The Lindeberg central limit theorem

1 Convergence in distribution

We denote by $𝒫({ℝ}^d)$ the collection of Borel probability measures on ${ℝ}^d$. Unless we say otherwise, we use the narrow topology on $𝒫({ℝ}^d)$: the coarsest topology such that for each $f\in C_b({ℝ}^d)$, the map

$$\mu \mapsto {\int }_{{ℝ}^d}f𝑑\mu$$

is continuous $𝒫({ℝ}^d)\to ℂ$. Because ${ℝ}^d$ is a Polish space it follows that $𝒫({ℝ}^d)$ is a Polish space. (In fact, its topology is induced by the Prokhorov metric.)

2 Characteristic functions

For $\mu \in 𝒫({ℝ}^d)$, we define its characteristic function $\tilde{\mu }:{ℝ}^d\to ℂ$ by

$$\tilde{\mu }(u)={\int }_{{ℝ}^d}{e}^{iu\cdot x}𝑑\mu (x).$$

Theorem 1.

If $\mu \in 𝒫(ℝ)$ has finite $k$th moment, $k\ge 0$, then, writing $\varphi =\tilde{\mu }$:

1. $\varphi \in C^k(ℝ)$.

2. ${\varphi }^{(k)}(v)={(i)}^k{\int }_{ℝ}{x}^k{e}^{ivx}𝑑\mu (x)$.

3. ${\varphi }^{(k)}$ is uniformly continuous.

4. $|{\varphi }^k(v)|\le {\int }_{ℝ}{|x|}^k𝑑\mu (x)$.

Proof.

For $0\le l\le k$, define $f_l:ℝ\to ℂ$ by

$$f_l(v)={\int }_{ℝ}{x}^l{e}^{ivx}𝑑\mu (x).$$

For $h\ne 0$,

$$\left|x^l{e}^{ivx}\frac{e^{ihx}-1}{h}\right|\le |x^l\cdot x|={|x|}^{l+1},$$

so by the dominated convergence theorem we have for $0\le l\le k-1$,

$\underset{h\to 0}{lim}{\displaystyle \frac{f_l(v+h)-f_l(v)}{h}}$ $=\underset{h\to 0}{lim}{\displaystyle {\int }_{ℝ}}{x}^l{e}^{ivx}{\displaystyle \frac{e^{ihx}-1}{h}}𝑑\mu (x)$

$={\displaystyle {\int }_{ℝ}}{x}^l{e}^{ivx}\left(\underset{h\to 0}{lim}{\displaystyle \frac{e^{ihx}-1}{h}}\right)𝑑\mu (x)$

$={\displaystyle {\int }_{ℝ}}i{x}^{l+1}{e}^{ivx}𝑑\mu (x).$

That is,

$$f_l^{\prime }=i{f}_{l+1}.$$

And, by the dominated convergence, for $ϵ>0$ there is some $\delta >0$ such that if then

hence if then

$|f_k(v)-f_k(u)|$ $=\left|{\displaystyle {\int }_{ℝ}}{x}^k{e}^{iux}(e^{i(v-u)x}-1)𝑑\mu (x)\right|$

$\le {\displaystyle {\int }_{ℝ}}{|x|}^k|e^{i(v-u)x}-1|𝑑\mu (x)$

showing that $f_k$ is uniformly continuous. As well,

$$|f_k(v)|\le {\int }_{ℝ}{|x|}^k𝑑\mu (x)$$

But $\varphi =f_0$, i.e. ${\varphi }^{(0)}=f_0$, so

$${\varphi }^{(1)}=f_0^{\prime }=i{f}_1,{\varphi }^{(2)}={(i{f}_1)}^{\prime }={(i)}^2{f}_2,\mathrm{\cdots },{\varphi }^{(k)}={(i)}^k{f}_k.$$

∎

If $\varphi \in C^k(ℝ)$, Taylor’s theorem tells us that for each $x\in ℝ$,

$\varphi (x)$ $={\displaystyle \sum _{l=0}^{k-1}}{\displaystyle \frac{{\varphi }^{(l)}(0)}{l!}}{x}^l+{\displaystyle {\int }_0^x}{\displaystyle \frac{{(x-t)}^{k-1}}{(k-1)!}}{\varphi }^{(k)}(t)𝑑t$

