In this section we prove the equivalence of the statements in theorem 3A by showing that each implies its successor. For ease of writing, on occasion we write \(F_{n}(\cdot)\) for \(F_{Z_{n}}(\cdot), \phi_{n}(\cdot)\) for \(\phi_{Z_{n}}(\cdot), F(\cdot)\) for \(F_{Z}(\cdot)\) , and \(\phi(\cdot)\) for \(\phi_{Z}(\cdot)\) .

It is immediate that (i) implies (ii), since the function \(g(z)=e^{i u z}\) is a bounded continuous function of \(z\) .

To prove that (ii) implies (iii), we make use of the basic formula (3.6) of Chapter 9 . For any \(d>0\) define the function \(g_{d}(\cdot)\) for any real number \(z\) by \begin{align} g_{d}(z) & = \begin{cases} 1, & \text{if } a \leq z \leq b \\ 1 - \left(\frac{a - z}{d}\right), & \text{if } a - d \leq z \leq a \\ 1 - \left(\frac{z - b}{d}\right), & \text{if } b \leq z \leq b + d \\ 0, & \text{otherwise.} \end{cases} \end{align} 

The function \(g_{d}(\cdot)\) is continuous and integrable. Its Fourier transform \(\gamma_{d}(\cdot)\) is given for any \(u\) by \[\gamma_{d}(u)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-i u z} g_{d}(z) d z=\frac{1}{2 \pi i u} \int_{-\infty}^{\infty} e^{-i u z} g_{d}^{\prime}(z) d z, \tag{5.1}\] Therefore, \begin{align} \gamma_{d}(u) & =\frac{1}{2 \pi i u d}\left(\int_{a-d}^{a} e^{-i u z}-\int_{b}^{b+d} e^{-i u z}\right) \tag{5.2} \\[3mm] & =\frac{1}{2 \pi u^{2} d}\left(e^{-i u a}-e^{-i u(a-d)}-e^{-i u(b+d)}+e^{-i u b}\right). \end{align} 

Thus we see that the Fourier transform \(\gamma_{d}(\cdot)\) is integrable. Consequently, from (3.6), of Chapter 9 we have. \[\int_{-\infty}^{\infty} g_{d}(z) d F_{Z_{n}}(z)-\int_{-\infty}^{\infty} g_{d}(z) d F_{Z}(z)=\int_{-\infty}^{\infty} d u \gamma_{d}(u)\left[\phi_{Z_{n}}(u)-\phi_{Z}(u)\right]. \tag{5.3}\] 

By letting \(n\) tend to \(\infty\) in (5.3) and using the hypothesis of statement (ii), we obtain for any \(d>0\) , as \(n\) tends to \(\infty\) , \[\int_{-\infty}^{\infty} g_{d}(z) d F_{Z_{n}}(z) \rightarrow \int_{-\infty}^{\infty} g_{d}(z) d F_{Z}(z). \tag{5.4}\] Next, define the function \(g_{d}^{*}(\cdot)\) for any \(z\) by \begin{align} g_{d}^{*}(z) & = \begin{cases} 1, & \text{if } a + d \leq z \leq b - d \\ \frac{z - a}{d}, & \text{if } a \leq z \leq a + d \\ \frac{b - z}{d}, & \text{if } b - d \leq z \leq b \\ 0, & \text{otherwise.} \end{cases} \end{align} 

By the foregoing argument, one may prove that (5.4) holds for \(g_{d}^{*}(\cdot)\) . Now, the expectations of the functions \(g_{d}(\cdot)\) and \(g_{d}^{*}(\cdot)\) clearly straddle the quantity \(F_{Z_{n}}(b)-F_{Z_{n}}(a)\) : \[\int_{-\infty}^{\infty} g_{a}^{*}(z) d F_{Z_{n}}(z) \leq F_{Z_{n}}(b)-F_{Z_{n}}(a) \leq \int_{-\infty}^{\infty} g_{d}(z) d F_{Z_{n}}(z) . \tag{5.5}\] From (5.5), letting \(n\) tend to \(\infty\) , we obtain \begin{align} \int_{-\infty}^{\infty} g_{d}^{*}(z) d F_{Z}(z) & \leq \liminf_{n} F_{Z_{n}}(b)-F_{Z_{n}}(a) \tag{5.6} \\ & \leq \limsup_{n} F_{Z_{n}}(b)-F_{Z_{n}}(a) \\ & \leq \int_{-\infty}^{\infty} g_{d}(z) d F_{Z}(z). \end{align} 

Now, let \(d\) tend to 0 in (5.6); since \begin{align} 0 \leq F_{Z}(b)-F_{Z}(a)&-\int_{-\infty}^{\infty} g_{d}^{*}(z) d F_{Z}(z) \leq F_{Z}(a+d) \\[3mm] &\qquad-F_{Z}(a)+F_{Z}(b)-F_{Z}(b-d) \rightarrow 0, \tag{5.7} \\[3mm] 0 \leq \int_{-\infty}^{\infty} g_{d}(z) d F_{Z}(z)&-\left[F_{Z}(b)-F_{Z}(a)\right] \leq F_{Z}(a) \\[3mm] &\quad -F_{Z}(a-d)+F_{Z}(b+d)-F_{Z}(b) \rightarrow 0. \end{align} as \(d\) tends to 0, it follows that (3.4) holds. Note that (5.7) would not hold if we did not require \(a\) and \(b\) to be points at which \(F_{Z}(\cdot)\) is continuous.

