Theorem 11

Let E be the set of all points on the x axis. Suppose that F_n (x) is a sequence of frequency distributions and that F(x) is a frequency distribution which is continuous for all $x \in E$. If

$$\lim_{n\to\infty} F_n (x) = F(x), \qquad \text{for } x \in E,$$

then F_n(x) converges uniformly to F(x) on E.

Proof: Since F(x) is a frequency distribution,

$$\begin{array} {r@{~}c@{~}l} \displaystyle \lim_{x \to-\infty... ... 0, \\ \displaystyle \lim_{x\to\infty} F(x) & = & 1.\end{array}$$

Then, given any $\epsilon \gt 0$, there exist X₁ and X₂ such that

$$\text{if } x < X_1, \quad \text{then } F(x) < \frac{\epsilon}3$$

$$\text{if } x \gt X_2, \quad \text{then } 1-F(x) < \frac{\epsilon}3$$

Furthermore, since

$$\begin{array} {r@{~}c@{~}l} \displaystyle \lim_{n\to\infty} ... ...splaystyle \lim_{n\to\infty} F_n (X_2) & = & F(X_2),\end{array}$$

there exist N₁ and N₂ such that

$$\text{if} \quad n \gt N_1, \quad \text{then} \quad \vert F_n (X_1) - F (X_1)\vert < \frac\epsilon3$$

$$\text{if} \quad n \gt N_2, \quad \text{then} \quad \vert F_n (X_2) - F (X_2) \vert < \frac\epsilon3$$

It follows from the monotonicity of F_n (x) and F(x) that

(a)

if n > N₁, x < X₁, then

$$\leq$$ |F_n (x) - F (x) | | F_n (X₁)| + | F(X₁)|

$$\leq$$ | F_n (x₁) - F (X₁) + F (X₁) | + F (X₁)

$$\leq \vert F_n (X_1) - F(X)\vert + 2F(X_1) < \frac\epsilon3 + \frac{2\epsilon}3 = \epsilon$$

(b)

if n > N₂, x > N₂, then

$$\left\vert [1-F(x)] - [1-F_n (x)]\right\vert$$ | F_n (x) - F (x)| =

$$\leq$$ | 1 - F (X₂)| + |1-F_n (X₂)|

$$\vert 1 - F (X_2)\vert + \left\vert [ 1 - F (X_2) ] - [F_n (X_2) - F(X_2) ] \right\vert$$ =

$$\leq 2 \vert 1 - F (X_2)\vert + \vert F_n (X_2) - F (X_2)\vert < \frac{2\epsilon}3 + \frac\epsilon3 = \epsilon$$

Thus, F_n (x) converges uniformly to F(x) for $X_2 < x \leq \infty$ and $-\infty \leq x < X_1$. But, by Theorem 10, F_n (x) also converges uniformly to F(x) for $x \in [X_1,X_2]$, and there exists N₃ such that if n > N₃ and $x \in [X_1,X_2]$, then

$$\vert F_n (x) - F(x) \vert < \epsilon.$$

Set $N = \max (N_1, N_2, N_3)$. Then, if n > N, it follows that

$$\vert F_n (x) - F(x)\vert < \epsilon$$

for every $x \in E$. Thus, F_n(x) converges uniformly to F(x) on E.