A Distribution Function is Determined on a Dense Set

Last updated Nov 1, 2022

Let $D \subset \mathbb{R}$ be Dense and let $F_{D}: D \to \mathbb{R}$. Then if we define

$$F(x) = \inf_{y \in D; y > x} F_{D}(y)$$ $F$ is a Distribution Function.

Furthermore, if $G$ is a Distribution Function so that $G {\big|}_{D} = F_D$, then $G = F$.

Observe that since $D$ is Dense and $\mathbb{R}$ is Archimedean, ${F_{D}(y): y \in D, y > x}$ is nonempty $\forall x \in \mathbb{R}$. So we must simply show that $F$ satisfies the definition of a Distribution Function. That is, we must show that $F$ is Non-Decreasing Function and Right-Continuous.

To see $F$ is non-decreasing, let $x,z \in \mathbb{R}$ s.t. $x \leq z$. Then we see

$$\begin{align*} F(x) &= \inf {F_{D}(y) : y \in D, y > x }\\ &\leq \inf{F_{D}(y) : y \in D, y > z \geq x}\\ &= F(z) \end{align*}$$

To see $F$ is right-continuous, let $x \in \mathbb{R}$, $\epsilon > 0$. We will find $\delta > 0$ s.t. $F(z) - F(x) < \epsilon$ $\forall z \in (x, x+\delta)$. Since Finite Extremums get arbitrarily close, we have that there exists $x < y \in D$ s.t. $F_{D}(y)- F(x) < \frac{\epsilon}{2}$. Set $\delta = y - x$. Then for any $z \in (x, x + \delta)$ we have that $z < y$ so $F_{D}(y) \geq F(z)$. But also $z > x$ so $F(z) \geq F(x)$ since $F$ is non-decreasing. Thus

$$F(z) - F(x) \leq F_{D}(y) - F(x) < \frac{\epsilon}{2} < \epsilon$$ showing that $F$ is right-continuous and that it is a distribution function.

Now let $G$ be a Distribution Function so that $G {\big|}{D} = F_D$. This means $F{D}$ is Non-Decreasing Function and $G$ is Right-Continuous. Let $x \in \mathbb{R}$. Since $D$ is Dense, we can construct a decreasing Sequence ${y_n}{n=1}^{\infty} \subset D$ s.t. $y{n} \in (x + \frac{1}{n+1}, x + \frac{1}{n})$ $\forall n \in \mathbb{N}$. Then $${y \in D : y > x} \supset {y_n}_{n=1}^{\infty}$$ so

$$\begin{align*} F(x) &= \inf {F_{D}(y) : y \in D, y > x } \\ &\leq \inf_{y_{n}} F_{D}(y_{n})\\ &= \lim\limits_{y_{n} \to x} G(y_{n})\\ &= G(x) \end{align*}$$ On the other hand, for every $y \in D$ s.t. $y > x$, $\exists n \in \mathbb{N}$ s.t. $$y_{n} < x + \frac{1}{n} < y$$ Thus, by monotonicity of $F_D$, $$F_{D}(y) \geq \lim\limits_{y_{n} \to x} F_{D}(y_{n})= G(x)$$ so $$F(x) = \inf {F_{D}(y) : y \in D, y > x } \geq G(x)$$

giving us $G(x) = F(x)$. $\blacksquare$

1. The first half of the proof does not actually require $D$ to be Dense or for $F_{D}$ to be Non-Decreasing Function. $D$ need only be Archimedean. This proof gives us a technique for constructing Distribution Functions from any function on an Archimedean subset of $\mathbb{R}$.
2. If $F$ and $G$ are any two Distribution Functions that agree on a Dense subset of $\mathbb{R}$, we must have that $F = G$, namely because they are both equal to the distribution function generated by their restrictions to the dense set.

1. 2022-02-24 - Resnick - A Probability Path - pg 248