We next prove that (iii) implies (iv). Let \(M\) be a positive number such that \(F(\cdot)\) is continuous at \(M\) and at \(-M\) . Then, for any real number \(a\) \begin{align} \left|F_{n}(a)-F(a)\right| \leq \mid F_{n}(a)-F_{n}(-M)-F(a) & +F(-M) \mid \tag{5.8} \\ & +F_{n}(-M)+F(-M). \end{align} Since statement (iii) holds, it follows that if \(a\) is a continuity point of \(F(\cdot)\) \[\limsup_{n} |F_n(a) - F(a)| \leq F(-M) + \limsup_{n} F_n(-M). \tag{5.9}\] Now, also by (iii), since \(F_{n}(M)-F_{n}(-M)\) tends to \(F(M)-F(-M)\) , \begin{align} \limsup_{n} F_{n}(-M) & \leq \limsup_{n} \left(1 - F_{n}(M) + F_{n}(-M)\right) \\[2mm] & \leq 1 - F(M) + F(-M). \tag{5.10} \end{align} Consequently, \[\limsup_{n} \left|F_{n}(a)-F(a)\right| \leq 2 F(-M)+1-F(M), \tag{5.11}\] which tends to \(0\) , as one lets \(M\) tend to \(\infty\) . The proof that (iii) implies (iv) is complete.

We next prove that (iv) implies (i). We first note that a function \(g(\cdot)\) , continuous on a closed interval, is uniformly continuous there; that is, for every positive number \(\epsilon\) there is a positive number, derioted by \(d(\epsilon)\) , such that \(\left|g\left(z_{1}\right)-g\left(z_{2}\right)\right| \leq \epsilon\) for any two points \(z_{1}\) and \(z_{2}\) in the interval satisfying. \(\left|z_{1}-z_{2}\right| \leq d(\epsilon)\) . Choose \(M\) so that \(F(\cdot)\) is continuous at \(M\) and \(-M\) . On the closed interval \([-M, M], g(\cdot)\) is continuous. Fix \(\epsilon>0\) , and let \(d(\epsilon)\) be defined as in the foregoing sentence. We may then choose \((K+1)\) real numbers \(a_{0}, a_{1}, \ldots, a_{K}\) having these properties: (i) \(-M=a_{0}, (ii) \(a_{k}-a_{k-1} \leq d(\epsilon)\) for \(k=1,2, \ldots, K\) , (iii) for \(k=1,2, \ldots, K, F(\cdot)\) is continuous at \(a_{k}\) . Then define a function \(g(\cdot ; \epsilon, M)\) : \begin{align} g(x ; \epsilon, M) & = \begin{cases} 0, & \text{if } |x| > M \\ g(a_{k}), & \text{if } a_{k-1} < x \leq a_{k}, \quad \text{for some } k = 1, 2, \ldots, K, \\ g(-M), & \text{if } x = -M. \end{cases} \end{align} It is clear that for \(|x| \leq M\) \[\quad|g(x)-g(x ; \epsilon, M)| \leq \epsilon. \tag{5.13}\] Now \[\left|\int_{-\infty}^{\infty} g(x) d F_{n}(x)-\int_{-\infty}^{\infty} g(x) d F(x)\right| \leq\left|I_{n}\right|+\left|J_{n}\right|+|I|, \tag{5.14}\] where \begin{align} I_{n} & =\int_{-\infty}^{\infty}[g(x)-g(x ; \epsilon, M)] d F_{n}(x) \\ I & =\int_{-\infty}^{\infty}[g(x)-g(x ; \epsilon, M)] d F(x) \\ J_{n} & =\int_{-\infty}^{\infty} g(x ; \epsilon, M) d F_{n}(x)-\int_{-\infty}^{\infty} g(x ; \epsilon, M) d F(x) \end{align} 

Let \(C\) be an upper bound for \(g(\cdot)\) ; that is, \(|g(x)| \leq C\) for all \(x\) . Then \[\left|J_{n}\right| \leq C \sum_{k=1}^{K}\left[\left|F_{n}\left(a_{k}\right)-F\left(a_{k}\right)\right|+\mid F_{n}\left(a_{k-1}\right)-F\left(a_{k-1}\right)|\right]. \tag{5.15}\] 

Next, we may write \(I_{n}\) as a sum of two integrals, one over the range \(|x| \leq M\) and the other over the range \(|x|>M\) . In view of (5.13), we then have \[\left|I_{n}\right| \leq \epsilon+C\left[1-F_{n}(M)+F_{n}(-M)\right].\] Similarly \[|I| \leq \epsilon+C[1-F(M)+F(-M)].\] In view of (5.14), (5.15), and the two preceding inequalities, it follows that \[\limsup_{n} \left|\int_{-\infty}^{\infty} g(x) \, dF_{n}(x) - \int_{-\infty}^{\infty} g(x) \, dF(x)\right| \tag{5.16}\] \[\leq 2 \epsilon+2 C[1-F(M)+F(-M)].\] 

Letting first \(\epsilon\) tend to 0 and then \(M\) tend to \(\infty\) , it follows that (5.13) will hold. The proof that (iv) implies (i) is complete.

The reader may easily verify that (v) is equivalent to the preceding statements.

Theoretical exercises