Uniqueness of a law from its characteristic function

Let $\mu ,\nu$ be the two laws. Each positive probability law, regarded as a complex measure, has total variation one: every measurable partition has sum of absolute masses equal to its total mass. Thus [F1] applies in dimension one under the stated AC. For every real $\xi$ and $x$, $e^{i(-2\pi \xi )x}=e^{-2\pi ix\xi }$, so the defining bounded integrands are identical. Integrating this pointwise identity against either law gives $\widehat{\mu }(\xi )={\phi }_{\mu }(-2\pi \xi )$ and $\widehat{\nu }(\xi )={\phi }_{\nu }(-2\pi \xi )$. Since the characteristic functions agree at every real argument, $\widehat{\mu }(\xi )=\widehat{\nu }(\xi )$ for every $\xi$. This also verifies the finite-variation hypotheses of Fourier uniqueness.

Apply F2 in dimension one to obtain $\mu =\nu$. AC is inherited from that proof: it supplies the Hahn/Jordan and Radon–Nikodym selections used in Gaussian smoothing and covers its regularity argument. No inversion result from this page is used.

For the final equivalence, F5 with $a=-1,b=0$ and F4 give ${\phi }_{-X}(t)={\phi }_X(-t)=\overline{{\phi }_X(t)}$. If ${\phi }_X$ is real-valued, the two characteristic functions agree and step 2.1 proves symmetry of the law. Conversely, symmetry means the two laws, hence their defining integrals, agree; the displayed identity then forces ${\phi }_X(t)=\overline{{\phi }_X(t)}$, so every value is real.