# Averages over infinite sets

1. Jun 3, 2013

### cragar

The set of all functions is larger than $2^{\aleph_0}$.
So lets say I wanted to average over all functions over some given region. that was
larger than $2^{\aleph_0}$ how would I do that.

2. Jun 3, 2013

### pwsnafu

Define what you mean by "average". The word is pretty much useless in mathematics.

3. Jun 3, 2013

### cragar

add them all up and then divide by the total number.

4. Jun 3, 2013

### Stephen Tashi

That doesn't define a particular mathematical procedure unless you can define "add" and "divide" in your context.

The problem of defining an "average" of all the real numbers seems conceptually simpler and I don't know of any useful definition for such an average.

Some people use the term "average" to mean "expectation". If you have a particular probability distribution on the set of all real numbers, the "expectation" of that distribution is defined. Some people use the term "average" to mean the "mean value" of a finite sample of data or the "expectation" of a probability distribution.

5. Jun 3, 2013

### Dmobb Jr.

Technically the set of all functions does not exist. It is to big to be a set.

If you mean real valued functions, while the above post is completely correct, I think the answer you want is f(x) = 0. If you sum every function, I believe you will get 0 because for every f(x) there exists g(x) = -f(x)

6. Jun 3, 2013

### Stephen Tashi

I don't think symmetry directs us to a particular answer. For every function h(x) = there is a function g(x) = 5 - h(x) so by the same reasoning, the answer would be the function f(x) = 5.

7. Jun 3, 2013

### Dmobb Jr.

Good point. I am pretty sure 0 is still going to be the best answer but I have to consider the problem more thoroughly.

Edit: The answer will probably be 0 assuming there is some sort of reasonable answer at all.