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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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> Given : abs(f) <= g, where g is an integrable
Dexter,
> function.
> Prove that abs(f) is also integrable.
>
Given a measurable map f, being integrable means int(f)<+oo
Furthermore, given two nonnegative 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. 0 Attachment
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 gabs(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 nonnegative 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  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 gabs(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 nonnegative 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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> But Sir, will this also be true if f and g are not
f is certainly nonnegative and if f<=g, then so is g.
> nonnegative?
>
I would personally be reluctant to write differences such that "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.
f", unless you know for sure the functions are realvalued
(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)"...
>
If f<=g, then int(abs(f))<=int(g) is true immediately as both f
> 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.
and g are nonnegative.
>
Again, I think (depending on the exact statement of your problem),
> OR i will use the definition of the integral of a
> nonnegative function and apply it to gabs(f)=> 0,
that a map such as "gf" may not be welldefined.
Noel. 0 Attachment
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
fsimple function, fgeneral measurable function.
Now, for fnonnegative 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,
> Let mu1, mu2 and v be measures such that
Yes, this is what you have to prove.
> 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
> fsimple function, fgeneral measurable function.
>
> Now, for fnonnegative measurable function :
> I have to show that
> int (f dv ) = sup { int ( fn dv):fn<=f}
> = int(f dmu1) + int (f dmu2 )
> Now,
Yes, this is correct
> 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
Your problem is that proving the "other inequality" will effectively
> me to conclude equality.
>
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 nonnegative and
measurable, let (fn) be a nondecreasing sequence of simple functions
converging simply to f. In particular, (fn) is a nondecreasing
sequence of nonnegative 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 d(mu1 + mu2) is "the integral of f w.r. to the measure
> int ( f dv ) = int ( f d(mu1 + mu2) )
> where v = mu1 + mu2
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. 0 Attachment
Thanks Sir Noel,
I have this solution. i dont know if this makes sense
.
Remember this problem ? : gintegrable, abs(f) <= g.
WTS that f is integrable.
My solution.
let k = abs(f).
so gk => 0.(thus nonnegative )
then,
int(gk) = sup { int( gn  kn) : gnkn<=gk}
<= sup{ int (gn): gn<=g }
 sup {int (kn):kn<=k }
= int(g)  int(k)
thus, 0<= int(gk) <= 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
> > fsimple function, fgeneral measurable function.
> >
> > Now, for fnonnegative 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
> nonnegative and
> measurable, let (fn) be a nondecreasing sequence of
> simple functions
> converging simply to f. In particular, (fn) is a
> nondecreasing
> sequence of nonnegative 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
 dexter cahoy <d_cahoy@...> wrote:> Thanks Sir Noel,
abs(f) <= g.
>
> I have this solution. i dont know if this makes
> sense
> .
> 1. Remember this problem ? : gintegrable,
> WTS that f is integrable.
probability
> My solution.
> let k = abs(f).
> so gk => 0.(thus nonnegative )
> then,
> int(gk) = sup { int( gn  kn) : gnkn<=gk}
>
> <= sup{ int (gn): gn<=g }
>  sup {int (kn):kn<=k }
> = int(g)  int(k)
> thus, 0<= int(gk) <= 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
> 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
> > > fsimple function, fgeneral measurable
> function.
> > >
> > > Now, for fnonnegative 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
> > nonnegative and
> > measurable, let (fn) be a nondecreasing sequence
> of
> > simple functions
> > converging simply to f. In particular, (fn) is a
> > nondecreasing
> > sequence of nonnegative 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 have this solution. i dont know if this makes sense
I think I have already pointed out that this route is not
> .
> Remember this problem ? : gintegrable, abs(f) <= g.
> WTS that f is integrable.
> My solution.
> let k = abs(f).
> so gk => 0.(thus nonnegative )
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 gk 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 realvalued....
What you want to prove is this:
Let g,k be two nonnegative 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,
It is not clear to me this should be true
> int(gk) = sup { int( gn  kn) : gnkn<=gk}
> <= sup{ int (gn): gn<=g }
>  sup {int (kn):kn<=k }
> = 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 ab or ba, 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. 0 Attachment
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 fmeasurable, gintegrable 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 nonnegative 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 nonnegative 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 fmeasurable, gintegrable 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 nonnegative and measurable maps such
> that k<=g. Then int(k)<=int(g)
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