$={\displaystyle \sum _{l=0}^k}{\displaystyle \frac{{\varphi }^{(l)}(0)}{l!}}{x}^l+{\displaystyle {\int }_0^x}{\displaystyle \frac{{(x-t)}^{k-1}}{(k-1)!}}({\varphi }^{(k)}(t)-{\varphi }^{(k)}(0))𝑑t$

$={\displaystyle \sum _{l=0}^k}{\displaystyle \frac{{\varphi }^{(l)}(0)}{l!}}{x}^l+R_k(x),$

and $R_k(x)$ satisfies

$$|R_k(x)|\le \left(\underset{0\le u\le 1}{sup}|{\varphi }^{(k)}(ux)-{\varphi }^{(k)}(0)|\right)\cdot \frac{{|x|}^k}{k!}.$$

Define ${\theta }_k:ℝ\to ℂ$ by ${\theta }_k(0)=0$ and for $x\ne 0$

$${\theta }_k(x)=\frac{k!}{x^k}\cdot R_k(x),$$

with which, for all $x\in ℝ$,

$$\varphi (x)=\sum _{l=0}^k\frac{{\varphi }^{(l)}(0)}{l!}{x}^l+\frac{1}{k!}{\theta }_k(x)x^k.$$

Because $R_k$ is continuous on $ℝ$, ${\theta }_k$ is continuous at each $x\ne 0$. Moreover,

$$|{\theta }_k(x)|\le \underset{0\le u\le 1}{sup}|{\varphi }^{(k)}(ux)-{\varphi }^{(k)}(0)|,$$

and as ${\varphi }^{(k)}$ is continuous it follows that ${\theta }_k$ is continuous at $0$. Thus ${\theta }_k$ is continuous on $ℝ$.

Lemma 2.

If $\mu \in 𝒫(ℝ)$ have finite $k$th moment, $k\ge 0$, and for $0\le l\le k$,

$$M_l={\int }_{ℝ}{x}^l𝑑\mu (x),$$

then there is a continuous function $\theta :ℝ\to ℂ$ for which

$$\tilde{\mu }(x)=\sum _{l=0}^k\frac{{(i)}^l{M}_l}{l!}{x}^l+\frac{1}{k!}\theta (x)x^k.$$

The function $\theta$ satisfies

$$|\theta (x)|\le \underset{0\le u\le 1}{sup}|{\tilde{\mu }}^{(k)}(ux)-{\tilde{\mu }}^{(k)}(0)|.$$

Proof.

From Theorem 1, $\tilde{\mu }\in C^k(ℝ)$ and

$${\tilde{\mu }}^{(l)}(0)={(i)}^l{\int }_{ℝ}{x}^l𝑑\mu (x)={(i)}^l{M}_l.$$

Thus from what we worked out above with Taylor’s theorem,

$$\tilde{\mu }(x)=\sum _{l=0}^k\frac{{(i)}^l{M}_l}{l!}{x}^l+\frac{1}{k!}{\theta }_k(x)x^k,$$

for which

$$|{\theta }_k(x)|\le \underset{0\le u\le 1}{sup}|{\tilde{\mu }}^{(k)}(ux)-{\tilde{\mu }}^{(k)}(0)|.$$

∎

For $a\in ℝ$ and $\sigma >0$, let

$$p(t,a,{\sigma }^2)=\frac{1}{\sigma \sqrt{2\pi }}\exp \left(-\frac{{(t-a)}^2}{2{\sigma }^2}\right),t\in ℝ.$$

Let ${\gamma }_{a,{\sigma }^2}$ be the measure on $ℝ$ whose density with respect to Lebesgue measure is $p(\cdot ,a,{\sigma }^2)$. We call ${\gamma }_{a,{\sigma }^2}$ a Gaussian measure. We calculate that the first moment of ${\gamma }_{a,{\sigma }^2}$ is $a$ and that its second moment is ${\sigma }^2$. We also calculate that

$${\tilde{\gamma }}_{a,{\sigma }^2}(x)=\exp \left(iax-\frac{1}{2}{\sigma }^2{x}^2\right).$$

Lévy’s continuity theorem is the following.

