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Let $\varphi (u)=\int e^{iuy}\mu (dy)$. Since $1-\cos (uy)\ge 0$, the Tonelli theorem giveswith the fraction defined to be one at zero. For $|x|\ge 1$, $\sin x/x\le \sin 1$. To verify the constant, the function is even; on $[1,\pi ]$ it decreases because $(x\cos x-\sin x{)}^{\prime }=-x\sin x<0$ on $(0,\pi )$ and $x\cos x-\sin x$ starts from zero. On $[\pi ,\infty )$, $\sin x/x\le 1/\pi <\sin 1$. Consequently the integrand is at least $1-\sin 1$ on $|y|\ge \lambda$, proving the characteristic-function tail boundFor pointwise convergence ${\varphi }_n\to \varphi$, where $\varphi$ is a characteristic function, the dominated convergence theorem applies on each finite integration interval, since the integrands lie between zero and two. HenceContinuity of $\varphi$ at zero and $\varphi (0)=1$ make the right side tend to zero as $\lambda \to \infty$. Given $\epsilon >0$, choose $\lambda$ so it is below $\epsilon /2$. The tail probability is then below $\epsilon$ for all sufficiently large $n$. Enlarge the compact interval to control the finitely many remaining measures, each of which has vanishing tails. This proves uniform tightness of the entire sequence, not just of its tail. It is the characteristic-function tightness bound principle with the sharper displayed constant.

For the integrable mean-zero summands, differentiation at zero is justified byThe dominated convergence theorem gives ${\varphi }^{\prime }(0)=i\mathbb{E}{X}_1=0$. DefineIt is continuous away from zero by continuity of the characteristic function, and continuous at zero by the zero derivative. Since $|\varphi (u)|\le 1$, the finite geometric series yields, for $n\ge 1$,and the case $n=0$ has both sides zero. Thus the requested uniform-in-$n$ estimate isThis is the characteristic-function derivative under an absolute first moment estimate.

Independence makes the characteristic function of $S_n/n=(X_1+\cdots +X_n)/n$ equal to $\varphi (u/n{)}^n$. Therefore $|1-\varphi (u/n{)}^n|\le |u|\psi (u/n)$. Applying the already proved characteristic-function tail bound to its distribution, for every $\epsilon >0$,The convergence follows because ${\sup }_{0\le u\le 1/\epsilon }\psi (u/n)\to 0$, and the remaining integral of $u$ is finite. Hence $\boxed{S_n/n\to 0\text{ in probability}}$, the mean-zero weak law of large numbers. This deduction uses the tail bound directly and does not presuppose the Lévy continuity theorem.