Let $\lambda$ be a Borel probability measure on $\mathbb{R}^n$ and $\phi$ be its characteristic function (that is, $\phi(u)=\int_{\mathbb{R}^n} e^{i \langle u,x \rangle} d\lambda(x)$), and assume that $\phi\in L^1(m)$.
Define $f(x)=\int_{\mathbb{R}^n} \phi(u)e^{-i \langle u,x \rangle} dm(u)$.
How do I show that $f$ is nonnegative “everywhere”?
(The only thing I can derive now is that $f$ is real and continuous and bounded)
Suppose that there exists $x \in \mathbb{R}^n$ such that $f(x)<0$. As $f$ is continuous, we can find a "good" rectangle $I$ and some $\epsilon>0$ such that $x \in I$ and
$$f(y) \leq -\epsilon <0 \qquad \text{for all $y \in I$}.$$
This would imply
$$\int_I f \, dm \leq - \epsilon m(I) <0$$
in contradiction to
$$\int_I f \, dm = \lambda(I) \geq 0.$$
Consequently we conclude that $f(x) \geq 0$ for all $x \in \mathbb{R}^n$. By choosing a sequence of "good" rectangles $I_k \uparrow \mathbb{R}^n$, we find from the monotone convergence theorem that
$$\int_{\mathbb{R}^n} f\, dm = \sup_{k \in \mathbb{N}} \int_{I_k} f \, dm = \sup_{k \in \mathbb{N}} \lambda(I_k) \leq 1,$$
and so $f \in L^1(m)$.