Theorem 3 (Lévy’s continuity theorem).

Let ${\mu }_n$ be a sequence in $𝒫({ℝ}^d)$.

1. If $\mu \in 𝒫({ℝ}^d)$ and ${\mu }_n\to \mu$, then for each ${\tilde{\mu }}_n$ converges to $\tilde{\mu }$ pointwise.

2. If there is some function $\varphi :{ℝ}^d\to ℂ$ to which ${\tilde{\mu }}_n$ converges pointwise and $\varphi$ is continuous at $0$, then there is some $\mu \in 𝒫({ℝ}^d)$ such that $\varphi =\tilde{\mu }$ and such that ${\mu }_n\to \mu$.

3 The Lindeberg condition, the Lyapunov condition, the Feller condition, and asymptotic negligibility

Let $(\mathrm{\Omega },ℱ,P)$ be a probability and let $X_n$, $n\ge 1$, be independent $L^2$ random variables. We specify when we impose other hypotheses on them; in particular, we specfify if we suppose them to be identically distributed or to belong to $L^p$ for $p>2$.

For a random variable $X$, write

$$\sigma (X)=\sqrt{\mathrm{Var}(X)}=\sqrt{E({|X-E(X)|}^2)}.$$

Write

$${\sigma }_n=\sigma (X_n),$$

and, using that the $X_n$ are independent,

$$s_n=\sigma \left(\sum _{j=1}^n{X}_j\right)={\left(\sum _{j=1}^n{\sigma }_j^2\right)}^{1/2}$$

and

$${\eta }_n=E(X_n).$$

For $n\ge 1$ and $ϵ>0$, define

$L_n(ϵ)$ $={\displaystyle \frac{1}{s_n^2}}{\displaystyle \sum _{j=1}^n}E({(X_j-{\eta }_j)}^2||X_j-{\eta }_j|\ge ϵs_n)$

$={\displaystyle \frac{1}{s_n^2}}{\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|x-{\eta }_j|\ge ϵs_n}}{(x-{\eta }_j)}^{2}d(X_j{}_{*}P)(x).$

We say that the sequence $X_n$ satisfies the Lindeberg condition if for each $ϵ>0$,

$$\underset{n\to \mathrm{\infty }}{lim}{L}_n(ϵ)=0.$$

For example, if the sequence $X_n$ is identically distributed, then $s_n^2=n{\sigma }_1^2$, so

$L_n(ϵ)$ $={\displaystyle \frac{1}{n{\sigma }_1^2}}{\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|x-{\eta }_1|\ge ϵn^{1/2}{\sigma }_1}}{(x-{\eta }_1)}^{2}d(X_1{}_{*}P)(x)$

$={\displaystyle \frac{1}{{\sigma }_1^2}}{\displaystyle {\int }_{|x-{\eta }_1|\ge ϵn^{1/2}{\sigma }_1}}{(x-{\eta }_1)}^{2}d(X_1{}_{*}P).$

But if $\mu$ is a Borel probability measure on $ℝ$ and $f\in L^1(\mu )$ and $K_n$ is a sequence of compact sets that exhaust $ℝ$, then

$${\int }_{ℝ\setminus K_n}|f|𝑑\mu \to 0,n\to \mathrm{\infty }.$$

Hence $L_n(ϵ)\to 0$ as $n\to \mathrm{\infty }$, showing that $X_n$ satisfies the Lindeberg condition.

We say that the sequence $X_n$ satisfies the Lyapunov condition if there is some $\delta >0$ such that the $X_n$ are $L^{2+\delta }$ and

$$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{s_n^{2+\delta }}\sum _{j=1}^{n}E({|X_j-{\eta }_j|}^{2+\delta })=0.$$

In this case, for $ϵ>0$, then $|x-\eta |\ge ϵs_n$ implies ${|x-\eta |}^{2+\delta }\ge {|x-\eta |}^2{(ϵs_n)}^{\delta }$ and hence

$L_n(ϵ)$ $\le {\displaystyle \frac{1}{s_n^2}}{\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|x-{\eta }_j|\ge ϵs_n}}{\displaystyle \frac{{|x-{\eta }_j|}^{2+\delta }}{{(ϵs_n)}^{\delta }}}d(X_j{}_{*}P)(x)$

