Proof of Kolmogorov's Maximal Inequality and the Convergence of Series of Independent Square-Integrable Random Variables

Each result cited below is universally quantified over the data in its own statement.

Throughout, expectations and integrals are taken with respect to $P$, and "random variable" and "event" are as in Probability Space, Event, and Random Variable. Elementary arithmetic and order in $\mathbb{R}$, finite sums, the limit laws, the order of limits, the Archimedean property and square roots are used through the background results of the real-number setting. We use repeatedly that for a nonnegative integrable random variable $f$ one has $f^{+}=f$ and $f^{-}=0$, so by Integrable Function and the Lebesgue Integral (together with the vanishing of the integral of a function that is zero off a null set, applied with the empty null set to $f^{-}=0$) its integral as an integrable function equals its integral as a nonnegative function; thus $\mathbb{E}[f]$ is unambiguous.

Step 0. Preliminaries and notation.

(a) Random variables. By claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions the zero function $T_{0}$ is a random variable, and by claim 2 of the same lemma each $T_{n}=\sum_{k=1}^{n}\xi_{k}$ ($n\in\mathbb{N}$) is a random variable. Each $\xi_{k}$ is square-integrable, so by Square-Integrable Random Variables and the Mean-Square Inner Product (closure of square-integrable variables under sums and scalar multiples, and integrability of products of square-integrable variables and of square-integrable variables themselves) every finite linear combination of the $\xi_{k}$, in particular every $T_{n}$ and every difference $T_{m}-T_{n}$, is square-integrable and integrable, and products of two such combinations are integrable. For $0\le n<m$ with $m\in\mathbb{N}$, the finite-sum rules give, pointwise,

$$T_{m}-T_{n}=\sum_{k=n+1}^{m}\xi_{k},$$

where for $n=0$ this is the definition of $T_{m}$. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral (applied finitely often to the integrable $\xi_{k}$) and $\mathbb{E}[\xi_{k}]=0$, we get $\mathbb{E}[T_{m}-T_{n}]=0$.

(b) Tails. Put $v_{k}=\mathbb{E}[\xi_{k}^{2}]$, a nonnegative real number since $\xi_{k}$ is square-integrable. Put $s_{0}=0$ and $s_{n}=\sum_{k=1}^{n}v_{k}$ for $n\in\mathbb{N}$, and let $S=\sum_{k=1}^{\infty}v_{k}$, which exists by hypothesis; by the definition of the sum of a series, $s_{n}\to S$. For $n\ge0$ put $\tau_{n}=S-s_{n}$. By induction on $n$, the series $\sum_{k=1}^{\infty}v_{n+k}$ converges and its sum is $\tau_{n}$: for $n=0$ this is the hypothesis, and if it holds for $n$, then the index-shift lemma applied to $a_{k}=v_{n+k}$, $b_{k}=v_{n+1+k}$ shows that $\sum_{k=1}^{\infty}v_{n+1+k}$ converges with sum $\tau_{n}-v_{n+1}=S-s_{n+1}=\tau_{n+1}$. Thus $\tau_{n}$ is the number written $\sum_{k=n+1}^{\infty}\mathbb{E}[\xi_{k}^{2}]$ in the statement. By domination by the sum, $s_{n}\le S$, so $\tau_{n}\ge0$; by the limit laws, $\tau_{n}\to0$ as $n\to\infty$; and for $0\le n<m$,

$$\tau_{n}-\tau_{m}=s_{m}-s_{n}=\sum_{k=n+1}^{m}v_{k},\qquad\text{so}\qquad\sum_{k=n+1}^{m}v_{k}\le\tau_{n}.$$

(c) Second moments of increments. Let $0\le n<m$ with $m\in\mathbb{N}$ and $D=T_{m}-T_{n}$. Put $r=m-n$ and $X_{i}=\xi_{n+i}$ for $1\le i\le r$. These form a finite subfamily of the independent sequence $(\xi_{k})$, hence are independent by Independence of Events and of Random Variables, and each has finite expectation and finite second moment. By the variance additivity in Expectation of a Product of Independent Random Variables,

$$\operatorname{Var}(X_{1}+\cdots+X_{r})=\sum_{i=1}^{r}\operatorname{Var}(X_{i}).$$

