Tightness from characteristic function equicontinuity at zero

The continuous nonnegative weight is supported on $[-\delta ,\delta ]$ and has integral $2{\delta }^{-1}{\int }_0^{\delta }(1-t\mathrm{/}\delta )dt=1$. Set $K_{\delta }(x)=\int w_{\delta }(t)\cos (tx)dt$. At $x=0$ it equals one. For $x\mathrm{\ne }0$, integration by parts on $[0,\delta ]$, with $u=1-t\mathrm{/}\delta$ and $v=\sin (tx)\mathrm{/}x$, gives $K_{\delta }(x)=\frac{2}{{\delta }^{2}x}{\int }_0^{\delta }\sin (tx)dt=\frac{2(1-\cos (\delta x))}{{\delta }^2{x}^2}\mathrm{.}$ All functions and their derivatives here are continuous on that interval; the primitive and integration bridge therefore apply. The formula gives $K_{\delta }\ge 0$, while its defining integral and $\cos \le 1$ give $K_{\delta }\le 1$. Also $K_{\delta }(x)\le 4\mathrm{/}({\delta }^2{x}^2)\le 1\mathrm{/}4$ when $\mathrm{\mid }x\mathrm{\mid }\ge 4\mathrm{/}\delta$, including equality in the cutoff.

The function $w_{\delta }(t)(1-\cos (tx))$ is jointly Borel, nonnegative, and has product integral at most $2\int w_{\delta }=2$: Lebesgue measure is sigma-finite and $\mu$ is finite. Fubini and the characteristic-function definition yield $\int w_{\delta }(t)(1-\mathrm{Re}{\varphi }_{\mu }(t))dt=\int (1-K_{\delta }(x))\mu (dx)\ge \frac{3}{4}\mu \{\mathrm{\mid }x\mathrm{\mid }\ge 4\mathrm{/}\delta \}\mathrm{.}$ Nonnegativity off the tail justifies discarding its complement. Rearrangement proves the quantitative assertion.

Given $\epsilon >0$, equicontinuity supplies $r>0$ with $\mathrm{\mid }1-{\varphi }_{\mu }(t)\mathrm{\mid }<3\epsilon \mathrm{/}8$ whenever $\mathrm{\mid }t\mathrm{\mid }<r$, uniformly in $\mu$. Choose $0<\delta <r$. The weight has mass one, so the integral in the bound is at most $3\epsilon \mathrm{/}8$. Thus $\mu (\mathbb{R}\setminus [-4\mathrm{/}\delta ,4\mathrm{/}\delta ])\le \epsilon \mathrm{/}2<\epsilon$ for every member. The interval is compact, proving tightness. For an empty family the empty compact set suffices. A singleton family and an atom at zero satisfy the same calculation (the latter has zero right-hand side). There is no assertion at $\delta =0$, where the weight is undefined.