$={\displaystyle \frac{1}{ϵs_n^{2+\delta }}}{\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|x-{\eta }_j|\ge ϵs_n}}{|x-{\eta }_j|}^{2+\delta }d(X_j{}_{*}P)(x)$

$={\displaystyle \frac{1}{ϵs_n^{2+\delta }}}{\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|X_j-{\eta }_j|\ge ϵs_n}}{|X_j-{\eta }_j|}^{2+\delta }𝑑P$

$\le {\displaystyle \frac{1}{ϵs_n^{2+\delta }}}{\displaystyle \sum _{j=1}^n}E({|X_j-{\eta }_j|}^{2+\delta })$

$\to 0.$

This is true for each $ϵ>0$, showing that if $X_n$ satisfies the Lyapunov condition then it satisfies the Lindeberg condition.

For example, if $X_n$ are identically distributed and $L^{2+\delta }$, then

$$\frac{1}{s_n^{2+\delta }}\sum _{j=1}^{n}E({|X_j-{\eta }_j|}^{2+\delta })=\frac{1}{n^{\delta /2}{\sigma }_1^{2+\delta }}E({|X_j-{\eta }_j|}^{2+\delta })\to 0,$$

showing that $X_n$ satisfies the Lyapunov condition.

Another example: Suppose that the sequence $X_n$ is bounded by $M$ almost surely and that $s_n\to \mathrm{\infty }$. $|X_n|\le M$ almost surely implies that

$$|{\eta }_n|=|E(X_n)|\le E(|X_n|)\le E(M)=M.$$

Therefore $|X_n-{\eta }_n|\le |X_n|+|{\eta }_n|\le 2M$ almost surely. Let $\delta >0$. Then, as $s_n^2=n{\sigma }_1^2$,

${\displaystyle \frac{1}{s_n^{2+\delta }}}{\displaystyle \sum _{j=1}^n}E({|X_j-{\eta }_j|}^{2+\delta })$ $\le {\displaystyle \frac{1}{s_n^{2+\delta }}}{\displaystyle \sum _{j=1}^N}E({|X_j-{\eta }_j|}^2){(2M)}^{\delta }$

$={\displaystyle \frac{{(2M)}^{\delta }}{s_n^{\delta }}}$

$\to 0,$

showing that $X_n$ satisfies the Lyapunov condition.

We say that a sequence of random variables $X_n$ satisfies the Feller condition when

$$\underset{n\to \mathrm{\infty }}{lim}\underset{1\le j\le n}{\max }\frac{{\sigma }_j}{s_n}=0,$$

where ${\sigma }_j=\sigma (X_j)=\sqrt{\mathrm{Var}(X_j)}$ and

$$s_n={\left(\sum _{j=1}^n{\sigma }_j^2\right)}^{1/2}.$$

We prove that if a sequence satisfies the Lindeberg condition then it satisfies the Feller condition.

Lemma 4.

If a sequence of random variables $X_n$ satisfies the Lindeberg condition, then it satisfies the Feller condition.

Proof.

Let $ϵ>0$¿ For $n\ge 1$ and $1\le k\le n$, we calculate

${\sigma }_k^2$ $={\displaystyle {\int }_{ℝ}}{(x-{\eta }_k)}^{2}d(X_k{}_{*}P)(x)$

$\le {(ϵs_n)}^2+{\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|x-{\eta }_j|\ge ϵs_n}}{(x-{\eta }_j)}^{2}d(X_j{}_{*}P)(x)$

$={ϵ}^2{s}_n^2+s_n^2{L}_n(ϵ).$

Hence

$$\underset{1\le k\le n}{\max }{\left(\frac{{\sigma }_k}{s_n}\right)}^2\le {ϵ}^2+L_n(ϵ),$$

and so, because the $X_n$ satisfy the Lindeberg condition,

$$\underset{n\to \mathrm{\infty }}{lim\; sup}\underset{1\le k\le n}{\max }{\left(\frac{{\sigma }_k}{s_n}\right)}^2\le {ϵ}^2.$$

