# Equality of integrals => equality of integrands

1. Jan 2, 2012

### Derivator

hi folks,

$\int_A f(x) dx = \int_A g(x) dx$ for arbirary A, thus f(x) = g(x), since the equaltiy of the Integrals holds for any domain A.

I don't see, why the argument "...for any domain A..." really justifies this conclusion.

Can someone explain this to me, please?

2. Jan 2, 2012

### LCKurtz

For continuous functions and argument might go like this. If f and g aren't identical, there is a point a where one of them, say f(a), is greater than g(a). So there is an interval I containing a where f(x) > g(x). But then $\int_I f(x)-g(x)\, dx > 0$ contradicting the assumption "for any domain A".

3. Jan 2, 2012

### micromass

And for discontinuous functions, the result might not even be true!! It needs to be:

If $\int_A f(x)dx = \int_A g(x)dx$ then $f=g$ almost everywhere.

4. Jan 2, 2012

### mathman

For the more general case the conclusion would be f(x)=g(x), except on a set of measure 0.

5. Jan 2, 2012

### resolvent1

I think you can do something like the following. Assuming the integrals of f and g are equal for every set A (and f and g are obviously measurable):

Consider the function $f(x) - g(x)$. Let E be the set of x's in A where this function is positive. Define

$E_{1/n} = \left{ x \in E : f(x) - g(x) > 1/n \right}$

Thus by Tschebyshev,

$m(E_{1/n}) \leq n \int_{ E_{1/n} } f - g = 0$

Therefore

$m(E) = \cup_{n=1}^\infty m(E_{1/n}) = 0$

ie, the set where f is strictly greater than g has measure zero.

You can do basically the same argument to show that the set where g is strictly greater than f has measure zero. So f and g are the same for ae x. It's not really a for any domain A type question, it just needs to be the case for these E_{1/n}'s.

Last edited: Jan 2, 2012