suppose we are in $\mathbb R^n$ and fix $E\subset \mathbb R^n$ and let $\chi_E$ be the indicator function of the set $E$. I'm reading an article that says that a vector field $X$ on $\mathbb R^n$, seen as a differential operator on $C^\infty(\mathbb R^n)$, has:
- distributional derivative $X\chi_E\geq 0$ if and only if $$-\int_{\mathbb R^n}\chi_E \,X\phi \,dx \geq 0 $$ for all $\phi\in C^\infty_c(\mathbb R^n)$ with $\phi\geq 0;$
- distributional derivative $X\chi_E=0$ if and only if $$\int_{\mathbb R^n}\chi_E \,X\phi \,dx = 0 $$ for all $\phi\in C^\infty_c(\mathbb R^n)$ with $\phi\geq 0.$
Now, my first doubt is why I require in both points above that the test function $\phi$ is non negative. Indeed, I think I've already seen definition 2., but, if I remember correctly, I've seen it with a general test function $\phi$ without requiring the non negativity. However, I can write any test function $\phi$ as $\phi=\phi^+-\phi^-$, thus, if $X\chi_E=0$ in the sense of 2., it also holds that $$\int_{\mathbb R^n}\chi_E \,X\phi \,dx = 0 $$ for all $\phi\in C^\infty_c(\mathbb R^n),$ right? But I can't do a similar reasoning for the point 1., so I think that for it I have to ask the non negativity of the test function $\phi$.
My second doubt is the following one: if $X\chi_E$ in the sense of distribution is a measure, then clearly $X\chi_E\geq 0$ in the sense of 2. Is the viceversa true? I mean, if I have that $X\chi_E\geq 0$ in the sense of 2., is the distributional derivative $X\chi_E$ a measure?
I hope I made myself clear.
Before starting my answer, let me remark that all your questions concern a general distribution $T$, and isn't really related to the fact that $T=X\chi_E$.
In point 2, the usual definition is indeed with any test function (it is a trivial definition: $T=0\Leftrightarrow(\forall\phi\in C^\infty_c, T(\phi)=0)$; then the specificity of $T=X\chi_E$ rely on the fact that $XS(\phi)=-S(X\phi)$ with $S=\chi_E$). However, your proof of this fact is incorrect as $\phi_+,\phi_-$ are not $C^\infty$. Though I trust the result, but you should find a proof of the fact : if $T$ satisfies $\forall \phi\geq 0, T(\phi)=0$, then $T=0$.
For point 1, you should require a sign for $\phi$. More generally, the definition of $T\geq 0$ when $T$ is a distribution is $$\forall \phi\in C^\infty_c, \phi\geq 0\Rightarrow T(\phi)\geq 0.$$ A simple reason for that is to notice that if $T(\phi)=\int f\phi$ for a locally integrable function, then the above definition is equivalent to $f\geq 0 \; a.e.$. If you didn't require a sign on $\phi$, then using $\pm\phi$, you would be lead to point 1!
About your last question: if $T$ is a positive (Radon) measure (depending on the setting you work in, you could consider non-positive measure), then it is a positive distribution. The converse is true: if $T$ is a positive distribution, then it is of order $0$, and therefore, according to Riesz representation theorem, can be represented with of positive (Radon) measure. The proof of that is a classical and short exercise (start on a compact set so that you don't have issue with the compact support of test functions)