This is true for all $ϵ>0$, which yields

$$\underset{n\to \mathrm{\infty }}{lim}\underset{1\le k\le n}{\max }{\left(\frac{{\sigma }_k}{s_n}\right)}^2=0,$$

namely, that the $X_n$ satisfy the Feller condition. ∎

We do not use the following idea of an asymptotically negligible family of random variables elsewhere, and merely take this as an excsuse to write out what it means. A family of random variables $X_{n,j}$, $n\ge 1$, $1\le j\le k_n$, is called asymptotically negligible if for each $ϵ>0$,

$$\underset{n\to \mathrm{\infty }}{lim}\underset{1\le j\le k_n}{\max }P(|X_{n,j}|\ge ϵ)=0.$$

A sequence of random variables $X_n$ converging in probability to $0$ is equivalent to it being asymptotically negligible, with $k_n=1$ for each $n$.

For example, suppose that $X_{n,j}$ are $L^2$ random variables each with $E(X_{n,j})=0$ and that they satisfy

$$\underset{n\to \mathrm{\infty }}{lim}\underset{1\le j\le k_n}{\max }\mathrm{Var}(X_{n,j})=0.$$

For $ϵ>0$, by Chebyshev’s inequality,

$$P(|X_{n,j}|\ge ϵ)\le \frac{1}{{ϵ}^2}E(|X_{n,j}{|}^2)=\frac{1}{{ϵ}^2}\mathrm{Var}(X_{n,j}),$$

whence

$$\underset{n\to \mathrm{\infty }}{lim}\underset{1\le j\le k_n}{\max }P(|X_{n,j}|\ge ϵ)\le \underset{n\to \mathrm{\infty }}{lim\; sup}\underset{1\le j\le k_n}{\max }\frac{1}{{ϵ}^2}\mathrm{Var}(X_{n,j})=0,$$

and so the random variables $X_{n,j}$ are asymptotically negligible.

Another example: Suppose that random variables $X_{n,j}$ are identically distributed, with $\mu =X_{n,j}{}_{*}P$. For $ϵ>0$,

$$P\left(\right|\frac{X_{n,j}}{n}|\ge ϵ)=P(|X_{n,j}|\ge nϵ)=\mu (A_n),$$

where $A_n=\{x\in ℝ:|x|\ge nϵ\}$. As $A_n\downarrow \mathrm{\varnothing }$, ${lim}_{n\to \mathrm{\infty }}\mu (A_n)=0$. Hence the random variables $\frac{X_{n,j}}{n}$ are asymptotic negligible.

The following is a statement about the characteristic functions of an asymptotically negligible family of random variables.

Lemma 5.

Suppose that a family $X_{n,j}$, $n\ge 1$, $1\le j\le k_n$, of random variables is asymptotically negligible, and write ${\mu }_{n,j}=X_{n,j}{}_{*}P$ and ${\varphi }_{n,j}={\tilde{\mu }}_{n,j}$. For each $x\in ℝ$,

$$\underset{n\to \mathrm{\infty }}{lim}\underset{1\le j\le k_n}{\max }|{\varphi }_{n,j}(x)-1|=0.$$

Proof.

For any real $t$, $|e^{it}-1|\le |t|$. For $x\in ℝ$, $ϵ>0$, $n\ge 1$, and $1\le j\le k_n$,

$|{\varphi }_{n,j}(x)-1|$ $=\left|{\displaystyle {\int }_{ℝ}}(e^{ixy}-1)𝑑{\mu }_{n,j}(y)\right|$

$\le ϵ|x|+2P(|X_{n,j}|\ge ϵ).$

Hence

$$\underset{1\le j\le k_n}{\max }|{\varphi }_{n,j}(x)-1|\le ϵ|x|+2\underset{1\le j\le k_n}{\max }P(|X_{n,j}|\ge ϵ).$$

Using that the family $X_{n,j}$ is asymptotically negligible,

$$\underset{n\to \mathrm{\infty }}{lim\; sup}\underset{1\le j\le k_n}{\max }|{\varphi }_{n,j}(x)-1|\le 2ϵ|x|.$$

But this is true for all $ϵ>0$, so

$$\underset{n\to \mathrm{\infty }}{lim\; sup}\underset{1\le j\le k_n}{\max }|{\varphi }_{n,j}(x)-1|=0,$$

proving the claim. ∎

4 The Lindeberg central limit theorem

We now prove the Lindeberg central limit theorem.

