Lévy's Continuity Theorem

Derives Lemma A18.5. The uniform bound. On \([0,1]\) the integrand is bounded by \(1\), so \(\abs{\int_{0}^{1}}\le1\). For \(Y\ge1\) integrate by parts (Equation (11.28)) with \(\sin s\,\dd s=-\dd\cos s\):

whose first term is at most \(1/Y+\cos1\le2\) in modulus and whose second is at most \(\int_{1}^{\infty}s^{-2}\dd s=1\). Hence \(\abs{S(Y)}\le1+2+1=4\) for \(Y\ge1\), and \(\abs{S(Y)}\le1\) for \(Y\le1\). The same estimate applied on \([Y,Y']\) with \(1\le Y<Y'\) gives

so \(S(Y)\) satisfies the Cauchy criterion (Theorem 11.8) and the limit \(D=\lim_{Y\rightarrow\infty}S(Y)\) exists.

The value. For \(\lambda>0\) put \(I(\lambda)=\int_{0}^{\infty}\ee^{-\lambda s}\left(\sin s\right)/s\,\dd s\), which converges absolutely since the integrand is bounded by \(\ee^{-\lambda s}\). Its \(\lambda\)-derivative is \(-\ee^{-\lambda s}\sin s\) in modulus at most \(\ee^{-\lambda_{0}s}\) for \(\lambda\ge\lambda_{0}>0\), an integrable bound independent of \(\lambda\), so the difference quotients converge uniformly on \([\lambda_{0},\infty)\) and differentiation under the integral sign is legitimate there:

Also \(\abs{I(\lambda)}\le\int_{0}^{\infty}\ee^{-\lambda s}\dd s=1/\lambda\rightarrow0\). Integrating Equation (A18.5) from \(\lambda\) to \(\infty\) therefore gives \(I(\lambda)=\tfrac{\pi}{2}-\arctan\lambda\).

The two agree in the limit. Fix \(Y\ge1\) and estimate \(I(\lambda)-D\) in three pieces. On \([0,Y]\), using \(\abs{\ee^{-\lambda s}-1}\le\lambda s\) and \(\abs{\sin s}\le1\),

The tail of \(D\) is at most \(2/Y\) by Equation (A18.4). For the tail of \(I\), integrate by parts with \(u=\ee^{-\lambda s}/s\):

the middle step bounding \(1/s\le1/Y\) in the \(\lambda\)-term and \(\ee^{-\lambda s}\le1\) in the other. Hence \(\abs{I(\lambda)-D}\le\lambda Y+5/Y\) for every \(Y\ge1\) and \(\lambda>0\). Given \(\varepsilon>0\) choose \(Y=10/\varepsilon\) and then \(\lambda<\varepsilon/(2Y)\): the right side is below \(\varepsilon\). So \(D=\lim_{\lambda\downarrow0}I(\lambda)=\tfrac{\pi}{2}-\arctan0 =\tfrac{\pi}{2}\).

The final statement is the substitution \(s=tu\) (Equation (11.27)), under which \(\left(\sin(tu)\right)/t\,\dd t=\left(\sin s\right)/s\,\dd s\) for \(u>0\); oddness of \(\sin\) in \(u\) supplies the case \(u<0\).

Derives Theorem A18.6. The diagonal argument. Enumerate the rationals, \(\Q=\set{q_{1},q_{2},\dots}\). The numbers \(F_{n}(q_{1})\) lie in the bounded set \([0,1]\), so by Bolzano–Weierstrass (Theorem 11.7) some subsequence, indexed by an infinite set \(N_{1}\subseteq\N\), has \(F_{n}(q_{1})\) convergent along \(N_{1}\); call the limit \(G(q_{1})\). Having chosen \(N_{1}\supseteq N_{2}\supseteq\dots\supseteq N_{m}\) with \(F_{n}(q_{i})\rightarrow G(q_{i})\) along \(N_{m}\) for every \(i\le m\), apply Theorem 11.7 again to the bounded sequence \(\left(F_{n}(q_{m+1})\right)_{n\in N_{m}}\) to get \(N_{m+1}\subseteq N_{m}\) and a limit \(G(q_{m+1})\). Let \(n_{k}\) be the \(k\)th element of \(N_{k}\), in increasing order and chosen larger than \(n_{k-1}\), which is possible because each \(N_{k}\) is infinite. For each fixed \(i\), all but the first \(i-1\) terms of \((n_{k})\) lie in \(N_{i}\), so

Each \(F_{n}\) is non-decreasing with values in \([0,1]\), so \(G\) inherits both properties on \(\Q\).