By Expectation, Variance, and Moments, $\operatorname{Var}(X_{i})=\mathbb{E}[(\xi_{n+i}-0)^{2}]=v_{n+i}$, and since $X_{1}+\cdots+X_{r}=D$ has $\mathbb{E}[D]=0$ by (a), $\operatorname{Var}(D)=\mathbb{E}[D^{2}]$. Hence

$$\mathbb{E}\bigl[(T_{m}-T_{n})^{2}\bigr]=\sum_{k=n+1}^{m}v_{k}=\tau_{n}-\tau_{m}\le\tau_{n}.\tag{0}$$

Step 1. Claim 1 (maximal inequality). Fix $n\ge0$, $m\in\mathbb{N}$ with $n<m$, and $\varepsilon>0$. Write $D=T_{m}-T_{n}$.

Events. For each integer $i$ with $n<i\le m$, $|T_{i}-T_{n}|$ is a random variable by claims 2 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and the interval $[\varepsilon,\infty)$ is a Borel set by Borel Sigma-Algebra on the Real Line; so $E_{i}=\{|T_{i}-T_{n}|\ge\varepsilon\}$ is an event, and so is its complement $\Omega\setminus E_{i}=\{|T_{i}-T_{n}|<\varepsilon\}$ by property 2 of Sigma-Algebra and Measurable Space. Hence $A=\bigcup_{n<i\le m}E_{i}$, a finite union of events, is an event, since by Sigma-Algebra and Measurable Space a $\sigma$-algebra is closed under finite unions. For $n<j\le m$ define

$$A_{j}=E_{j}\cap\bigcap_{n<i<j}(\Omega\setminus E_{i}),$$

where the empty intersection (for $j=n+1$) is $\Omega$; each $A_{j}$ is an event, being a finite intersection of events, under which $\mathcal{F}$ is closed by Sigma-Algebra and Measurable Space. The $A_{j}$ are pairwise disjoint: if $j<j'$ then $A_{j}\subseteq E_{j}$ while $A_{j'}\subseteq\Omega\setminus E_{j}$. Their union is $A$: each $A_{j}\subseteq E_{j}\subseteq A$, and if $\omega\in A$, let $j$ be the least integer in $\{n+1,\dots,m\}$ with $\omega\in E_{j}$; then $\omega\in A_{j}$. By finite additivity (after relabelling the indices $n+1,\dots,m$ as $1,\dots,m-n$),

$$P(A)=\sum_{j=n+1}^{m}P(A_{j}).\tag{1}$$

The key estimate. Fix $j$ with $n<j\le m$, and put $U_{j}=(T_{j}-T_{n})\mathbf{1}_{A_{j}}$ and $W_{j}=T_{m}-T_{j}$ (the zero function when $j=m$). Since $T_{j}-T_{n}$ and $D$ are random variables by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions (as established above), claims 1 and 3 of the same lemma show that $U_{j}$ and $D^{2}\mathbf{1}_{A_{j}}$ are random variables. Pointwise $U_{j}^{2}\le(T_{j}-T_{n})^{2}$, so by the monotonicity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral $U_{j}$ is square-integrable; $W_{j}$ is square-integrable by Step 0(a). Hence by Square-Integrable Random Variables and the Mean-Square Inner Product, $U_{j}$ and $W_{j}$ are integrable and $U_{j}W_{j}$ is integrable. Also $0\le D^{2}\mathbf{1}_{A_{j}}\le D^{2}$ pointwise, so $\mathbb{E}[D^{2}\mathbf{1}_{A_{j}}]\le\mathbb{E}[D^{2}]<\infty$ by the same monotonicity, and $D^{2}\mathbf{1}_{A_{j}}$ is integrable by Integrable Function and the Lebesgue Integral.

