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## I need a help/hint on this problem

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• Hello everyone, i have this problem. i hope you can help me how to work it out. Given : abs(f)
Message 1 of 20 , Nov 3, 2001
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Hello everyone,

i have this problem. i hope you can help me how to
work it out.

Given : abs(f) <= g, where g is an integrable
function.
Prove that abs(f) is also integrable.

Hints would be ok.

thnaks,
dexter

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• ... Dexter, Given a measurable map f, being integrable means int(|f|)
Message 2 of 20 , Nov 5, 2001
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> Given : abs(f) <= g, where g is an integrable
> function.
> Prove that abs(f) is also integrable.
>

Dexter,

Given a measurable map f, being integrable means int(|f|)<+oo
Furthermore, given two non-negative measurabke maps, f and g such
that f<=g, we have int(f)<=int(g).

So for your question:

int(|f|)<=int(g)<+oo

which shows that |f| (and f) is integrable

Noel.
• Thanks Sir NOel, But Sir, will this also be true if f and g are not nonnegative? This is my idea : i hope taht this will make sense. 1. i set this up : g -
Message 3 of 20 , Nov 5, 2001
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Thanks Sir NOel,

But Sir, will this also be true if f and g are not
nonnegative?

This is my idea : i hope taht this will make sense.

1. i set this up : g - abs(f) => 0. since abs(f) will
always be positive.

2. i dcompose abs(f) into abs(f) = (f+) + (f-)
likewise g.

3. integrate and manipulate suhc that youll arrive
into this expression : int(abs(f)) <= int(g) <
+infinity.

OR i will use the definition of the integral of a
nonnegative function and apply it to g-abs(f)=> 0,
i.e., ill consider a sequence of nonnegative and
nondecreasing simple functions.

It seems that im not convinced with this proof. is
there no otehr proof aside from this?

thanks,
dexter

--- vaillant@... wrote:
> > Given : abs(f) <= g, where g is an integrable
> > function.
> > Prove that abs(f) is also integrable.
> >
>
> Dexter,
>
> Given a measurable map f, being integrable means
> int(|f|)<+oo
> Furthermore, given two non-negative measurabke maps,
> f and g such
> that f<=g, we have int(f)<=int(g).
>
> So for your question:
>
> int(|f|)<=int(g)<+oo
>
> which shows that |f| (and f) is integrable
>
> Noel.
>
>
>

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• Thanks Sir NOel, But Sir, will this also be true if f and g are not nonnegative? This is my idea : i hope taht this will make sense. 1. i set this up : g -
Message 4 of 20 , Nov 5, 2001
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Thanks Sir NOel,

But Sir, will this also be true if f and g are not
nonnegative?

This is my idea : i hope taht this will make sense.

1. i set this up : g - abs(f) => 0. since abs(f) will
always be positive.

2. i dcompose abs(f) into abs(f) = (f+) + (f-)
likewise g.

3. integrate and manipulate suhc that youll arrive
into this expression : int(abs(f)) <= int(g) <
+infinity.

OR i will use the definition of the integral of a
nonnegative function and apply it to g-abs(f)=> 0,
i.e., ill consider a sequence of nonnegative and
nondecreasing simple functions.

It seems that im not convinced with this proof. is
there no otehr proof aside from this?

thanks,
dexter

--- vaillant@... wrote:
> > Given : abs(f) <= g, where g is an integrable
> > function.
> > Prove that abs(f) is also integrable.
> >
>
> Dexter,
>
> Given a measurable map f, being integrable means
> int(|f|)<+oo
> Furthermore, given two non-negative measurabke maps,
> f and g such
> that f<=g, we have int(f)<=int(g).
>
> So for your question:
>
> int(|f|)<=int(g)<+oo
>
> which shows that |f| (and f) is integrable
>
> Noel.
>
>
>

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• ... I would personally be reluctant to write differences such that g- ... (excluding he value +oo). In integration theory, it is very common to work with maps
Message 5 of 20 , Nov 5, 2001
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> But Sir, will this also be true if f and g are not
> nonnegative?

|f| is certainly non-negative and if |f|<=g, then so is g.