The limit function. Define \(F(x)=\inf\set{G(q)\mid q\in\Q,\ q>x}\). It is non-decreasing and \([0,1]\)-valued because \(G\) is. It is right-continuous: given \(x\) and \(\varepsilon>0\), pick a rational \(q>x\) with \(G(q)<F(x)+\varepsilon\); every \(y\in(x,q)\) then has \(F(y)\le G(q)<F(x)+\varepsilon\), and monotonicity gives \(F(y)\downarrow F(x)\) as \(y\downarrow x\).

Two inequalities are used repeatedly and follow at once from the definition and the monotonicity of \(G\): for rationals \(r<s\),

The first holds because \(s>r\) makes \(G(s)\) one of the numbers whose infimum is \(F(r)\); the second because every rational \(q>s\) has \(G(q)\ge G(r)\), so the infimum defining \(F(s)\) is at least \(G(r)\).

Convergence at continuity points. Let \(x\) be a continuity point of \(F\) and let \(\varepsilon>0\). By left-continuity of \(F\) at \(x\) there is \(y<x\) with \(F(y)>F(x)-\varepsilon\); choose rationals \(r,s\) with \(y<r<s<x\). Then \(F_{n_{k}}(x)\ge F_{n_{k}}(s)\rightarrow G(s)\), and by Equation (A18.7) twice, \(G(s)\ge G(r)\ge F(y)\), whence

For the other side, continuity gives \(z>x\) with \(F(z)<F(x)+\varepsilon\); choose a rational \(q\in(x,z)\). Then \(F_{n_{k}}(x)\le F_{n_{k}}(q)\rightarrow G(q)\), and \(G(q)\le F(z)\) by the second inequality of Equation (A18.7), so \(\limsup_{k}F_{n_{k}}(x)\le F(z)<F(x)+\varepsilon\). As \(\varepsilon>0\) was arbitrary, \(F_{n_{k}}(x)\rightarrow F(x)\).

Derives Lemma A18.8. Given \(\varepsilon>0\) take \(K\) as in Definition A18.7. A monotone function has at most countably many discontinuities — the open intervals \(\left(F(x^{-}),F(x^{+})\right)\) belonging to distinct jumps are disjoint and each contains a rational — so continuity points are dense and we may pick continuity points \(x>K\) and \(x'<-K\). Then \(F(x)=\lim_{k}F_{n_{k}}(x)\ge\liminf_{k}\Pr(\abs{X_{n_{k}}}\le K) \ge1-\varepsilon\) and \(F(x')=\lim_{k}F_{n_{k}}(x')\le\varepsilon\). Monotonicity extends both to all larger, respectively smaller, arguments.

Derives Lemma A18.9. Write \(\norm{g}_{\infty}=\sup\abs{g}\) and let \(\varepsilon>0\). Since \(F\) is a distribution function and its continuity points are dense, choose continuity points \(a<b\) with \(F(a)<\varepsilon\) and \(1-F(b)<\varepsilon\); then for \(n\) large \(F_{n}(a)<2\varepsilon\) and \(1-F_{n}(b)<2\varepsilon\). The contributions to \(\avg{g(X_{n})}\) and \(\avg{g(X)}\) from outside \((a,b]\) are therefore at most \(2\varepsilon\norm{g}_{\infty}\) and \(\varepsilon\norm{g}_{\infty}\) in modulus.

