n. Look at sets A1,
A2 up to An - subsets of X; look at their indicator functions and take a linear combination
of them - ai times the indicator function of Ai; such a function is called a simple
function on X. Such a function is called a simple function on X.
Our claim is that a simple function is measurable if and only if each one of the Ais belongs
to S.
We want to prove the simple function S which is sigma ai indicator function of Ai, i equal
to 1 to n is measurable if and only if.... Note: to check measurability, we have to look
at s inverse of an interval I. What is that going to be? The function s takes values small
ais on the set Ai; this is the main thing to be observed - a finite linear combination
of the indicator functions is a function which takes only finite member of values, namely,
a1, a2 up to an and the value small ai is taken on the set capital Ai.
What will be s inverse of I? That will be union of those sets Ai union over i such that
ai belongs to the interval I. Clear? Let us once again observe that s takes values A1,
A2 up to An. Look at the inverse image of an interval I; look at those is such that
ai belongs to the interval I;. the pull-back of this will be the set Ai. Look at the unions
of these Ais; so, s inverse of I is union of Ais.
If each Ai belongs to S for every i, this will imply that s inverse of intervals belongs
to S. It implies that s is measurable because the inverse image of every interval belongs
to I. This interval is in extended real numbers ; plus infinity and minus infinity are included
in this; so, it is measurable. Conversely, if s is measurable, then look at s inverse
of singleton ai; that will be equal to Ai. Hence, measurability implies this belongs
to S. Of course, here, one has to take slight care; we can assume that all the Ais are distinct,
because if they are not distinct we can put together those Ais into one box. That says
that a simple function is measurable if and only if all the sets involved in the representation
ai times chi of Ai are all measurable.
As observed just now, we have given a simple function s; we can also write it in the form,
we can represent as, summation ai indicator function of Ai where all the ais are distinct
and these capital Ais are disjoint, because if sets are not disjoint we can put them together;
if the same value is taken on two distinct sets, then we can put them together in one
box and call that set as a new Ai. This is sometimes called a standard representation
of a simple function where the ais are distinct and these capital Ais form a partition of
the whole space X. A simple function is nothing but a finite linear combination of indicator
functions; that is an example of a measurable function.
We will study some more properties of this class of simple measurable functions. Let
us start; let s, s1 and s2 be simple measurable functions and alpha be a real number. First
of all, we want to observe that every constant function is a simple measurable function.
What is a constant function? A constant function is nothing but a function which takes a single
value everywhere on the set. We can think of it as this: if the constant value taken
is c, then it is c times the indicator function of the whole space X. Every constant function
is simple measurable, because it is a constant multiple of the indicator function. Alpha
times a simple function is also a simple measurable function because of the fact that if...
If a simple function s is equal to summation ai chi Ai, i equal to 1 to n, then alpha times
s is equal to sigma alpha ai times chi of Ai, i equal to 1 to n. alpha s is again a
simple function and only its values have changed, but the sets on which these values are taken
remain the same. Clearly, it indicates that if s is measurable then alpha s is also a
measurable set. The next property we want to check is that
if s1 and s2 are two simple measurable functions, then s1 plus s2 is also a simple measurable
function. Let us take a function s1 which is sigma ai chi Ai, i equal to 1 to n. Let
us say s1 has the representation sigma ai chi Ai and s2 has the representation j equal
to 1 to m bj chi of Bj. Whenever one is dealing with more than one simple function, the idea
is to try to bring the sets involved in the representation to be the same.
We have a standard representation that union Ai is equal to X; here, union Bjs is also
equal to X. Then you can write s1 as.... Each Ai can be decomposed into a union of the Bjs.
You can write i equal to 1 to n ai chi of Ai intersection Bj and union over js. Each
Ai can be intersected with union of Bjs. Here is an observation: if you have two sets A and B and they are disjoint, then
A union B is equal to chi of A plus chi of B; this we leave for you to verify: the indicator
function of the union of two sets is equal to sum of the indicator functions whenever
the sets are disjoint.
