What extra conditions are needed to state $f = g, \mu-$a.e. if $\int_{X}fd\mu=\int_{X}gd\mu$

Question

I am aware that:

$g = f, \mu-$a.e. $\Rightarrow \int_{X}fd\mu=\int_{X}gd\mu$

But I am interested in the conditions need to state that

$\int_{X}fd\mu=\int_{X}gd\mu\Rightarrow f = g, \mu-$a.e.

Do such conditions exist at all?

Answer 1

You are probably after the following theorem.

Theorem. For two integrable functions $f, g:\mathrm{X} \to \bar{\mathbf{R}} =[-\infty, \infty]$ to be equal for almost every point of $\mathrm{X},$ it is necessary and sufficient that for every measurable subset $\mathrm{A}$ of $\mathrm{X}$ the two integrals coincide $\int\limits_\mathrm{A} f = \int\limits_\mathrm{A} g.$

Proof. You probably already know the necessity, as for the sufficency consider $\mathrm{A}_n = \{f > g + \frac{1}{n}\}$ and $\mathrm{B}_n = \{f + \frac{1}{n} < g\}.$ Is easy to see each of these measurable sets has measure zero, hence so their union and thus, $f = g$ for almost every point, as was to be shown.

Answer 2

For two integrable functions $f,g$ on $X = \mathbb{R}^n$, if you know not only that $\int_X f d\mu = \int_X g d\mu$, but also $\int_Q f d\mu = \int_Q g d \mu$ for every cube $Q = [a_1,b_1] \times ... \times [a_n,b_n] \subset X$, then you can conclude that $f = g$ almost everywhere. This is proven by seeing that this is property is equivalent to $f-g$ being orthogonal to indicator functions of all cubes in the $L^2$-scalar product, and with arguments about the density of the linear span of indicator functions, one can then deduce that this means $f-g$ must be zero almost everywhere. (There are some details to be filled in, but in general, that should be the idea.)

EDIT: In fact, a result like that holds, by a more general, measure-theoretic proof, even on arbitrary measure spaces, see here. Had not thought of that at first.

Of course, that is a much stronger requirement than what you started with, but what you started with is pretty weak to begin with :) It is like asking "If two shapes have the same area, what are the conditions that they are identical?", which as you can imagine, has a boatload of different answers.