On the compact interval \([a,b]\) the function \(g\) is uniformly continuous (Theorem 11.25), so there is \(\delta>0\) with \(\abs{g(x)-g(y)}<\varepsilon\) whenever \(\abs{x-y}<\delta\) inside \([a,b]\). Choose continuity points \(a=x_{0}<x_{1}<\dots<x_{m}=b\) of \(F\) with \(x_{j}-x_{j-1}<\delta\) — possible because continuity points are dense. Then for any distribution function \(H\),

because on each \((x_{j-1},x_{j}]\) the integrand differs from \(g(x_{j})\) by less than \(\varepsilon\) and the increments of \(H\) sum to at most \(1\). Applying this to \(H=F_{n}\) and to \(H=F\), and using \(F_{n}(x_{j})\rightarrow F(x_{j})\) at each of the finitely many partition points, gives

Since \(\varepsilon>0\) was arbitrary, the limit is \(0\).

Derives Theorem A18.10. Write \(J_{T}\) for the integral on the left. For real \(u,v\),

so with \(u=X-a\), \(v=X-b\) the integrand of \(J_{T}\), written as \(\left(\ee^{\ii t(X-a)}-\ee^{\ii t(X-b)}\right)/(\ii t)\) under the expectation, is bounded by the constant \(b-a\). Fubini's theorem (Theorem A18.3) — the first of the two quoted uses — therefore applies:

Now \(\ee^{\ii tu}/(\ii t)=\left(\sin tu\right)/t -\ii\left(\cos tu\right)/t\), and the cosine terms are odd in \(t\), so they cancel over the symmetric interval. Hence the inner integral is

which by Lemma A18.5 is bounded by \(4C_{0}\le16\) uniformly in \(T\) and in \(X\), and converges as \(T\rightarrow\infty\) to \(\pi\left[\sgn(X-a)-\sgn(X-b)\right]\) pointwise. Dominated convergence (Theorem A18.2) — the second quoted use — gives

The bracket takes the value \(0\) when \(X<a\) or \(X>b\), the value \(2\) when \(a<X<b\), and the value \(1\) when \(X=a\) or \(X=b\). So

which, using \(F(x^{-})=\Pr(X<x)\) and \(F(x)=\Pr(X\le x)\), is exactly the right-hand side of Equation (A18.9).

Derives Corollary A18.11. Let \(F_{1},F_{2}\) have the common characteristic function \(\varphi\). The left-hand side of Equation (A18.9) depends only on \(\varphi\), so \(F_{1}(b)-F_{1}(a)=F_{2}(b)-F_{2}(a)\) whenever \(a,b\) are continuity points of both. Continuity points of either are dense (the argument of Lemma A18.8), so the continuity points common to both are dense as well, their complement being a union of two countable sets. Letting \(a\rightarrow-\infty\) through such points gives \(F_{1}(b)=F_{2}(b)\) at every common continuity point \(b\); and every \(x\in\R\) is the limit from the right of such points, so right-continuity of both functions gives \(F_{1}=F_{2}\) everywhere.

Derives Lemma A18.12. The integrand \(1-\cos(tX)\) is bounded by \(2\), so Fubini's theorem (Theorem A18.3) — the third quoted use — allows the expectation and the \(t\)-integral to be exchanged:

the inner integral being elementary and the quotient being read as \(2\) when \(X=0\). Put \(y=uX\). The function \(1-\left(\sin y\right)/y\) is non-negative for every real \(y\), since \(\abs{\sin y}\le\abs{y}\); and when \(\abs{y}\ge2\) the bound \(\abs{\sin y}\le1\) gives \(1-\left(\sin y\right)/y\ge1-1/\abs{y}\ge\tfrac12\), so that \(2\left(1-\left(\sin y\right)/y\right)\ge1\) there. Hence the integrand of the expectation dominates \(\indic{\abs{uX}\ge2}\) pointwise, and taking expectations gives Equation (A18.14).