>
> This is my idea : i hope taht this will make sense.
>
> 1. i set this up : g - abs(f) => 0. since abs(f) will
> always be positive.

I would personally be reluctant to write differences such that "g-
|f|", unless you know for sure the functions are real-valued
(excluding he value +oo). In integration theory, it is very common to
work with maps with values in [0,+oo]. It is not meaningful to write
something like "+oo - (+oo)"...

>
> 2. i dcompose abs(f) into abs(f) = (f+) + (f-)
> likewise g.
>
> 3. integrate and manipulate suhc that youll arrive
> into this expression : int(abs(f)) <= int(g) <
> +infinity.

If |f|<=g, then int(abs(f))<=int(g) is true immediately as both |f|
and g are non-negative.

>
> OR i will use the definition of the integral of a
> nonnegative function and apply it to g-abs(f)=> 0,

Again, I think (depending on the exact statement of your problem),
that a map such as "g-|f|" may not be well-defined.

Noel.
• Hello Everybody, i have another problem. Let mu1, mu2 and v be measures such that v(E) = mu1(E) +mu2(E). show that f is an integrable function wrt both mu1 and
Message 6 of 20 , Nov 6, 2001
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Hello Everybody,

i have another problem.

Let mu1, mu2 and v be measures such that
v(E) = mu1(E) +mu2(E). show that f is an integrable
function wrt both mu1 and mu2 then f is integrable
with respect to v.

I already have this solution : i showed it for
f-simple function, f-general measurable function.

Now, for f-nonnegative measurable function :
I have to show that
int (f dv ) = sup { int ( fn dv):fn<=f}
= int(f dmu1) + int (f dmu2 )

Now,
int(f dv) = sup { int (fn dv) : fn<=f }
= sup {int ( fn dmu1 ) + int (fn dmu2) : fn<=f }
<= sup { int(fn dmu1 : fn<= f } + sup { int (fn dmu2
}

My problem now is how to show the other inequality for
me to conclude equality.

Another one, is it right to write this?
int ( f dv ) = int ( f d(mu1 + mu2) )
where v = mu1 + mu2

hoping for your help.

dex

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• Hi Dexter, ... Yes, this is what you have to prove. ... Yes, this is correct ... Your problem is that proving the other inequality will effectively be as
Message 7 of 20 , Nov 7, 2001
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Hi Dexter,

> Let mu1, mu2 and v be measures such that
> v(E) = mu1(E) +mu2(E). show that f is an integrable
> function wrt both mu1 and mu2 then f is integrable
> with respect to v.
>
> I already have this solution : i showed it for
> f-simple function, f-general measurable function.
>
> Now, for f-nonnegative measurable function :
> I have to show that
> int (f dv ) = sup { int ( fn dv):fn<=f}
> = int(f dmu1) + int (f dmu2 )

Yes, this is what you have to prove.

> Now,
> int(f dv) = sup { int (fn dv) : fn<=f }
> = sup {int ( fn dmu1 ) + int (fn dmu2) : fn<=f }
> <= sup { int(fn dmu1 : fn<= f } + sup { int (fn dmu2
> }

Yes, this is correct

> My problem now is how to show the other inequality for
> me to conclude equality.
>

Your problem is that proving the "other inequality" will effectively
be as hard as proving the "Monotone Convergence Theorem"(MON).
Although, you approach may be successful in the end, it is infinitely
simpler to make use of MON, as follows:

Having established that:

int fn dv = int fn dmu1 + int fn dmu2 (*)

is true whenever fn is a simple function, given f non-negative and
measurable, let (fn) be a non-decreasing sequence of simple functions
converging simply to f. In particular, (fn) is a non-decreasing
sequence of non-negative measurable maps converging simply to f. From
MON, we have:

int fn dv -> int f dv
int fn dmu1 -> int f dmu1
int fn dmu2 -> int f dmu2

as n-> +oo. Taking the limit in (*) as n tends to +oo allows us to
replace "fn" by "f" in (*) and you re done !

