Characteristic Functions Inversion and Continuity

Write $\phi (t)=\int e^{itx}\mu (dx)$ with $\mu (\mathbb{R})=1$. At zero the integrand is one, so $\phi (0)=1$. The triangle inequality gives $\mathrm{\mid }\phi (t)\mathrm{\mid }\le \int \mathrm{\mid }{e}^{itx}\mathrm{\mid }d\mu =1$.

Writing the integral componentwise, $\phi (-t)=\int \cos (tx)d\mu -i\int \sin (tx)d\mu =\overline{\phi (t)}$. Linearity and exponential addition further give $\mathrm{\mid }\phi (t+h)-\phi (t)\mathrm{\mid }\le \int \mathrm{\mid }{e}^{itx}(e^{ihx}-1)\mathrm{\mid }d\mu =\int \mathrm{\mid }{e}^{ihx}-1\mathrm{\mid }d\mu \mathrm{.}$

For real $u$, apply the mean value theorem separately to sine and cosine between $0$ and $u$. Their derivatives have absolute value at most one by Euler's formula, so $\mathrm{\mid }\sin u\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$ and $\mathrm{\mid }\cos u-1\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$. Thus $\mathrm{\mid }{e}^{iu}-1\mathrm{\mid }\le \min (2,2\mathrm{\mid }u\mathrm{\mid })$. For each positive integer $n$, define $g_n(x)=\min (2,2\mathrm{\mid }x\mathrm{\mid }\mathrm{/}n)$. This measurable sequence tends pointwise to zero and is dominated by the integrable constant $2$, so DCT gives $\int g_{n}d\mu \to 0$. Whenever $\mathrm{\mid }h\mathrm{\mid }\le 1\mathrm{/}n$, step 1.2 bounds $\mathrm{\mid }\phi (t+h)-\phi (t)\mathrm{\mid }$ by $\int g_{n}d\mu$, for every $t$. Given $\epsilon >0$, take the least positive integer $n$ with this integral less than $\epsilon$. Then $\delta =1\mathrm{/}n$ proves uniform continuity. This uses a prescribed sequence and no choice of a sequence of counterexamples.

The addition law gives $e^{it(aX+b)}=e^{itb}{e}^{i(at)X}$. Both random exponentials are bounded and integrable. Pulling out the constant $e^{itb}$ gives the affine identity, including $a=0$ and $b=0$.

For $n\ge 1$ put $c_j=\cos (t{X}_j)$ and $s_j=\sin (t{X}_j)$. Expand $e^{it{\sum }_j{X}_j}={\prod }_j(c_j+i{s}_j)={\sum }_{A\subseteq \{1,\dots ,n\}}{i}^{\mathrm{\mid }A\mathrm{\mid }}{\prod }_{j\in A}{s}_j{\prod }_{j\mathrm{\notin }A}{c}_j\mathrm{.}$ Each real factor is a bounded Borel function of its own coordinate, so the real factorization theorem applies to each of these finitely many products. Complex linearity then gives $\mathbb{E}{e}^{it{\sum }_j{X}_j}={\sum }_A{i}^{\mathrm{\mid }A\mathrm{\mid }}{\prod }_{j\in A}\mathbb{E}{s}_j{\prod }_{j\mathrm{\notin }A}\mathbb{E}{c}_j={\prod }_j(\mathbb{E}{c}_j+i\mathbb{E}{s}_j)={\prod }_j{\phi }_{X_j}(t)\mathrm{.}$

For $n=0$ the sum is zero, its exponential is one, and the empty product is one. For $n=1$ the asserted identity is the defining expectation itself.

Fix $n\ge 1$, real $t_j$ and complex $z_j$. Set $Z(x)={\sum }_{j=1}^n{z}_j{e}^{i{t}_{j}x}$. By the unit-modulus formula in the characteristic-function definition, $\mathrm{\mid }Z(x)\mathrm{\mid }\le {\sum }_j\mathrm{\mid }{z}_j\mathrm{\mid }$, so $\mathrm{\mid }Z{\mathrm{\mid }}^2$ is integrable against the probability law $\mu$. Moreover $\overline{e^{i{t}_{k}x}}=e^{-i{t}_{k}x}$, so $\mathrm{\mid }Z(x){\mathrm{\mid }}^2={\sum }_{j,k}{z}_j\overline{z_k}{e}^{i(t_j-t_k)x}\mathrm{.}$

Integrating this finite sum gives ${\sum }_{j,k}{z}_j\overline{z_k}{\phi }_{\mu }(t_j-t_k)=\int \mathrm{\mid }Z(x){\mathrm{\mid }}^2\mu (dx)\ge 0.$ The integral is real because its integrand is real and nonnegative. This verifies every quadratic form required by the definition.

