Characteristic Functions

At $t=0$, $\varphi_X(0)=\E[1]=1$, and $\abs{\varphi_X(t)}=\abs{\E[e^{itX}]}\le\E\abs{e^{itX}}=1$. For uniform continuity, $\abs{\varphi_X(t+h)-\varphi_X(t)}=\abs{\E[e^{itX}(e^{ihX}-1)]}\le\E\abs{e^{ihX}-1}$, a bound independent of $t$, and $\abs{e^{ihX}-1}\le 2$ with $e^{ihX}-1\to 0$ pointwise as $h\to 0$, so the dominated convergence theorem sends $\E\abs{e^{ihX}-1}\to 0$, which is uniform continuity. For the product rule, independence makes $e^{itX}$ and $e^{itY}$ independent, so the factorisation of expectations applied to real and imaginary parts gives $\varphi_{X+Y}(t)=\E[e^{itX}e^{itY}]=\E[e^{itX}]\,\E[e^{itY}]=\varphi_X(t)\varphi_Y(t)$.

The difference quotient $(e^{i(t+h)X}-e^{itX})/h$ converges to $iXe^{itX}$ as $h\to 0$ and is bounded in modulus by $\abs X$, since $\abs{e^{ihX}-1}\le\abs{hX}$. When $\E\abs X<\infty$ the dominated convergence theorem lets the derivative pass inside, $\varphi_X'(t)=\E[iXe^{itX}]$. Iterating $k\le n$ times, with dominator $\abs X^k$ integrable by hypothesis, gives $\varphi_X^{(k)}(t)=\E[(iX)^k e^{itX}]$, continuous in $t$ by dominated convergence, and at $t=0$ equal to $i^k\E[X^k]$. The expansion Equation (1) is Taylor's theorem applied to the $n$ times differentiable $\varphi_X$ at the origin with these derivatives.

Write $\varphi_X(t)=\int_\R e^{itx}\,d\P_X(x)$ and insert it into the integral. The integrand is bounded by $b-a$ on $[-T,T]\times\R$, since $\abs{(e^{-ita}-e^{-itb})/(it)}=\abs{\int_a^b e^{-its}\,ds}\le b-a$, so the finite-measure Fubini theorem permits exchanging the integrals,

The inner expression collapses to a sine integral because, after division by $it$, the cosine parts of $e^{it(x-a)}-e^{it(x-b)}$ carry a factor $1/t$ that is odd in $t$ and integrate to zero over the symmetric range, while the sine parts are even and survive. The Dirichlet integral $\int_0^T\frac{\sin ct}{t}\,dt\to\frac{\pi}{2}\operatorname{sgn}(c)$ as $T\to\infty$, with the partial integrals bounded uniformly in $c$ and $T$. So the inner expression is bounded and converges to $\tfrac12(\operatorname{sgn}(x-a)-\operatorname{sgn}(x-b))$, which equals $1$ on $(a,b)$, $\tfrac12$ at $x\in\{a,b\}$, and $0$ outside $[a,b]$. The bounded convergence theorem sends the right side of Equation (3) to $\P(a<X<b)+\tfrac12(\P(X=a)+\P(X=b))$, which under the continuity hypothesis is $\P(a<X\le b)$.

The formula Equation (2) determines $\P(a<X\le b)$ from $\varphi_X$ for every $a,b$ that are not atoms. The non-atoms are all but countably many points, hence dense, so the distribution function is determined at a dense set of points and, being right-continuous, everywhere. The law is determined by its distribution function.

By change of variables and Fubini,

the inner integral being $\frac1u(2u-2\sin(ux)/x)=2(1-\sin(ux)/(ux))$. The integrand is nonnegative, and where $\abs{ux}\ge 2$ it satisfies $1-\frac{\sin ux}{ux}\ge 1-\frac1{\abs{ux}}\ge\tfrac12$. Restricting the integral to $\{\abs x\ge 2/u\}$ and using this bound gives the right side at least $2\cdot\tfrac12\P(\abs X \ge 2/u)=\P(\abs X\ge 2/u)$.

First, tightness. Fix $\varepsilon>0$. Since $\varphi(0)=\lim\varphi_n(0)=1$ and $\varphi$ is continuous at $0$, choose $u>0$ with $\frac1u\int_{-u}^u(1-\Re\varphi(t))\,dt<\varepsilon/2$. The integrands $1-\Re \varphi_n$ are bounded by $2$ and converge pointwise to $1-\Re\varphi$, so by bounded convergence the integrals converge and $\frac1u\int_{-u}^u(1-\Re\varphi_n(t))\,dt<\varepsilon$ for all large $n$. By Lemma 6, $\P(\abs{X_n}\ge 2/u)<\varepsilon$ for those $n$, and enlarging $2/u$ to cover the finitely many remaining $n$ makes the family tight.

Now take any subsequence. By Helly's selection theorem, every sequence of distribution functions has a further subsequence converging at continuity points to a nondecreasing right-continuous limit $G$, obtained by a diagonal extraction of convergent values on the rationals followed by right-continuous interpolation. Tightness forces $G$ to be a genuine distribution function, with no mass lost to $\pm\infty$. Along that subsequence $X_{n_k}\Rightarrow\nu$ for the law $\nu$ of $G$, and convergence in distribution with the uniformly bounded continuous integrands $e^{itx}$ gives $\varphi_{n_k}(t)\to\varphi_ \nu(t)$. But $\varphi_{n_k}(t)\to\varphi(t)$ by hypothesis, so $\varphi_\nu=\varphi$, and by Corollary 5 every subsequential limit is the same law $\mu$ with characteristic function $\varphi$. A sequence all of whose subsequences have a further subsequence converging to the same limit converges to that limit, so $X_n\Rightarrow\mu$.