Theorem 6 (Lindeberg central limit theorem).

If $X_n$ is a sequence of independent $L^2$ random variables that satisfy the Lindeberg condition, then

$$S_n{}_{*}P\to {\gamma }_1,$$

where

$$S_n=\frac{1}{s_n}\sum _{j=1}^n(X_j-{\eta }_j)=\frac{{\sum }_{j=1}^n(X_j-E(X_j))}{\sigma (X_1+\mathrm{\cdots }+X_n)}.$$

Proof.

The sequence $Y_n=X_n-E(X_n)$ are independent $L^2$ random variables that satisfy the Lindeberg condition and $\sigma (Y_n)=\sigma (X_n)$. Proving the claim for the sequence $Y_n$ will prove the claim for the sequence $X_n$, and thus it suffices to prove the claim when $E(X_n)=0$, i.e. ${\eta }_n=0$.

For $n\ge 1$ and $1\le j\le n$, let

$${\mu }_{n,j}={\left(\frac{X_j}{s_n}\right)}_{*}P\mathit{\hspace{1em}\hspace{1em}}\text{and}\mathit{\hspace{1em}\hspace{1em}}{\tau }_{n,j}=\frac{{\sigma }_j}{s_n}.$$

The first moment of ${\mu }_{n,j}$ is

$${\int }_{ℝ}xd\left({\left(\frac{X_j}{s_n}\right)}_{*}P\right)(x)={\int }_{\mathrm{\Omega }}\frac{X_j}{s_n}𝑑P=\frac{1}{s_n}E(X_j)=0,$$

and the second moment of ${\mu }_{n,j}$ is

$${\int }_{ℝ}{x}^{2}d\left({\left(\frac{X_j}{s_n}\right)}_{*}P\right)(x)={\int }_{\mathrm{\Omega }}{\left(\frac{X_j}{s_n}\right)}^2𝑑P=\frac{1}{s_n^2}E(X_j^2)=\frac{{\sigma }_j^2}{s_n^2}={\tau }_{n,j}^2,$$

for which

$$\sum _{j=1}^n{\tau }_{n,j}^2=\frac{1}{s_n^2}\sum _{j=1}^n{\sigma }_j^2=1.$$

For $\mu \in 𝒫(ℝ)$ with first moment ${\int }_{ℝ}x𝑑\mu (x)=0$ and second moment , Lemma 2 tells us that

$$\tilde{\mu }(x)=M_0+i{M}_{1}x-\frac{M_2}{2}{x}^2+\frac{1}{2}{\theta }_2(x)x^2=1-\frac{{\sigma }^2}{2}{x}^2+\frac{1}{2}\theta (x)x^2,$$

with

$$|\theta (x)|\le \underset{0\le u\le 1}{sup}|{\tilde{\mu }}^{\prime \prime }(ux)-{\tilde{\mu }}^{\prime \prime }(0)|.$$

But by Lemma 1,

$${\tilde{\mu }}^{\prime \prime }(ux)=-{\int }_{ℝ}{y}^2{e}^{iuxy}𝑑\mu (y),$$

so

$|\theta (x)|$ $\le \underset{0\le u\le 1}{sup}\left|{\displaystyle {\int }_{ℝ}}{y}^2(-e^{iuxy}+1)𝑑\mu (y)\right|$

$\le \underset{0\le u\le 1}{sup}{\displaystyle {\int }_{ℝ}}{y}^2|e^{iuxy}-1|𝑑\mu (y).$

For $0\le u\le 1$, $|e^{iuxy}-1|\le |uxy|\le |xy|$, so for $x\in ℝ$ and $ϵ>0$, with $\delta =\min \{ϵ,\frac{ϵ}{|x|}\}$, when and $0\le u\le 1$ we have . Thus

