If $p(x)$ is a function such that $\int_{\mathbb{R}^m} p(x) dx$ then can I say it is the density of a probability measure $\mu$ on $\mathbb{R}^m$?

Question

(I am aware this is a bit imprecise. I will make it more precise with time)

Suppose $\mu$ is a probability measure on $\mathbb{R}^m$ and that $\lambda$ is the Lebesgue measure on the same space. I have found a function $p(x)\geq 0$ such that $$ \int_{\mathbb{R}^m} p(x) \lambda(x) = 1. $$ Can I say that $p(x)$ is the Radon-Nikodym derivative of $\mu$ with respect to $\lambda$?

I am aware of the definition of Randon-Nikodym derivative. One requires $\mu\ll \lambda$ and then the RN derivative is a measurable function $f$ such that $\mu(A) = \int f(x) \lambda (x)$ for every measurable set $A$. However this is somehow the reverse of that statement.