> Another one, is it right to write this?
> int ( f dv ) = int ( f d(mu1 + mu2) )
> where v = mu1 + mu2

int f d(mu1 + mu2) is "the integral of f w.r. to the measure
mu1+mu2". Such notation is perfectly legitimate (mu1+mu2 is indeed a
measure), and furthermore since v=mu1+mu2, such notation also refer
to int f dv. So yes, you are right to write what you wrote i think.

Regards. Noel.
• Thanks Sir Noel, I have this solution. i dont know if this makes sense . Remember this problem ? : g-integrable, abs(f)
Message 8 of 20 , Nov 7, 2001
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Thanks Sir Noel,

I have this solution. i dont know if this makes sense
.
Remember this problem ? : g-integrable, abs(f) <= g.
WTS that f is integrable.
My solution.
let k = abs(f).
so g-k => 0.(thus nonnegative )
then,
int(g-k) = sup { int( gn - kn) : gn-kn<=g-k}
<= sup{ int (gn): gn<=g }
- sup {int (kn):kn<=k }
= int(g) - int(k)
thus, 0<= int(g-k) <= int(g) - int(k)
int(g) => int (k) which implies that
int(abs(f)) is integrable which fuirther
implies that int(f) exists.

But I have another problem.

find the int(sin(x) d mu) where mu is a probability
measure with a corresponding distribution function
which has jumps at (pi/6), pi/4, pi/3 with
corresponding magnitudes 1/2, 1/4, 1/4.

i have a solution but im not convinced/satisfied.

thanks,
dexter
==============================================

--- vaillant@... wrote:
> Hi Dexter,
>
> > Let mu1, mu2 and v be measures such that
> > v(E) = mu1(E) +mu2(E). show that f is an
> integrable
> > function wrt both mu1 and mu2 then f is integrable
> > with respect to v.
> >
> > I already have this solution : i showed it for
> > f-simple function, f-general measurable function.
> >
> > Now, for f-nonnegative measurable function :
> > I have to show that
> > int (f dv ) = sup { int ( fn dv):fn<=f}
> > = int(f dmu1) + int (f dmu2 )
>
>
> Yes, this is what you have to prove.
>
>
> > Now,
> > int(f dv) = sup { int (fn dv) : fn<=f }
> > = sup {int ( fn dmu1 ) + int (fn dmu2) : fn<=f
> }
> > <= sup { int(fn dmu1 : fn<= f } + sup { int (fn
> dmu2
> > }
>
> Yes, this is correct
>
>
> > My problem now is how to show the other inequality
> for
> > me to conclude equality.
> >
>
> Your problem is that proving the "other inequality"
> will effectively
> be as hard as proving the "Monotone Convergence
> Theorem"(MON).
> Although, you approach may be successful in the end,
> it is infinitely
> simpler to make use of MON, as follows:
>
>
> Having established that:
>
> int fn dv = int fn dmu1 + int fn dmu2 (*)
>
> is true whenever fn is a simple function, given f
> non-negative and
> measurable, let (fn) be a non-decreasing sequence of
> simple functions
> converging simply to f. In particular, (fn) is a
> non-decreasing
> sequence of non-negative measurable maps converging
> simply to f. From
> MON, we have:
>
> int fn dv -> int f dv
> int fn dmu1 -> int f dmu1
> int fn dmu2 -> int f dmu2
>
> as n-> +oo. Taking the limit in (*) as n tends to
> +oo allows us to
> replace "fn" by "f" in (*) and you re done !
>
>
>
> > Another one, is it right to write this?
> > int ( f dv ) = int ( f d(mu1 + mu2) )
> > where v = mu1 + mu2
>
>
> int f d(mu1 + mu2) is "the integral of f w.r. to the
> measure
> mu1+mu2". Such notation is perfectly legitimate
> (mu1+mu2 is indeed a
> measure), and furthermore since v=mu1+mu2, such
> notation also refer
> to int f dv. So yes, you are right to write what you
> wrote i think.
>
>
> Regards. Noel.
>
>
>