$|\theta (x)|$

$\le ϵ{\sigma }^2+2{\displaystyle {\int }_{|y|\ge \delta }}{y}^2𝑑\mu (y).$

Let $x\in ℝ$ and $ϵ>0$, and take $\delta =\min \{ϵ,\frac{ϵ}{|x|}\}$. On the one hand, for $n\ge 1$ and $1\le j\le n$, because the first moment of ${\mu }_{n,j}$ is $0$ and its second moment is ${\tau }_{n,j}^2$,

$${\tilde{\mu }}_{n,j}(x)=1-\frac{{\tau }_{n,j}^2}{2}{x}^2+\frac{1}{2}{\theta }_{n,j}(x)x^2,$$

with, from the above,

$$|{\theta }_{n,j}(x)|\le ϵ{\tau }_{n,j}^2+2{\int }_{|y|\ge \delta }{y}^2𝑑{\mu }_{n,j}(y).$$

On the other hand, the first moment of the Gaussian measure ${\gamma }_{0,{\tau }_{n,j}^2}$ is $0$ and its second moment is ${\tau }_{n,j}^2$. Its characteristic function is

$${\tilde{\gamma }}_{0,{\tau }_{n,j}^2}(x)=\exp \left(-\frac{{\tau }_{n,j}^2}{2}{x}^2\right)=1-\frac{{\tau }_{n,j}^2}{2}{x}^2+\frac{1}{2}{\psi }_{n,j}(x)x^2,$$

with, from the above,

$$|{\psi }_{n,j}(x)|\le ϵ{\tau }_{n,j}^2+2{\int }_{|y|\ge \delta }{y}^2𝑑{\gamma }_{0,{\tau }_{n,j}^2}(x).$$

In particular, for all $x\in ℝ$,

$${\tilde{\mu }}_{n,j}(x)-{\tilde{\gamma }}_{0,{\tau }_{n,j}^2}(x)=\frac{x^2}{2}\left({\theta }_{n,j}(x)-{\psi }_{n,j}(x)\right).$$

For $k\ge 1$ and for $a_l,b_l\in ℂ$, $1\le l\le k$,

$$\prod _{l=1}^k{a}_l-\prod _{l=1}^k{b}_l=\sum _{l=1}^k{b}_1\mathrm{\cdots }{b}_{l-1}(a_l-b_l)a_{l+1}\mathrm{\cdots }{a}_k.$$

If further $|a_l|\le 1$, $|b_l|\le 1$, then

$$\left|\prod _{l=1}^k{a}_l-\prod _{l=1}^k{b}_l\right|\le \sum _{l=1}^k|a_l-b_l|.$$ (1)

Because the $X_n$ are independent, the distribution of

$$S_n=\sum _{j=1}^n\frac{X_j}{s_n}$$

is the convolution of the distributions of the summands:

$${\mu }_{n,1}*\mathrm{\cdots }*{\mu }_{n,n},$$

whose characteristic function is

$${\varphi }_n=\prod _{j=1}^n{\tilde{\mu }}_{n,j},$$

since the characteristic function of a convolution of measures is the product of the characteristic functions of the measures. Using ${\sum }_{j=1}^n{\tau }_{n,j}^2=1$ and (1), for $x\in ℝ$ we have

$|{\varphi }_n(x)-e^{-\frac{x^2}{2}}|$ $=\left|{\displaystyle \prod _{j=1}^n}{\tilde{\mu }}_{n,j}(x)-{\displaystyle \prod _{j=1}^n}{e}^{-\frac{1}{2}{\tau }_{n,j}^2{x}^2}\right|$

$\le {\displaystyle \sum _{j=1}^n}\left|{\tilde{\mu }}_{n,j}(x)-e^{-\frac{1}{2}{\tau }_{n,j}^2{x}^2}\right|$

$={\displaystyle \sum _{j=1}^n}\left|{\tilde{\mu }}_{n,j}(x)-{\tilde{\gamma }}_{0,{\tau }_{n,j}^2}(x)\right|$

$={\displaystyle \frac{x^2}{2}}{\displaystyle \sum _{j=1}^n}\left|{\theta }_{n,j}(x)-{\psi }_{n,j}(x)\right|.$

Therefore, for $x\in ℝ$, $ϵ>0$, and $\delta =\min \{ϵ,\frac{ϵ}{|x|}\}$,

We calculate

$L_n(\delta )$ $={\displaystyle \frac{1}{s_n^2}}{\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|y|\ge \delta s_n}}{y}^{2}d(X_j{}_{*}P)(y)$