We claim $\mathbb{E}[U_{j}W_{j}]=0$. If $j=m$ then $U_{j}W_{j}$ is the zero function and its integral is $0$ by the null-integral clause. Let $j<m$, put $r=m-n$, $p=j-n$, $q=m-j$, so $p,q\ge1$ and $p+q=r$, and let $V_{i}=\xi_{n+i}$ for $1\le i\le r$; these are independent random variables, as in Step 0(c). On $\mathbb{R}^{p}$ with the product Borel $\sigma$-algebra $\mathcal{B}_{p}$ of Joint Distribution, Expectations, and Block Independence for Independent Random Variables, let $\pi_{1},\dots,\pi_{p}$ be the coordinate projections; they are jointly Borel by claim 4 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables (applied with $p$ in place of $r$ and the independent variables $\xi_{n+1},\dots,\xi_{j}$). For $1\le i\le p$ put $\sigma_{i}=\pi_{1}+\cdots+\pi_{i}$; by claims 2 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, applied on the measurable space $(\mathbb{R}^{p},\mathcal{B}_{p})$, each $\sigma_{i}$ and each $|\sigma_{i}|$ is measurable, so (as above, using that $[\varepsilon,\infty)$ is Borel) the set

$$H=\{|\sigma_{p}|\ge\varepsilon\}\cap\bigcap_{1\le i<p}\{|\sigma_{i}|<\varepsilon\}$$

belongs to $\mathcal{B}_{p}$ (the empty intersection being $\mathbb{R}^{p}$), and by claims 1 and 3 of the same lemma $\varphi=\sigma_{p}\mathbf{1}_{H}:\mathbb{R}^{p}\to\mathbb{R}$ is jointly Borel. For $\omega\in\Omega$ and $1\le i\le p$ we have $\sigma_{i}(V_{1}(\omega),\dots,V_{p}(\omega))=\sum_{k=n+1}^{n+i}\xi_{k}(\omega)=T_{n+i}(\omega)-T_{n}(\omega)$, so $(V_{1}(\omega),\dots,V_{p}(\omega))\in H$ exactly when $\omega\in A_{j}$, and therefore $\varphi(V_{1},\dots,V_{p})=U_{j}$. Let $\psi:\mathbb{R}^{q}\to\mathbb{R}$ be the addition map, jointly Borel by claim 4 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables (applied with $q$ in place of $r$ and the independent variables $\xi_{j+1},\dots,\xi_{m}$); then $\psi(V_{p+1},\dots,V_{r})=\sum_{k=j+1}^{m}\xi_{k}=W_{j}$. Since $\{1,\dots,p\}$ and $\{p+1,\dots,r\}$ are disjoint nonempty subsets of $\{1,\dots,r\}$, claim 3 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables shows that $U_{j}$ and $W_{j}$ are independent. Both have finite expectation, so Expectation of a Product of Independent Random Variables gives $\mathbb{E}[U_{j}W_{j}]=\mathbb{E}[U_{j}]\,\mathbb{E}[W_{j}]=0$, because $\mathbb{E}[W_{j}]=\mathbb{E}[T_{m}-T_{j}]=0$ by Step 0(a).

Now compare pointwise. On $A_{j}$ we have $D=(T_{j}-T_{n})+W_{j}$ and $U_{j}=T_{j}-T_{n}$, so $D^{2}=U_{j}^{2}+2U_{j}W_{j}+W_{j}^{2}$ there; off $A_{j}$ both $D^{2}\mathbf{1}_{A_{j}}$ and $U_{j}$ vanish. Hence, pointwise on $\Omega$,

$$D^{2}\mathbf{1}_{A_{j}}=U_{j}^{2}+2U_{j}W_{j}+W_{j}^{2}\mathbf{1}_{A_{j}}\ \ge\ U_{j}^{2}+2U_{j}W_{j}.$$

All functions on the right are integrable, so claim 2 of Linearity and Monotonicity of the Lebesgue Integral (linearity and monotonicity) gives

$$\mathbb{E}[D^{2}\mathbf{1}_{A_{j}}]\ge\mathbb{E}[U_{j}^{2}]+2\,\mathbb{E}[U_{j}W_{j}]=\mathbb{E}[U_{j}^{2}].$$

