Let $(X, \mathcal{A}, \mu)$ be a measure space and $k \colon X \times X \to \mathbb{R}$ be a measurable positive-definite kernel such that $\int_{X} \int_X k^2(x,y) \,\mathrm{d}\mu(x) \, \mathrm{d}\mu(y) < \infty$. Then it is well known that the operator $T \colon L^2(\mu) \to L^2(\mu)$ defined by $$ Tf(x) := \int_X k(x,y) f(y) \, \mathrm{d}\mu(y) $$ is a Hilbert-Schmidt operator.
Since $k(x,y) = k(y,x)$, it is easy to see from Fubini's theorem that $$ \langle Tf, g \rangle = \int_X \left( \int_X k(x,y) f(y) \, \mathrm{d} \mu(y) \right) g(x) \, \mathrm{d} \mu(x) = \int_X \left( \int_X k(y,x) g(x) \, \mathrm{d} \mu(x) \right) f(y) \, \mathrm{d} \mu(y) = \langle f, Tg \rangle $$ and hence $T$ is self-adjoint.
How do I show that $T$ is positive?
That is, I want to show that for every $f \in L^2(\mu)$ $$ \langle Tf, f \rangle = \int_X \left( \int_X k(x,y) f(y) \, \mathrm{d} \mu(y) \right) f(x) \, \mathrm{d} \mu(x) \ge 0 $$
We need to somehow use the positive-definiteness of $k$, but I am unable to do that unless I assume $\mu$ is something simple like a counting measure. Ultimately, I want to be able to say that the eigenvalues of $T$ are such that $\lambda_1 \ge \lambda_2 \ge \cdots > 0$.