# Expand integration and undesirable points

1. Feb 6, 2008

### husseinshimal

The purpose behinde this post is an attempt to expand integration space to include functions with undesirable points where the function could be undefined.I need help to understand if this post makes sense.The main question here, is this way could be more general than using the measurement set theory to expand the integration?

Assume the function F(x),

F:[a,b]→R
We will put the interval[a,b] as acombination of subsets, UGk,

UGk=GNUGQUGQ̀....etc.(U,here stands for combination symbol) which stands for natural set,rational set ,irrational set,….etc.

Lets define the subsets,

өi={gk:F(xi)≥gk≥0},

xiЄ[a,b],

Let,(pi) be apartitoning of Gk in [a,b] and (pөi) apartitioning of(өi),

(Mөi=SUPөi), and (mөi=infөi),

(UFөi,pi)= Σi Mөi(xi-xi-1), (UFөi,pi)=upper darboux sum.

(LFөi,pi)= Σi mөi(xi-xi-1),(LFөi,pi)=lowe darboux sum.

(UFөi,pi,Pөi)=inf{UFөi:piPөi,(Pөi) partioning of(өi), (pi) partitioning of (Gk)},

(LFөi,pi,Pөi)=sup{LFөi:piPөi,(Pөi) partioning of(өi),(pi) partitioning (Gk)},

now, we put the integration in the form,

∫Fөi,over,Gk={0,UFөi}={0,LFөi}=the subsets,sөi,

∫F,over,[a,b]=Usөi

for example;

f(x):[0,1]→R,

f(x)=x , xЄ irrational numbers=Q̀, within[0,1],

f(x)=1,xЄ rational numbers=Q, within[0,1],

∫F,over,[o,1]={0,1\2}Q̀,U{0,1}Q={0,1\2}R,U,{1\2,1}Q , ,i.e,the integration would involve the real numbers within {0,1\2} plus the rational numbers within {1\2,1},(Q̀,Q and,R,are suffixes her and, U, stands for combination symbol.)

one might say it seems to be similar to the difference between Riemann integration and Lebesgue integration.

in Lebesgue integration,the above example would be, μ(0,1\2)inQ`set+μ(0,1)inQset=1\2+0=1\2,iam suggesting to keep integration in form of combination of subsets regardless they were countable or not.wouldnot this be more genral?

2. Feb 6, 2008

### HallsofIvy

Staff Emeritus
Frankly, I'm not at all clear what you are doing. You seem to be staying with intervals with the exception of "undesirable points". If so, your integration is much weaker than Lebesque integration in which we can integrate over all measurable sets. Other than the fact that all countable sets have measure 0 and so are trivial in Lebesque integration, I don't see what "countable or not" has to do with it.

The Lebesque integral of your f (f(x)= x if x is irrational, 0 if rational, 0< x< 1) is trivially 1/2. I have no idea why you would worry about any "undesirable points".