On $A_{j}$, $|U_{j}|=|T_{j}-T_{n}|\ge\varepsilon$, so $U_{j}^{2}\ge\varepsilon^{2}$; off $A_{j}$, $U_{j}^{2}=0$. Thus $U_{j}^{2}\ge\varepsilon^{2}\mathbf{1}_{A_{j}}$ pointwise, and by claim 1 of Linearity and Monotonicity of the Lebesgue Integral together with The Integral of an Indicator Function is the Measure of the Set,

$$\mathbb{E}[D^{2}\mathbf{1}_{A_{j}}]\ \ge\ \mathbb{E}[U_{j}^{2}]\ \ge\ \varepsilon^{2}P(A_{j}).\tag{2}$$

Summation. Since the $A_{j}$ are pairwise disjoint with union $A$, pointwise $\sum_{j=n+1}^{m}D^{2}\mathbf{1}_{A_{j}}=D^{2}\mathbf{1}_{A}\le D^{2}$. By the additivity and monotonicity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral (additivity applied finitely often), (1), (2) and (0),

$$\varepsilon^{2}P(A)=\sum_{j=n+1}^{m}\varepsilon^{2}P(A_{j})\le\sum_{j=n+1}^{m}\mathbb{E}[D^{2}\mathbf{1}_{A_{j}}]=\mathbb{E}[D^{2}\mathbf{1}_{A}]\le\mathbb{E}[D^{2}]=\sum_{k=n+1}^{m}\mathbb{E}[\xi_{k}^{2}].$$

Dividing by $\varepsilon^{2}>0$ proves claim 1.

Step 2. Claim 2 (almost sure convergence). For $n\ge0$ and $r\in\mathbb{N}$ let $B_{n,r}$ be the set of $\omega$ for which there is an integer $j>n$ with $|T_{j}(\omega)-T_{n}(\omega)|\ge1/r$. For $M\in\mathbb{N}$ let $A_{n,M,r}$ be the set $A$ of claim 1 for the data $n$, $m=n+M$, $\varepsilon=1/r$. Each $A_{n,M,r}$ is an event, $A_{n,M,r}\subseteq A_{n,M+1,r}$, and $B_{n,r}=\bigcup_{M\in\mathbb{N}}A_{n,M,r}$, so $B_{n,r}$ is an event, as a countable union of events (property 3 of Sigma-Algebra and Measurable Space). By claim 1 and Step 0(b), $P(A_{n,M,r})\le r^{2}\sum_{k=n+1}^{n+M}v_{k}\le r^{2}\tau_{n}$ for all $M$. These probabilities are real and bounded above, so by continuity from below $P(B_{n,r})$ is their least upper bound, whence

$$P(B_{n,r})\le r^{2}\tau_{n}.\tag{3}$$

For $r\in\mathbb{N}$ let $G_{r}=\bigcap_{n\in\mathbb{N}}B_{n,r}$, an event, since by Sigma-Algebra and Measurable Space a $\sigma$-algebra is closed under countable intersections. For every $n\in\mathbb{N}$, monotonicity and (3) give $0\le P(G_{r})\le r^{2}\tau_{n}$; since $r^{2}\tau_{n}\to0$ as $n\to\infty$ (Step 0(b) and the limit laws), the order of limits gives $P(G_{r})\le0$, so $P(G_{r})=0$. Let $G=\bigcup_{r\in\mathbb{N}}G_{r}$, an event by property 3 of Sigma-Algebra and Measurable Space; by countable subadditivity $P(G)\le\sum_{r}P(G_{r})=0$, so $P(G)=0$.

