What is the relationship between integrals and measures in measure theory?

  • Thread starter Thread starter cappadonza
  • Start date Start date
  • Tags Tags
    Integral Measure
cappadonza
Messages
26
Reaction score
0
I just beginning to study measure theory. so far from what i understand so far , can we say in general, an integral is a measure, (ie it is nothing but a set function. a mapping F : \mathcal{F} \rightarrow \mathbb{R} where \mathcal{F} is a family of sets.

does it make sense to say in general an integral of a function F, is \int_{A} F d\mu is the measure of the image of the F over some set A using the measure \mu. with the condition the image of F over A must be measurable using the measure \mu

so for example the two that i know are lebesgue-integral, lebesgue-stieltjes integral, are basically are general integrals using different measures.

if we could say the above then where would the riemann integral fit into this.
sorry if this is a bit vague, I'm trying to get my head around this stuff
 
Last edited:
Physics news on Phys.org
No, I would say your description is wrong. Why not wait a couple more weeks in the course, then try to fomulate it again?
 
I'm not actually doing a course, i work full-time, its something I'm trying to learn through self-study. i go back re-learn what i thought i knew and re-formulate it in the next week or so
thanks
 
okay here is my second attempt:
A measure is a set function \mathcal{F} \to \Re. where \mathcal{F} is a sigma-algebra. the invariants it must satisfy are it countable additive and the measure of a null-set is zero.
Now integral \int_{B} f d\mu is nothing but a special case of a measure, where it calculates the measure of the set described by f over the set B. The arbitrary measure \mu is use to calculate the measure of this set.
Since we viewing integral is as measure it must satisfy the invariants of a measure such as being countable additive. The Riemann-integral only satisfies this condition only a small class of functions, this why the Lebesgue integral is introduced, to over come some of these shor-coming
 
I'd say you're on the right track.

Even though the integral of a real-valued (measurable) function over a subset D of the domain on which the function is defined is not in general a measure, it is certainy possible (and easy) to define functions whose integral will be a measure.

E.g. a strictly positive real-valued function will have an integral that is a measure, but the intgral of a sine-wave will not fulfill the axioms of a measure.

You can, in fact, view any real-valued positive function as a quotient between two different measures on the same sigma-algebra: see the "Radon–Nikodym theorem", it was of great help for me understanding measures in relation to measurable functions.
 
Namaste & G'day Postulate: A strongly-knit team wins on average over a less knit one Fundamentals: - Two teams face off with 4 players each - A polo team consists of players that each have assigned to them a measure of their ability (called a "Handicap" - 10 is highest, -2 lowest) I attempted to measure close-knitness of a team in terms of standard deviation (SD) of handicaps of the players. Failure: It turns out that, more often than, a team with a higher SD wins. In my language, that...
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Back
Top