Let $\mu ,\nu$ be the two laws. For every real $\xi$, F1 gives $\widehat{\mu }(\xi )={\phi }_{\mu }(-2\pi \xi )={\phi }_{\nu }(-2\pi \xi )=\widehat{\nu }(\xi )$. Each positive probability law, regarded as a complex measure, has total variation one: every measurable partition has sum of absolute masses equal to its total mass. Thus the finite-variation hypotheses of Fourier uniqueness hold.

Apply F2 in dimension one to obtain $\mu =\nu$. AC is inherited from that proof: it supplies the Hahn/Jordan and Radon–Nikodym selections used in Gaussian smoothing and covers its regularity argument. No inversion result from this page is used.

For the final equivalence, F5 with $a=-1,b=0$ and F4 give ${\phi }_{-X}(t)={\phi }_X(-t)=\overline{{\phi }_X(t)}$. If ${\phi }_X$ is real-valued, the two characteristic functions agree and step 2.1 proves symmetry of the law. Conversely, symmetry means the two laws, hence their defining integrals, agree; the displayed identity then forces ${\phi }_X(t)=\overline{{\phi }_X(t)}$, so every value is real.

The derivative of sine at zero makes $\sin u\mathrm{/}u\to 1$. MVT and $\mathrm{\mid }\cos u\mathrm{\mid }\le 1$ give $\mathrm{\mid }\sin u\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$, so the extended quotient is continuous and bounded by one on $[0,1]$. It has a proper integral on every bounded interval. AC supplies the countable choice needed to identify these with Lebesgue integrals using F3 (and the compact integration interface F15).

For $B>A\ge 1$ and $\epsilon \ge 0$, put $w(u)=e^{-\epsilon u}\mathrm{/}u$. This is positive, decreasing, and continuously differentiable on $[A,B]$, with ${\int }_A^B\mathrm{\mid }w\mathrm{\prime }\mathrm{\mid }=w(A)-w(B)$. Integration by parts against $\sin u=(-\cos u)\mathrm{\prime }$ gives $\mid {\int }_A^{B}w(u)\sin udu\mid \le w(A)+w(B)+{\int }_A^B\mathrm{\mid }w\mathrm{\prime }\mathrm{\mid }=2w(A)\le 2\mathrm{/}A\mathrm{.}$ At $\epsilon =0$ this proves the Cauchy property of $S(T)$ as $T\to \mathrm{\infty }$ and the bound $\mathrm{\mid }S(T)\mathrm{\mid }\le 3$ for all $T\ge 0$. For positive damping it also bounds the infinite tail by $2\mathrm{/}A$.

Fix $\epsilon >0$. FTC gives $\sin u\mathrm{/}u={\int }_0^1\cos (su)ds$, including $u=0$. The double absolute integral of $e^{-\epsilon u}\cos (su)$ on $[0,\mathrm{\infty })\times [0,1]$ is at most $1\mathrm{/}\epsilon$, so Fubini applies. Differentiating $\frac{e^{-\epsilon u}(-\epsilon \cos (su)+s\sin (su))}{{\epsilon }^2+s^2}$ gives $e^{-\epsilon u}\cos (su)$; its limit at infinity is zero and its value at zero is $-\epsilon \mathrm{/}({\epsilon }^2+s^2)$. Consequently $J_{\epsilon }:={\int }_0^{\mathrm{\infty }}{e}^{-\epsilon u}\frac{\sin u}{u}du={\int }_0^1\frac{\epsilon }{{\epsilon }^2+s^2}ds=\arctan (1\mathrm{/}\epsilon )\mathrm{.}$

On $[0,A]$, dominated convergence gives convergence of the damped integral to $S(A)$ as $\epsilon \downarrow 0$. The two tails, damped and undamped, are each at most $2\mathrm{/}A$ by step 2.1. Thus, first taking $\epsilon \downarrow 0$ and then $A\to \mathrm{\infty }$, $J_{\epsilon }\to {\lim }_{T}S(T)$. The increasing inverse arctangent has limit $\pi \mathrm{/}2$ at infinity: its values are below $\pi \mathrm{/}2$, and for every $v<\pi \mathrm{/}2$ in its range, $y>\tan v$ implies $\arctan y>v$. Hence $S(T)\to \pi \mathrm{/}2$.