We show $C=\Omega\setminus G$. Let $\omega\in C$ with limit $L=\lim_{i}T_{i}(\omega)$, and let $r\in\mathbb{N}$. By Limit of a Sequence of Real Numbers there is $N\in\mathbb{N}$ with $|T_{i}(\omega)-L|<1/(2r)$ for all $i\ge N$; then for every $j>N$, $|T_{j}(\omega)-T_{N}(\omega)|\le|T_{j}(\omega)-L|+|L-T_{N}(\omega)|<1/r$, so $\omega\notin B_{N,r}$ and hence $\omega\notin G_{r}$. As $r$ was arbitrary, $\omega\notin G$. Conversely let $\omega\notin G$ and let $\varepsilon>0$. By the Archimedean property choose $r\in\mathbb{N}$ with $2/r<\varepsilon$. Since $\omega\notin G_{r}$, there is $n\in\mathbb{N}$ with $\omega\notin B_{n,r}$, that is, $|T_{j}(\omega)-T_{n}(\omega)|<1/r$ for all $j>n$, and trivially also for $j=n$. For all integers $i,j\ge n$ this gives $|T_{i}(\omega)-T_{j}(\omega)|\le|T_{i}(\omega)-T_{n}(\omega)|+|T_{n}(\omega)-T_{j}(\omega)|<2/r<\varepsilon$. So $(T_{i}(\omega))_{i\in\mathbb{N}}$ is a Cauchy sequence, which converges by Every Cauchy Sequence of Real Numbers Converges; thus $\omega\in C$.

Hence $C=\Omega\setminus G\in\mathcal{F}$ by property 2 of Sigma-Algebra and Measurable Space, and by the differences clause $P(C)=P(\Omega)-P(G)=1$. This proves claim 2. In particular $\Omega\setminus C=G$ is a null set.

Step 3. Claim 3 (mean-square convergence).

(a) $T$ is a random variable. For $i\in\mathbb{N}$ let $f_{i}=T_{i}\mathbf{1}_{C}$, a random variable by claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions since $C\in\mathcal{F}$. At every $\omega\in C$, $f_{i}(\omega)=T_{i}(\omega)\to T(\omega)$; at every $\omega\notin C$, $f_{i}(\omega)=0\to0=T(\omega)$. By claim 5 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, $T$ is measurable, i.e. a random variable.

(b) The upper bound. Fix $n\ge0$. For $m\in\mathbb{N}$ let $g_{m}=(T_{n+m}-T_{n})^{2}$ and $h=(T-T_{n})^{2}$; these are nonnegative random variables (claims 2 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions). By (0), $\mathbb{E}[g_{m}]\le\tau_{n}$ for every $m$. Let $\omega\in C$. Given $\delta>0$ there is $N$ with $|T_{i}(\omega)-T(\omega)|<\delta$ for $i\ge N$, hence for $i=n+m$ whenever $m\ge N$; so $T_{n+m}(\omega)\to T(\omega)$ as $m\to\infty$, and by the limit laws $g_{m}(\omega)\to h(\omega)$. We check that then $\bigl(\liminf_{m}g_{m}\bigr)(\omega)=h(\omega)$, with $\liminf$ as defined in Fatou's Lemma. Write $c_{m}=g_{m}(\omega)\ge0$ and $L=h(\omega)$, and let $\delta>0$; choose $K\in\mathbb{N}$ with $|c_{m}-L|<\delta$ for $m\ge K$. For every $k\in\mathbb{N}$, with $k'=\max(k,K)$, we have $\inf_{m\ge k}c_{m}\le c_{k'}<L+\delta$; and $\inf_{m\ge K}c_{m}\ge L-\delta$ since $L-\delta$ is a lower bound of $\{c_{m}:m\ge K\}$. Hence $\sup_{k}\inf_{m\ge k}c_{m}$ lies in $[L-\delta,L+\delta]$ for every $\delta>0$, so it equals $L$. Thus $\liminf_{m}g_{m}$ and $h$ agree on $C$, i.e. off the null set $\Omega\setminus C$. Both are measurable (the former by Fatou's Lemma), so by almost-everywhere comparison and Fatou's Lemma,

$$\mathbb{E}[h]=\int_{\Omega}\Bigl(\liminf_{m}g_{m}\Bigr)\,dP\le\liminf_{m}\mathbb{E}[g_{m}]\le\tau_{n},$$

the last inequality because $\inf_{m\ge k}\mathbb{E}[g_{m}]\le\mathbb{E}[g_{k}]\le\tau_{n}$ for every $k$. Thus

$$\mathbb{E}\bigl[(T-T_{n})^{2}\bigr]\le\tau_{n}<\infty\qquad(n\ge0).\tag{4}$$

Taking $n=0$ (so $T-T_{0}=T$) shows that $T$ is square-integrable. Consequently, by Square-Integrable Random Variables and the Mean-Square Inner Product, each $T-T_{n}$ is square-integrable, and products of any two of the variables $T-T_{m}$, $T_{m}-T_{n}$ are integrable.