Derives Lemma A18.13. First, \(\varphi(0)=\lim_{n}\varphi_{n}(0)=1\) by Proposition 15.43(i). Let \(\varepsilon>0\). Continuity of \(\varphi\) at \(0\) gives \(u>0\) with \(\abs{1-\varphi(t)}<\varepsilon/4\) for \(\abs{t}\le u\), whence

The functions \(1-\Re\varphi_{n}\) are bounded by \(2\) on \([-u,u]\) and converge pointwise to \(1-\Re\varphi\), so dominated convergence (Theorem A18.2) — the fourth and last quoted use, here applied to the uniform measure on \([-u,u]\) — gives

so there is \(N\) with the left side below \(\varepsilon\) for \(n\ge N\). By Lemma A18.12, \(\Pr(\abs{X_{n}}\ge2/u)\le \varepsilon\) for \(n\ge N\). Each of the finitely many remaining variables \(X_{1},\dots,X_{N-1}\) is individually tight, its distribution function tending to \(0\) and \(1\) at the two ends, so there is \(K_{0}\) with \(\Pr(\abs{X_{n}}>K_{0})\le\varepsilon\) for \(n<N\). Take \(K=\max(2/u,K_{0})\).

Proof of Theorem A18.1. Derives Theorem A18.1. By Lemma A18.13 the family \(\set{X_{n}}\) is tight.

Every subsequence has a convergent further subsequence with the same limit. Let \((n_{j})\) be any subsequence. By Helly's theorem (Theorem A18.6) it has a further subsequence \((n_{j_{k}})\) along which \(F_{n_{j_{k}}}\rightarrow F\) at every continuity point of some non-decreasing right-continuous \(F:\R\rightarrow[0,1]\); by Lemma A18.8 and tightness, \(F\) is a genuine distribution function. Let \(X\) be a random variable with distribution function \(F\). Applying the Helly–Bray lemma (Lemma A18.9) to the bounded continuous functions \(x\mapsto\cos(tx)\) and \(x\mapsto\sin(tx)\), for each fixed \(t\), gives

But by hypothesis \(\varphi_{n_{j_{k}}}(t)\rightarrow\varphi(t)\). Limits in \(\C\) are unique, so \(\varphi=\varphi_{X}\): the limit function is a characteristic function, which is the first assertion of the theorem. Moreover, by Corollary A18.11 the distribution function \(F\) is determined by \(\varphi\) alone, so every subsequential Helly limit obtained this way is the same function \(F\).

Hence the whole sequence converges. Let \(x\) be a continuity point of \(F\) and suppose \(F_{n}(x)\) does not converge to \(F(x)\). Then there are \(\eta>0\) and a subsequence \((n_{j})\) with \(\abs{F_{n_{j}}(x)-F(x)}\ge\eta\) for every \(j\). By the previous paragraph applied to \((n_{j})\) there is a further subsequence along which the distribution functions converge, at every continuity point of \(F\) and in particular at \(x\), to \(F\) — contradicting the standing inequality. Therefore \(F_{n}(x)\rightarrow F(x)\) at every continuity point of \(F\), which is \(X_{n}\rightarrow X\) in distribution.

Derives Corollary A18.14. The first claim is Equation (A18.15) read for the full sequence, the Helly–Bray lemma applying directly since \(F_{n}\rightarrow F_{X}\) at continuity points by hypothesis. For continuity of \(\varphi_{X}\): Equation (A18.10) read with \(u=X\), \(v=0\) and \(t=h\) gives \(\abs{\ee^{\ii hX}-1}\le\abs{h}\abs{X}\), while the modulus of a difference of two unit complex numbers is at most \(2\), so \(\abs{\ee^{\ii(t+h)X}-\ee^{\ii tX}}=\abs{\ee^{\ii hX}-1} \le\min(\abs{hX},2)\) and hence \(\abs{\varphi_{X}(t+h)-\varphi_{X}(t)}\le\avg{\min(\abs{hX},2)}\), a bound independent of \(t\) that tends to \(0\) with \(h\) — by the same truncation as in Lemma 15.44: split the expectation at \(\abs{X}=K\), bound the inner part by \(\abs{h}K\) and the outer by \(2\Pr(\abs{X}>K)\), and choose \(K\) then \(h\). So \(\varphi_{X}\) is in fact uniformly continuous.