Why does current conservation involve an arbitrary function? In section 6.1 of Peskin's quantum field theory introduction, right after equation 6.3, the four current density $j^{\mu}$ is said to be conserved because for any function $f \left( x \right)$ that falls off at infinity, we have
$$
\int f \left( x \right) \partial _{\mu} j^{\mu} \left( x \right) \mathrm{d}^4 x = 0
$$
I am just so confused on the fact that there is a function $f \left( x \right)$ involved in this evaluation. Is current conservation not just $\int \partial _{\mu} j^{\mu} \left( x \right) \mathrm{d}^4 x = 0$?
Thanks in advance!
 A: Current conservation is $\partial_\mu j^\mu=0$, not integrated over ANYTHING (note the zeroth component of $j$ is the current density, so this is equivalent to the usual statement of current conservation $\dot\rho=-\vec\nabla\cdot\vec j$). So, Peskin wants to show $\partial_\mu j^\mu$ is zero. He does it by showing its integral over any test function $f(x)$ is zero. If $g(x)$ is a function such that $\int f(x)g(x)=0$ for all test functions $f(x)$, then $g(x)=0$. 
A: I don't know Peskin's book, so I guess. Maybe the author has defined a quantum field as an operator-valued distribution? Then it would make sense to insert a test function, since to say $T=0$ if $T$ is a distribution means $T(f)=0$ for any test function $f$ and we physicists are used to read $T(f)$ as
$$\int\!T(x)\,f(x)\, {\rm d}^4x.$$
Rather meaningless in several cases$\dots$ E.g. $D$ defined by
$D(f)=(df/dx)_{x=0}$ is a tempered distribution on $\Bbb R$ but in
order to write it as an integral you have to resort to
$$D(f) = -\!\int\!\delta'(x)\,f(x)\, {\rm d}x.$$