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• I have this solution for number 2. does this make sense ? int (sin(x) ) = sum( sin(xi)*P(X = xi ) = sin(pi/6)*(1/2) + sin(pi/4)*1/4 + sin(pi/3)*(1/4) dexter
Message 9 of 20 , Nov 7, 2001
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I have this solution for number 2. does this make
sense ?

int (sin(x) ) = sum( sin(xi)*P(X = xi )
= sin(pi/6)*(1/2) + sin(pi/4)*1/4 + sin(pi/3)*(1/4)

dexter
--- dexter cahoy <d_cahoy@...> wrote:
> Thanks Sir Noel,
>
> I have this solution. i dont know if this makes
> sense
> .
> 1. Remember this problem ? : g-integrable,
abs(f) <= g.
> WTS that f is integrable.
> My solution.
> let k = abs(f).
> so g-k => 0.(thus nonnegative )
> then,
> int(g-k) = sup { int( gn - kn) : gn-kn<=g-k}
>
> <= sup{ int (gn): gn<=g }
> - sup {int (kn):kn<=k }
> = int(g) - int(k)
> thus, 0<= int(g-k) <= int(g) - int(k)
> int(g) => int (k) which implies that
> int(abs(f)) is integrable which fuirther
> implies that int(f) exists.
>
> But I have another problem.
>
> 2. find the int(sin(x) d mu) where mu is a
probability
> measure with a corresponding distribution function
> which has jumps at (pi/6), pi/4, pi/3 with
> corresponding magnitudes 1/2, 1/4, 1/4.
>
> i have a solution but im not convinced/satisfied.
>
> thanks,
> dexter
> ==============================================
>
> --- vaillant@... wrote:
> > Hi Dexter,
> >
> > > Let mu1, mu2 and v be measures such that
> > > v(E) = mu1(E) +mu2(E). show that f is an
> > integrable
> > > function wrt both mu1 and mu2 then f is
> integrable
> > > with respect to v.
> > >
> > > I already have this solution : i showed it for
> > > f-simple function, f-general measurable
> function.
> > >
> > > Now, for f-nonnegative measurable function :
> > > I have to show that
> > > int (f dv ) = sup { int ( fn dv):fn<=f}
> > > = int(f dmu1) + int (f dmu2 )
> >
> >
> > Yes, this is what you have to prove.
> >
> >
> > > Now,
> > > int(f dv) = sup { int (fn dv) : fn<=f }
> > > = sup {int ( fn dmu1 ) + int (fn dmu2) :
> fn<=f
> > }
> > > <= sup { int(fn dmu1 : fn<= f } + sup { int (fn
> > dmu2
> > > }
> >
> > Yes, this is correct
> >
> >
> > > My problem now is how to show the other
> inequality
> > for
> > > me to conclude equality.
> > >
> >
> > Your problem is that proving the "other
> inequality"
> > will effectively
> > be as hard as proving the "Monotone Convergence
> > Theorem"(MON).
> > Although, you approach may be successful in the
> end,
> > it is infinitely
> > simpler to make use of MON, as follows:
> >
> >
> > Having established that:
> >
> > int fn dv = int fn dmu1 + int fn dmu2 (*)
> >
> > is true whenever fn is a simple function, given f
> > non-negative and
> > measurable, let (fn) be a non-decreasing sequence
> of
> > simple functions
> > converging simply to f. In particular, (fn) is a
> > non-decreasing
> > sequence of non-negative measurable maps
> converging
> > simply to f. From
> > MON, we have:
> >
> > int fn dv -> int f dv
> > int fn dmu1 -> int f dmu1
> > int fn dmu2 -> int f dmu2
> >
> > as n-> +oo. Taking the limit in (*) as n tends to
> > +oo allows us to
> > replace "fn" by "f" in (*) and you re done !
> >
> >
> >
> > > Another one, is it right to write this?
> > > int ( f dv ) = int ( f d(mu1 + mu2) )
> > > where v = mu1 + mu2
> >
> >
> > int f d(mu1 + mu2) is "the integral of f w.r. to
> the
> > measure
> > mu1+mu2". Such notation is perfectly legitimate
> > (mu1+mu2 is indeed a
> > measure), and furthermore since v=mu1+mu2, such
> > notation also refer
> > to int f dv. So yes, you are right to write what
> you
> > wrote i think.
> >
> >
> > Regards. Noel.
> >
> >
> >
>
>
> __________________________________________________
> Do You Yahoo!?
> Find a job, post your resume.
> http://careers.yahoo.com
>