For $z=0$ the symmetric integral is zero. For $z\mathrm{\ne }0$, evenness in $t$ and substitution $u=\mathrm{\mid }z\mathrm{\mid }t$ give ${\int }_{-T}^T\frac{\sin (tz)}{t}dt=2\mathrm{sgn}(z)S(T\mathrm{\mid }z\mathrm{\mid })\mathrm{.}$ Its absolute value is at most six, and for each fixed nonzero $z$ its limit is $\pi \mathrm{sgn}(z)$. The uniform bound, but not uniform convergence in z, is asserted.

Let $\mu ={\mathbb{P}}_X$ and define $q(t)={\int }_a^b{e}^{-ity}dy$. Applying FTC to the sine and cosine components gives the stated quotient for $t\mathrm{\ne }0$, while $q(0)=b-a$. The integral expression shows $\mathrm{\mid }q(t)\mathrm{\mid }\le b-a$ and continuity at zero by dominated convergence on $[a,b]$. F7 identifies the compact integrals with Lebesgue integrals; AC covers its assumption and F2.

For $T>0$ the joint integrand $q(t)e^{itx}$ is Borel and its absolute integral against $dt\mu (dx)$ on $[-T,T]\times \mathbb{R}$ is at most $2T(b-a)$. Fubini gives ${\int }_{-T}^{T}q(t){\phi }_X(t)dt={\int }_{\mathbb{R}}(R_T(x-a)-R_T(x-b))\mu (dx),R_T(z)={\int }_{-T}^T\frac{\sin (tz)}{t}dt\mathrm{.}$ Indeed expand the exponentials after multiplication by $e^{itx}$: the imaginary part is an odd function of t and integrates to zero, leaving the two displayed real sine integrals.

F2 bounds the difference of sine kernels uniformly in x and T, and its limit is $\pi (\mathrm{sgn}(x-a)-\mathrm{sgn}(x-b))$. This equals $2\pi$ when $a<x<b$, $\pi$ when $x=a$ or $x=b$, and zero when $x<a$ or $x>b$. Since $\mu$ has mass one, dominated convergence applies to the right-hand side of step 2.1. Division by $2\pi$ proves every term of the stated formula, including the half endpoint atoms.

Put $C=(2\pi {)}^{-1}\int \mathrm{\mid }{\phi }_X\mathrm{\mid }<\mathrm{\infty }$. The defining integral is absolutely convergent and $\mathrm{\mid }f(x)\mathrm{\mid }\le C$. For $x_n\to x$, the integrands converge pointwise and are dominated by $\mathrm{\mid }{\phi }_X\mathrm{\mid }$, so DCT gives $f(x_n)\to f(x)$. Under the assumed AC the sequential criterion proves continuity. Conjugating the integral and substituting $s=-t$, conjugate symmetry yields $\overline{f(x)}=f(x)$; hence f is real.

By F1 and absolute convergence, for every $a<b$ $\mu ((a,b))+\frac{1}{2}\mu (\{a,b\})=\frac{1}{2\pi }\int q_{a,b}(t){\phi }_X(t)dt,q_{a,b}(t)={\int }_a^b{e}^{-ity}dy\mathrm{.}$ Its modulus is at most $C(b-a)$. Taking $a=x-1\mathrm{/}n,b=x+1\mathrm{/}n$, the point x lies in the open interval, so positivity and monotonicity give $\mu (\{x\})\le \mu ((a,b))\le C(b-a)=2C\mathrm{/}n$ for every n, hence every singleton has mass zero. Also the double absolute integral of $e^{-ity}{\phi }_X(t)$ on $(a,b)\times \mathbb{R}$ is $(b-a)\mathrm{\parallel }{\phi }_X{\mathrm{\parallel }}_1$, so Fubini gives $\mu ((a,b))={\int }_a^{b}f(y)dy$.