Using that, we can write s1 as summation i equal to 1 to n ai summation over j equal
to 1 to m chi of Ai intersection Bj. This is the same as summation over i summation
over j ai chi of Ai intersection Bj. Similarly, for the second simple function s2 which had
the representation bj chi of Bj, j equal to 1 to m, we can write this as summation over
i summation over j bj chi of Ai intersection Bj.
What we are saying is that whenever we are given two or a finite number of simple functions,
we can assume without loss of generality that the indicator functions involved are of same
sets. s1 is equal to summation over i summation over j ai times indicator function of Ai intersection
Bj. Similarly, s2 can be written as summation over i summation over j bj indicator function
of Ai intersection Bj. Then, what is s1 plus s2? s1 plus s2 is nothing but summation over
i summation over j of ai plus bj indicator function of Ai intersection Bj.
That should be clear, because if I take a point x, then if x belongs to Ai intersection
Bj, then s1 gives the value ai and s2 gives the value bj; sum will give the value ai plus
bj; outside, the value is 0; so, one does not have to bother. s1 plus s2 can be given
the representation summation over i summation over j ai plus bj of this. Since Ai belongs
to the sigma algebra and Bjs belong to the sigma algebra, that implies Ais intersection
Bjs also belong to the sigma algebra. So, s1 plus s2 is written as a linear combination
of indicator function of sets which are in the sigma algebra; that implies s1 plus s2
is measurable.
This proves the property that the class of simple measurable functions is closed under
addition. The first property said it is closed under scalar multiplication; this says it
is closed under addition. Next, let us take any fixed set - any set E in the sigma algebra
- and multiply S with the indicator function of E; then, the claim is this is also a simple
measurable function; that comes from a very simple observation.
Let us take a set E belonging to S and s is a simple function which is sigma ai indicator
function of Ai. Then, s times the indicator function of E is nothing but summation i equal
to 1 to n ai indicator function of Ai times indicator function of E. Here is an observation:
the product of indicator functions is nothing but the indicator function of the intersection;
the product of indicator functions is equal to indicator function of the intersected set.
If we use this, then the function s times indicator function of E can be written as
sigma ai indicator function of Ai intersection E. It is again a linear combination of indicator
functions of sets Ais intersection E. Since Ais belong to the sigma algebra and E belongs
to the sigma algebra, this belongs to the sigma algebra . The function s multiplied
by the indicator function is a linear combination of characteristic functions of sets which
are in the sigma algebra; this implies s into indicator function of E is measurable; that
proves our next property.
Using this, it is easy to check that a product of simple measurable functions is also a simple
measurable function.
For that, let us take s1 is sigma ai indicator function of Ai and s2 is sigma j equal to
1 to m bj chi of Bj. Then, s1 multiplied with s2 is nothing but this; we can do distributive
law; 1 to n ai of chi Ai summation bj 1 to m chi of Bj. We can write this as summation over i 1 to n ai summation
over j 1 to m chi of Ai chi of Bj into that constant bj; let us write that bj here
Anyway, we need not have done that much; we could have just said that is chi, indicator
function, of Ai times s2; each one of them is a simple. Anyway, this can be written as
summation over i 1 to n summation over j equal to 1 to m ai bj chi of Ai intersection Bj.
Once again, s1 into s2 is a linear combination of indicator function of sets where Ai belongs
to S because s1 is measurable, Bjs belongs to the sigma algebra S because s2 is measurable;
the intersection is measurable; so, s1 into s2 is measurable and product of simple measurable functions is again measurable.
Let us go a step further. Given two simple functions s1 and s2, let us define what is
called maximum of these two function s1 V s2. What is s1 V s2? This is the function
whose value at a point x is defined as the maximum of the numbers s1 of x and s2 of x.
At every point x, compare the values of s1 and s2; whichever is higher, define the value
to be that number. The claim is that s1 V s2 is also a simple measurable function. Once
again, the technique is same as for the sum.
Let us write s1 is equal to sigma ai chi of Ai, i equal to 1 to n and let us assume s1
is simple. That means all the Ais are in the sigma algebra S. Similarly, s2 is measurable.
