Gaussian smoothing of finite measures

Put $v=\mathrm{\mid }\mu \mathrm{\mid }$. The inequality $\mathrm{\mid }\mu (E)\mathrm{\mid }\le v(E)$ gives $\mu \ll v$; v is finite, hence sigma-finite. F1 gives $\mu =uv$ with $u\in L^1(v)$, and F2 gives $v(E)={\int }_E\mathrm{\mid }u\mathrm{\mid }dv$, in particular $\int \mathrm{\mid }u\mathrm{\mid }dv=v({\mathbb{R}}^n)$. For any bounded measurable H, $\int Hd\mu =\int Hudv$: first verify this for simple H using the density formula on sets, then approximate bounded H uniformly by quantizing its real and imaginary values. F3 bounds the error on the left by $v({\mathbb{R}}^n)\mathrm{\parallel }H-H_m{\mathrm{\parallel }}_{\mathrm{\infty }}$, and the right error by $\mathrm{\parallel }u{\mathrm{\parallel }}_1\mathrm{\parallel }H-H_m{\mathrm{\parallel }}_{\mathrm{\infty }}$. AC enters F1 through the signed RN/Hahn/Jordan existence selections; it also covers the existence assumption omitted in the older F2 proof.

Consequently $h_t(x)=\int k_t(x-y)u(y)dv(y)$, absolutely at every x because k_t is bounded and u integrable. The joint function is Borel measurable. By F4, F5 and translation invariance, its double absolute integral is $\int \mathrm{\mid }u(y)\mathrm{\mid }[\int k_t(x-y)dx]dv(y)=v({\mathbb{R}}^n)$. Fubini therefore supplies a measurable integrable h_t and its stated norm bound. With the additional modulus-one Fourier factor the same double bound applies. F4 and F7 give ${\widehat{h}}_t(\xi )={\widehat{k}}_t(\xi )\int e^{-2\pi iy\cdot \xi }u(y)dv(y)={\widehat{k}}_t(\xi )\widehat{\mu }(\xi )$ by step 1.1.

For a compactly supported continuous $\phi$, the double absolute integral after multiplication by $\phi (x)$ is at most $\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}v({\mathbb{R}}^n)$. F4 exchanges the integrals. Since k_t is even, the inner test integral is $(k_t\ast \phi )(y)$, and F6 gives uniform convergence to $\phi (y)$. Thus the difference from $\int \phi udv$ has modulus at most $\mathrm{\parallel }{k}_t\ast \phi -\phi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^1(v)}\to 0$. Step 1.1 identifies the limit with $\int \phi d\mu$.