$={\displaystyle \frac{1}{s_n^2}}{\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|X_j|\ge \delta s_n}}{X}_j^2𝑑P$

$={\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{\left|\frac{X_j}{s_n}\right|\ge \delta }}{\left({\displaystyle \frac{X_j}{s_n}}\right)}^2𝑑P$

$={\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|y|\ge \delta }}{y}^{2}d\left({\left({\displaystyle \frac{X_j}{s_n}}\right)}_{*}P\right)(y)$

$={\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|y|\ge \delta }}{y}^2𝑑{\mu }_{n,j}(y).$

Hence, the fact that the $X_n$ satisfy the Lindeberg condition yields

$$\underset{n\to \mathrm{\infty }}{lim\; sup}|{\varphi }_n(x)-e^{-\frac{x^2}{2}}|\le ϵx^2+x^2\underset{n\to \mathrm{\infty }}{lim\; sup}\sum _{j=1}^n{\int }_{|y|\ge \delta }{y}^2𝑑{\gamma }_{0,{\tau }_{n,j}^2}(y).$$ (2)

Write

$${\alpha }_n=\underset{1\le j\le n}{\max }{\tau }_{n,j}=\underset{1\le j\le n}{\max }\frac{{\sigma }_j}{s_n}.$$

We calculate

${\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|y|\ge \delta }}{y}^2𝑑{\gamma }_{0,{\tau }_{n,j}^2}(y)$ $={\displaystyle \sum _{j=1}^n}{\displaystyle {\int }_{|y|\ge \delta }}{y}^2{\displaystyle \frac{1}{{\tau }_{n,j}\sqrt{2\pi }}}\exp \left(-{\displaystyle \frac{y^2}{2{\tau }_{n,j}^2}}\right)𝑑y$

$={\displaystyle \sum _{j=1}^n}{\tau }_{n,j}^2{\displaystyle {\int }_{|u|\ge \delta /{\tau }_{n,j}}}{u}^2𝑑{\gamma }_{0,1}(u)$

$\le {\displaystyle \sum _{j=1}^n}{\tau }_{n,j}^2{\displaystyle {\int }_{|u|\ge \delta /{\alpha }_n}}{u}^2𝑑{\gamma }_{0,1}(u)$

$={\displaystyle {\int }_{|u|\ge \delta /{\alpha }_n}}{u}^2𝑑{\gamma }_{0,1}(u).$

Because the sequence $X_n$ satisfies the Lindeberg condition, by Lemma 4 it satisfies the Feller condition, which means that ${\alpha }_n\to 0$ as $n\to \mathrm{\infty }$. Because ${\alpha }_n\to 0$ as $n\to \mathrm{\infty }$, $\delta /{\alpha }_n\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$, hence

$${\int }_{|u|\ge \delta /{\alpha }_n}{u}^2𝑑{\gamma }_{0,1}(u)\to 0$$

as $n\to \mathrm{\infty }$. Thus we get

$$\sum _{j=1}^n{\int }_{|y|\ge \delta }{y}^2𝑑{\gamma }_{0,{\tau }_{n,j}^2}(y)\to 0$$

as $n\to \mathrm{\infty }$. Using this with (2) yields

$$\underset{n\to \mathrm{\infty }}{lim\; sup}|{\varphi }_n(x)-e^{-\frac{x^2}{2}}|\le ϵx^2.$$

This is true for all $ϵ>0$, so

$$\underset{n\to \mathrm{\infty }}{lim}|{\varphi }_n(x)-e^{-\frac{x^2}{2}}|=0,$$

namely, ${\varphi }_n$ (the characteristic function of $S_n{}_{*}P$) converges pointwise to $e^{-\frac{x^2}{2}}$. Moreover, $e^{-\frac{x^2}{2}}$ is indeed continuous at $0$, and $e^{-\frac{x^2}{2}}={\tilde{\gamma }}_{0,1}(x)$. Therefore, Lévy’s continuity theorem (Theorem 3) tells us that $S_n{}_{*}P$ converges narrowly to ${\gamma }_{0,1}$, which is the claim. ∎