If $f(x_0)<0$, continuity supplies an interval about $x_0$ on which $f<f(x_0)\mathrm{/}2<0$, contradicting the nonnegative interval mass in step 2.1. Thus $f\ge 0$, and F4 defines a Borel measure $\nu (B)={\int }_{B}f$. Applying continuity from below to $(-n,n)\uparrow \mathbb{R}$ for both measures gives $\nu (\mathbb{R})=\mu (\mathbb{R})=1$. Bounded open intervals together with the empty set form a pi-system generating the Borel sets: their rational-endpoint subfamily is a countable base for the real topology. Step 2.1 and F5 therefore imply $\nu =\mu$. AC is inherited from F1 and covers the sequential continuity use in step 1.1.

For $0\le j\le k$, $\mathrm{\mid }x{\mathrm{\mid }}^j\le 1+\mathrm{\mid }x{\mathrm{\mid }}^k$, so $G_j(t)=\mathbb{E}[(iX{)}^j{e}^{itX}]$ exists. Applying MVT to sine and cosine gives $\mathrm{\mid }{e}^{iv}-e^{iu}\mathrm{\mid }\le \min (2,2\mathrm{\mid }v-u\mathrm{\mid })$. For $\mathrm{\mid }h\mathrm{\mid }\le 1\mathrm{/}n$ it follows that $\mathrm{\mid }{G}_j(t+h)-G_j(t)\mathrm{\mid }\le \mathbb{E}[\mathrm{\mid }X{\mathrm{\mid }}^j\min (2,2\mathrm{\mid }X\mathrm{\mid }\mathrm{/}n)]\mathrm{.}$ This prescribed nonnegative sequence tends pointwise to zero and is dominated by $2\mathrm{\mid }X{\mathrm{\mid }}^j$. DCT makes the bound tend to zero, proving continuity of every $G_j$ without selecting an arbitrary sequence of frequencies.

For real x and nonzero h put $Q_h(x)=(e^{ihx}-1)\mathrm{/}h-ix$. MVT applied to $\cos (hx)-1$ and $\sin (hx)$ yields numbers between zero and hx with $\mid \frac{\cos (hx)-1}{h}\mid \le \mathrm{\mid }x\mathrm{\mid }\min (1,\mathrm{\mid }hx\mathrm{\mid }),\mid \frac{\sin (hx)}{h}-x\mid \le \mathrm{\mid }x\mathrm{\mid }\min (2,\mathrm{\mid }hx\mathrm{\mid })\mathrm{.}$ Here the first bound uses $\mathrm{\mid }\sin v\mathrm{\mid }\le \min (1,\mathrm{\mid }v\mathrm{\mid })$ and the second $\mathrm{\mid }\cos v-1\mathrm{\mid }\le \min (2,\mathrm{\mid }v\mathrm{\mid })$, each obtained from the same derivative bounds. Therefore $\mathrm{\mid }{Q}_h(x)\mathrm{\mid }\le \min (4\mathrm{\mid }x\mathrm{\mid },2\mathrm{\mid }h\mathrm{\mid }\mathrm{\mid }x{\mathrm{\mid }}^2)$, including x=0. For $1\le j\le k$, linearity and exponential addition give $\mid \frac{G_{j-1}(t+h)-G_{j-1}(t)}{h}-G_j(t)\mid \le \mathbb{E}[\mathrm{\mid }X{\mathrm{\mid }}^{j-1}\min (4\mathrm{\mid }X\mathrm{\mid },2\mathrm{\mid }X{\mathrm{\mid }}^2\mathrm{/}n)]$ whenever $0<\mathrm{\mid }h\mathrm{\mid }\le 1\mathrm{/}n$, interpreting the integrand as zero at X=0. The right side tends to zero by DCT, dominated by $4\mathrm{\mid }X{\mathrm{\mid }}^j$. Hence $G_{j-1}\mathrm{\prime }=G_j$.

Starting from $G_0={\phi }_X$, the derivative identities in step 2.1 and continuity in step 1.1 establish the assertion through order k. If k=0 only the continuity conclusion of step 1.1 is required. At t=0 the formula becomes ${\phi }_X^{(j)}(0)=i^j\mathbb{E}[X^j]$. All limits used prescribed majorants indexed by positive integers; this proof introduces no selection axiom.