Let us write s2 as j equal to 1 to m and bj chi of Bj where Bjs belong to S. Now, we bring
them to the common sets as before. Let us write s1 as sigma over i sigma over j ai chi
of Ai intersection Bj and s2 equal to sigma over i sigma over j bj chi of Ai intersection
Bj. Then, s1 maximum s2 at any point x.... We
want to define what will be at any point x the value of this . Look at the point x; it
will be in either one of the sets Ai intersection Bj. Then, s1 will give the value ai and s2
will give the value bj and the maximum of that has to be put. It is maximum of ai, bj
if x belongs to Ai intersection Bj. So, s1 V s2 is nothing but summation over i summation
over j of this . This is once again a finite linear combination
of characteristic function where the sets involved are in the sigma algebra. This will
imply s1 wedge s2 belong to S . A similar argument will imply that the corresponding
minimum of the two simple measurable functions is also a measurable function. What is the
minimum function?
s1 wedge s2 at a point x is defined as the minimum of s1 of x, s2 of x; that is called
the minimum of the two functions. We want to show that also is a simple measurable function.
Once again, if s1 is defined as this and s2 is defined as this, then what is s1 wedge
s2? This can be written as simply sigma over i sigma over j minimum of ai, bj into indicator
function of Ai intersection Bj. Once we write it that way, it becomes clear that the minimum
also is a... This implies that s1 wedge s2 is a measurable function whenever s1 and s2
are measurable functions.
Not only the maximum but the minimum also is a simple measurable function whenever s1
and s2 are measurable functions. Let us finally prove that if s is simple measurable, then
mod s is also a simple measurable function. There are many ways of looking at this.
If s is equal to sigma ai chi of Ai 1 to n, then what is mod s? Mod s is a function defined
at x to be equal to mod of s of x. Mod s is nothing but sigma mod of ai, i equal to 1
to n into indicator function of Ai; this also is measurable, because if s is measurable,
each Ai is a measurable set and mod s is a linear combination of indicator function of
sets which are measurable.
At this point, it is worth noting a few things about mod of a function . Let us take any
function f from X to R or R star. Let us define f plus of x to be a function on x as follows.
It is equal to f of x if f of x is greater than or equal to 0; it is 0 if f of x is less
than 0. What we are saying is look at the value of the function f of x; if it is bigger
than or equal to 0, then you keep the value of function as it is; as soon as it goes below,
you cut it off by the value 0. If this is the function f of x, then what
is f plus? When it goes below, you keep the value to be 0 because is going below; because
it is up, you keep it as it is. Now, it is going below and you keep the value to be 0;
now, it is going up. This is the function f plus; this is called the positive part of
the function; this is called the positive part of the function.
Similarly, we can define what is called the negative part of the function to be as follows.
Given a function f from X to R star, the negative part of the function f of x is defined as
0 if f of x is bigger than 0. As soon as it becomes positive, we make it 0; we make it
equal to minus of f of x if f of x is less than or equal to 0; keep in mind the negative.
If this is the graph of the function, then what do we do?
We look at the graph; as soon as it is below, we keep it as it is; it is 0 if f of x is
positive; so, on the positive part we keep it here; when it is below, we reflect it up.
So, it is this, this, this, this and so on . This is called the negative part of f. Let
us observe that the function f is written as the positive part minus the negative part.
Every function can be represented as the positive part and the negative part; both these functions
are nonnegative functions; mod of f can be written as f plus plus f minus; that is the
mod f.
You can also think of the positive part f plus as the maximum of f and the constant
function 0; f minus can be thought of as maximum of minus f and 0; this is another way of looking
at it. For a simple function, saying that mod f is measurable can also be looked at
because if s is measurable, simple function is measurable, the maximum of simple function
and 0 is measurable; the positive part is measurable; the negative part is measurable;
hence, mod f will be also measurable. We will continue properties of measurable
functions in our next lecture. In the next lecture, we will prove an important theorem;
namely, we will look at how sequences of measurable functions behave whether the limits of sequences
of measurable functions are measurable or not. Thank you.