(c) Equality. Fix $n\ge0$ and let $m\in\mathbb{N}$ with $m>n$. Pointwise $T-T_{n}=(T-T_{m})+(T_{m}-T_{n})$, so

$$(T-T_{n})^{2}=(T-T_{m})^{2}+(T_{m}-T_{n})^{2}+2(T-T_{m})(T_{m}-T_{n}),$$

with all terms integrable. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral and (0),

$$\mathbb{E}\bigl[(T-T_{n})^{2}\bigr]-\tau_{n}=\Bigl(\mathbb{E}\bigl[(T-T_{m})^{2}\bigr]-\tau_{m}\Bigr)+2\,\mathbb{E}\bigl[(T-T_{m})(T_{m}-T_{n})\bigr].$$

By (4) for $m$, $0\le\mathbb{E}[(T-T_{m})^{2}]\le\tau_{m}$, so the first bracket has absolute value at most $\tau_{m}$. By claim 1 of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm, (4) and (0), and the monotonicity of the nonnegative square root,

$$\bigl|\mathbb{E}[(T-T_{m})(T_{m}-T_{n})]\bigr|\le\lVert T-T_{m}\rVert_{2}\,\lVert T_{m}-T_{n}\rVert_{2}\le\sqrt{\tau_{m}}\,\sqrt{\tau_{n}}.$$

Hence $\bigl|\mathbb{E}[(T-T_{n})^{2}]-\tau_{n}\bigr|\le\tau_{m}+2\sqrt{\tau_{m}}\sqrt{\tau_{n}}$ for every $m>n$. As $m\to\infty$, $\tau_{m}\to0$ by Step 0(b), and $\sqrt{\tau_{m}}\to0$ (given $\delta>0$, $\tau_{m}<\delta^{2}$ for large $m$, and then $\sqrt{\tau_{m}}<\delta$ by monotonicity of squares on nonnegative numbers); so the right-hand side tends to $0$ by the limit laws, and by the order of limits the nonnegative constant on the left is $\le0$, hence $0$. Therefore

$$\mathbb{E}\bigl[(T-T_{n})^{2}\bigr]=\tau_{n}=\sum_{k=n+1}^{\infty}\mathbb{E}[\xi_{k}^{2}]\qquad(n\ge0).$$

(d) Consequences. Taking $n=0$ gives $\mathbb{E}[T^{2}]=\tau_{0}=\sum_{k=1}^{\infty}\mathbb{E}[\xi_{k}^{2}]$, and $\mathbb{E}[(T-T_{n})^{2}]=\tau_{n}\to0$ by Step 0(b). For $\mathbb{E}[T]$: the constant function $\mathbf{1}=\mathbf{1}_{\Omega}$ is a random variable with $\mathbb{E}[\mathbf{1}^{2}]=P(\Omega)=1$ by The Integral of an Indicator Function is the Measure of the Set, so it is square-integrable with $\lVert\mathbf{1}\rVert_{2}=1$. For $n\in\mathbb{N}$, $T$ and $T_{n}$ are integrable and $\mathbb{E}[T_{n}]=\mathbb{E}[T_{n}-T_{0}]=0$ by Step 0(a), so by claim 2 of Linearity and Monotonicity of the Lebesgue Integral and claim 1 of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm,

$$\bigl|\mathbb{E}[T]\bigr|=\bigl|\mathbb{E}[T-T_{n}]\bigr|=\bigl|\mathbb{E}[(T-T_{n})\mathbf{1}]\bigr|\le\lVert T-T_{n}\rVert_{2}\,\lVert\mathbf{1}\rVert_{2}=\sqrt{\tau_{n}}.$$

Since $\sqrt{\tau_{n}}\to0$ as in (c), the order of limits gives $|\mathbb{E}[T]|\le0$, so $\mathbb{E}[T]=0$. This proves claim 3.