Fix $t\in \mathbb{R}$. The real functions $x\mapsto \cos (tx)$ and $x\mapsto \sin (tx)$ are continuous and bounded by one, so weak convergence gives convergence of each of their integrals against ${\mu }_n$ to the corresponding integral against $\mu$.

By the componentwise definition, the cosine integrals are the real parts of the characteristic functions and the sine integrals their imaginary parts. Combining the two convergences gives the asserted complex limit. At t=0 these two integral sequences are constantly one and zero respectively. Since t was arbitrary the result holds at every frequency.

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.

Normalization and the pointwise limit give $\psi (0)=1$ and $\mathrm{\mid }\psi \mathrm{\mid }\le 1$. Its real and imaginary parts are Borel as pointwise limits of continuous real functions. For a fixed $\delta >0$, put $I_n(\delta )=\int w_{\delta }(1-\mathrm{Re}{\phi }_n)$ and $I(\delta )=\int w_{\delta }(1-\mathrm{Re}\psi )$. The integrands converge pointwise and lie between zero and $2w_{\delta }$, an integrable majorant of integral two. Hence $I_n(\delta )\to I(\delta )$. Given $\epsilon >0$, continuity at zero permits $\delta >0$ so small that $\mathrm{\mid }1-\psi (t)\mathrm{\mid }<3\epsilon \mathrm{/}16$ on $[-\delta ,\delta ]$. Then $I(\delta )\le 3\epsilon \mathrm{/}16$, and for every sufficiently large $n$, $I_n(\delta )<3\epsilon \mathrm{/}8$. The quantitative tail bound gives ${\mu }_n\{\mathrm{\mid }x\mathrm{\mid }\ge 4\mathrm{/}\delta \}<\epsilon \mathrm{/}2$ for those $n$. No equicontinuity of the sequence has been assumed.

For each of the finitely many earlier indices, ${\mu }_n([-m,m])\uparrow 1$ as $m\to \mathrm{\infty }$. Taking the maximum of $4\mathrm{/}\delta$ and finitely many radii therefore gives $R$ with ${\mu }_n(\mathbb{R}\setminus [-R,R])<\epsilon$ for every $n$. This interval is compact, so the whole sequence is tight. The argument also covers the case of no exceptional early indices. The real line is complete, and its countable dense rational subset makes it Polish. Prokhorov now provides a subsequence ${\mu }_{n_j}\Rightarrow \mu$, with $\mu$ a Borel probability law.

For each real $t$, forward continuity gives ${\phi }_{\mu }(t)={\lim }_j{\phi }_{n_j}(t)=\psi (t)$. Uniqueness of laws with a given characteristic function shows that this $\mu$ is unique. Every subsequence of the original sequence is tight by the same compact bounds and hence has a further weakly convergent subsequence; its limit has characteristic function $\psi$ by exactly the preceding equality and therefore equals $\mu$.

Fix a bounded continuous real $f$. If $\int fd{\mu }_n$ failed to converge to $\int fd\mu$, there would be $a>0$ and infinitely many indices whose errors are at least $a$. Enumerate them in increasing order, taking the least next index at every stage. Step 3.1 gives a further subsequence converging weakly to $\mu$, contradicting this fixed error bound for $f$. Thus every such test converges and ${\mu }_n\Rightarrow \mu$. At frequency zero all characteristic functions and $\psi$ equal one, so a zero-mass limit is excluded. Constant sequences and point masses need no separate nondegeneracy condition. AC here is inherited from Prokhorov (compact selections and its subsequence supplier), Fourier uniqueness, and the integration bridge; the least-index test argument uses no additional choice.

If ${\mu }_n\Rightarrow \mu$, the forward continuity theorem gives the right-hand side at every frequency, including zero, where all values are one.

Conversely suppose the right-hand side. The specified target $\mu$ has a characteristic function continuous at zero. Apply the converse theorem with $\psi ={\phi }_{\mu }$: it gives a unique law $\nu$ with this characteristic function and ${\mu }_n\Rightarrow \nu$. The law $\mu$ itself satisfies the characterizing property, so that uniqueness gives $\nu =\mu$. AC is inherited through the converse theorem's Prokhorov, Fourier-uniqueness and integration-bridge uses. Point masses and constant sequences are included, and no nonzero variance, moment, or density is required.