# Interesting math theorem in measure theory

1. Oct 1, 2006

### jbusc

Sorry if this is kind of vague, but the other day, one of my math profs told me about a theorem which he thought was particularly interesting. I might be missing or getting a condition wrong, but here goes:

Suppose $$I(f, d)$$ is a real-valued function, where f is a real-valued function always defined on a region d, and that region d. if:

1. $$I(f_1 + f_2, d) = I(f_1, d) + I(f_2, d)$$
2. If $$d = d_1 \cup d_2$$, and $$d_1 \cap d_2 = \emptyset$$, then $$I(f, d) = I(f, d_1) + I(f, d_2)$$
3. If $a$ is a real constant, then $$I(af, d) = aI(f, d)$$

Then, $$I(f, d) = \int_d f$$

I asked about where I could find this in a book or online and he said a graduate-level book on measure theory (which unfortunately I don't have any). I did look through Spivak's calculus on manifolds, but nothing came close even.

Does anyone recognize this?

Last edited: Oct 1, 2006
2. Oct 1, 2006

### matt grime

As stated it cannot be true: the functional I(f,d) = k*int_d d for instance satisfies the same criterion as above, as do many others: you need some kind of normalization process.

What is true is that you're looking at the Riesz Representation theorem. Note I am very unsure of the spelling: it could be Riezche, Rietzche, or who knows what. It says that every linear functional (yes, that is the key word) is given by integrating with respect to something.

http://en.wikipedia.org/wiki/Riesz_representation_theorem

There are two related eaiser to understand ideas. The latter is the riesz theorem for hilbert spaces.

1. Given the space of polynomials on one variable, R[x], define an operator D on R[x] that is linear, sends x to 1, and some other things and you can prove that D must be differentiation. Can't recall the exact statement.

2. If H is a hilbert space (possibly with some conditions), and if f is a linaer functional then there is a y such that f(x) =<x,y> for all x.