__________________________________________________
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• ... I think I have already pointed out that this route is not satisfactory for the following reason: 1. Either you do not know for a fact that g and k will not
Message 10 of 20 , Nov 8, 2001
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> I have this solution. i dont know if this makes sense
> .
> Remember this problem ? : g-integrable, abs(f) <= g.
> WTS that f is integrable.
> My solution.
> let k = abs(f).
> so g-k => 0.(thus nonnegative )

I think I have already pointed out that this route is not
satisfactory for the following reason:

1. Either you do not know for a fact that g and k will not take +oo
as value, in which case writing anything like g-k does not make sense.

2. Or you do know that g and k have values in R only, in which case
you are still embarking into a proof which only deals with that case,
which is a shame since the result is more general, and you should not
therefore rely on the fact that k and g are real-valued....

What you want to prove is this:

Let g,k be two non-negative and measurable maps such that k<=g.
Then int(k)<=int(g)

The proof is an immediate consequence of the definition of the
lebesgue integral (the one you keep using) and the fact that:

{s simple function: s<=k} is a subset of {s simple function: s<=g}
and the fact that if A<B then supA <= supB

This is all that should be said about this, I think.

> then,
> int(g-k) = sup { int( gn - kn) : gn-kn<=g-k}
> <= sup{ int (gn): gn<=g }
> - sup {int (kn):kn<=k }

It is not clear to me this should be true

> = int(g) - int(k)

The fact that you did not use int(g)<+oo and int(k)<+oo and yet still
wrote this difference, shows there is a flaw in your derivations.
The thing to remember is:

if a,b are in [0,+oo], do not write a-b or b-a, unless you know for
sure that the situation +oo - (+oo) (meaningless!) does not arise.

Dexter, it is not necessary to resend entire posts when asking a new
question. It is advisable to 'edit' the message you are sending and
delete all the things which are not relevant to your new message.

Regards. Noel.
• Hello Sir Noel, im sorry for resending entire posts. i just want to know if the problem that you want me to prove is the same as my problem. My Problem : Given
Message 11 of 20 , Nov 8, 2001
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Hello Sir Noel,

im sorry for resending entire posts.

i just want to know if the problem that you want me to
prove is the same as my problem.

My Problem : Given f-measurable, g-integrable wrt
mu ,and abs(f) <= g.
WTS that f is integrable.

What you want to prove is this:

Let g,k be two non-negative and measurable maps such
that k<=g. Then int(k)<=int(g)

The proof is an immediate consequence of the
definition of the lebesgue integral (the one you keep
using) and the fact that:

{s simple function: s<=k} is a subset of {s simple
function: s<=g}and the fact that if A<B then supA <=
supB This is all that should be said about this, I
think.

thanks,

dexter

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• Hi Dexter, The two problems are different, but solving mine allows us to solve yours immediately. Suppose i have prove that for g,k non-negative with k
Message 12 of 20 , Nov 8, 2001
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Hi Dexter,

The two problems are different, but solving 'mine' allows us to
solve 'yours' immediately.

Suppose i have prove that for g,k non-negative with k<=g we have:
int(k)<=int(g).
Then if |f|<=g, we have int(|f|)<=int(g), and if g is integrable, we
have int(g)<+oo and consequently int(|f|)<+oo. So f is itself
integrable.

Regards. Noel.

P.S. Not sure what 'WTS' stands for...

> i just want to know if the problem that you want me to
> prove is the same as my problem.
>
> My Problem : Given f-measurable, g-integrable wrt
> mu ,and abs(f) <= g.
> WTS that f is integrable.
>
> Your Problem :
>
> What you want to prove is this:
>
> Let g,k be two non-negative and measurable maps such
> that k<=g. Then int(k)<=int(g)
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