This might be a foolish question but it's bothering me today. Suppose we have some expression of the form
$$g(x_{\mu})-f(x_{\mu})=0$$ which holds everywhere on some domain D. If we make the above expression an integrand and say:
$$\intop_{D}\left[g(x_{\mu})-f(x_{\mu})\right]dD$$
Is it still equal to zero, or a constant? Logic tells me if I integrate the RHS of the first equation a constant results; However it also seems like the left hand side when considered as the limit of a series (like a riemann sum integral), is just the summation of infinitely many zeros and hence zero itself.
This started because in the calculus of variations you get a Lagrange type equation when you consider the variation to be zero.
$$\delta S=\intop\left[g(x_{\mu})-f(x_{\mu})\right]=0$$
In my texts (yes physics texts) this is taken to imply:
$$g(x_{\mu})-f(x_{\mu})=0$$
Does it go the other way (are the two logically equivalent?) Maybe I just need some more coffee EDIT: Here's the specific example that's at the heart of my issue. In considering energy and momentum in physics, the statement of it's conservation is written (“normal” derivatives, not covariant ones here)
$$\nabla_{\mu}T^{\mu\nu}=0$$
Where $T^(\mu\nu)$ is the stress energy tensor. It is a second rank 4x4 tensor symmetric in it's two indices. The fact that I'm talking about tensors shouldn't alter the issue (hopefully). The Vanishing divergence of this quantity is taken as the definitive statement of conservation of it's integral quantities. Using Guass's theorem, it is often written as an expression for energy momentum:
$$\intop_{M}\nabla_{\mu}T^{\mu\nu}dM=\intop_{\partial M}T^{\mu\nu}n_{\nu}d(\partial M)$$
choosing a spacelike surface (with a timelike normal $\nu=0$), we have conserved quantities, each of which correspond to components of the momentum four-vector.
$$\intop_{\partial M}T^{\mu0}d(\partial M)=P_{\mu}=momentumconstant$$ But how can this be if the initial integrand above vanishes identically everywhere?
You seem to be confused between definite and indefinite integrals.
If a function $f=c$ is constant on a domain $D=[a,b]$, then $$ \int_D f=\int_D c=c(b-a) $$ Set $c=0$ to get that your integral is indeed 0.
In fact this direction of the implication is much easier than the other; in order to have $$ \int_Df=0\implies f\equiv 0 $$ on $D$, we need something much stronger, like continuity and nonnegativity of $f$, arbitrary domain $D$ in some larger set $\Omega$ etc.