Sufficient criteria to know if a complex measure is the zero measure I'm trying to see if the following condition is enough to determine if a complex measure in $\mathbb{R}^n$ is the zero measure, however I dont see a clear way to handle the question. Suppose that $\int p\mathop{}\!d \mu=0$ for any complex polynomial $p$, can we say that $\mu$ is the zero measure? Some help will be appreciated.

An idea could be to see if the integral of any simple function is zero knowing that the integral of any polynomial is zero, however it doesn't seem true that a simple function is the limit of a sequence of polynomials.

If the measure $\mu$ has compact support (or rather its variation $|\mu|$ has compact support), then the answer should be in the positive as the Stone-Weierstrass would show.
The result in general is not true. This is related to the moment problem in probability (Stieltjes moment problem in the $(0,\infty)$, Hamburger moment problem for $\mathbb{R}$)
Here is a counterexample by Heyde:
Consider the lognormal distribution whose density is
$$f_0(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{x}\exp(-\frac12 \log^2x)\mathbb{1}_{(0,\infty)}(x)$$
for $|a|<1$ let
$$f_a(x)=f_0(x)\big(1+a\sin(2\pi\log x)\big)$$
It is not difficult to check that
$$\int^\infty_0 f_0(x)x^n \sin(2\pi\log x)\,dx=0,\qquad n=0,1,\ldots$$
So, the measure $\mu(dx)=(f_0(x)-f_a(x))\,dx$ is not $0$ and $\mu(p)=0$  for any polynomial.
There are other nice counterexamples. The book of counterexamples in Probability by Stoyan is a good place to see more of them.
