Theorem 12

Let n be any positive integer greater than one. Let x₀ be the greatest lower bound of the values of x satisfying the inequality

 

$$1 - F (x+0) \leq \frac1n \leq 1 - F (x-0).$$ (44)

Then, x₀ also satisfies the inequality.

Proof: Since F(x) is monotonic increasing from $F(-\infty) = 0$ to $F(+\infty) = 1$, it follows that there is at least one value of x satisfying Eq. (44). Assume that x₀ is the greatest lower bound of such x and that x₀ does not satisfy Eq. (44). It will be shown that this leads to a contradiction. If x₀ is the greatest lower bound and does not satisfy Eq. (44), it follows that

 

$$\frac1n < 1 - F (x_0 + 0).$$ (45)

By Definition 8, every frequency distribution is right-continuous. Hence, F (x₀ + 0) = F (x₀), and Eq. (45) becomes

$$\frac1n < 1-F (x_0) \quad \text{or} \quad F(x_0) < 1 - \frac1n.$$

Choose B such that

$$F(x_0) < B < 1 - \frac1n.$$

Since F(x) is right-continuous at x = x₀, there exists a $\delta \gt 0$ such that for $\vert x_0 - x_1 \vert < \delta$, x₁ > x₀,

$$\vert F (x_1) - F (x_0)\vert < B-F (x_0) \quad \text{or} \quad F(x_1) < B < 1 - \frac1n.$$

It follows that

$$\frac1n < 1 - F (x_1).$$

But, x₁ is any value such that x₁ > x₀ and $\vert x_1 - x_0\vert < \delta$. It follows that x₀ is not the greatest lower bound of the values of x satisfying Eq. (44). This contradiction establishes the theorem, since certainly

$$F (x_0-0) < F(x-0) \quad \text{and} \quad \frac1n \leq 1 - F (x-0) < 1-F (x_0-0).$$

Combining results gives

$$1 -F (x_0+0) \leq \frac1n \leq 1 - F (x_0-0).$$

The theorem would not be true if the words ``greatest lower bound'' are replaced by ``least upper bound'', since this would require left continuity, a property which a frequency distribution does not necessarily possess at every x value.