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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.

For probability measures on a metric space, weak convergence of probability measures ${\mu }_n\Rightarrow \mu$ means $\int fd{\mu }_n\to \int fd\mu$ for every bounded continuous real function $f$. On the real line it is equivalently convergence of the cumulative distribution functions at every continuity point of the limiting cumulative distribution function. Bounded continuous upper and lower approximations to interval indicators give one direction; a partition at continuity points, uniform approximation on a compact interval, and control of the tails give the reverse direction.

For a probability measure $\mu$ on $\mathbb{R}$, its characteristic function is ${\phi }_{\mu }(t)=\int e^{itx}\mu (dx)$. The Lévy continuity theorem states: ${\mu }_n\Rightarrow \mu$ implies ${\phi }_{{\mu }_n}(t)\to {\phi }_{\mu }(t)$ for every real $t$; conversely, if these characteristic functions have a pointwise limit $\phi$ which is continuous at zero, there is a unique probability measure $\mu$ with characteristic function $\phi$, and ${\mu }_n\Rightarrow \mu$. In particular, for a specified limiting probability measure, weak convergence is equivalent to pointwise convergence of its characteristic functions.

The forward implication follows by testing cosine and sine. The dominated convergence theorem shows that every characteristic function is continuous at zero and has value one there. For the converse, first establish tightness of probability measures. By Fubini's theorem, for $h>0$,The integrand on the right is at least $1/2$ when $|x|\ge 2/h$. This proves the characteristic-function tightness boundThe integrals converge for each $h$ by dominated convergence, and continuity of $\phi$ at zero makes the limit as small as desired by choosing $h$ small. Thus all sufficiently late measures have a common tail bound; the finitely many earlier measures also have small tails once the radius is increased. This is uniform tightness of probability measures.

We give the required compactness argument explicitly. For any subsequence, choose a further diagonal subsequence whose cumulative distribution functions $F_n$ converge at every rational $q$, to $f(q)$. DefineThis function is nondecreasing and right-continuous: given a rational $q>x$ with $f(q)$ close to $F(x)$, the same bound holds for $F(y)$ when $x<y<q$. Tightness gives $F(-\infty )=0$ and $F(+\infty )=1$, so $F$ is a cumulative distribution function of a probability measure $\nu$. For rationals $r<x<s$, $F_n(r)\le F_n(x)\le F_n(s)$. The limits bracket $F(x-)$ and $F(x)$; at a continuity point of $F$ they coincide. Thus $F_n(x)\to F(x)$ at all such points. To obtain weak convergence, choose a compact interval with small tails and partition it using continuity points of $F$. A bounded continuous function is uniformly continuous there, so step-function approximations and convergence of interval probabilities prove convergence of its integral. This proves Helly selection for distribution functions and supplies a weak subsequential limit $\nu$.

Its characteristic function is $\phi$ by the already proved forward implication. We also prove uniqueness rather than assume it. If two probability measures $\nu ,\rho$ have the same characteristic function, convolve each with a centered normal distribution of variance ${\epsilon }^2>0$. Their densities are, by the Gaussian integral and Fubini's theorem,and likewise for $\rho$. The integrable Gaussian factor justifies exchanging the integrals, so these densities agree. For every bounded continuous $f$, letting $\epsilon \downarrow 0$ in $\int \mathbb{E}[f(x+\epsilon Z)]\nu (dx)$, with $Z$ a standard normal random variable, gives $\int fd\nu$ by bounded convergence theorem. Hence the two measures have equal integrals against all these test functions, and approximating interval indicators shows $\nu =\rho$. This proves the uniqueness theorem for characteristic functions.

Every subsequence therefore has a further subsequence weakly converging to the same unique $\mu$. If the original sequence failed weak convergence, some bounded continuous test function would have a subsequence of integrals staying a fixed distance from its integral under $\mu$, contradicting that further convergence. This completes both directions of the theorem. Continuity at zero is essential: the laws $N(0,n)$ have pointwise limiting characteristic function zero away from zero and one at zero, and are not tight.