Left Continuous Inverse

Last updated Nov 6, 2022

Let $F: \mathbb{R} \to \mathbb{R}$ be a Distribution Function. The Left Continuous Inverse of $F$ is defined

$$F^{\leftarrow}(x) = \inf{s \in \mathbb{R} : F(s) \geq x}$$

1. For fixed $x \in \mathbb{R}$, write $A = {s \in \mathbb{R} : F(s) \geq x}$. Then $A$ is Closed. Thus, $F^{\leftarrow}(x) \in A$ when $F^{\leftarrow}(x) \in \mathbb{R}$ since There Exists a Sequence Converging to Extremum.

   Proof: Recall that A Set is Closed in a Metric Space iff it contains all its Sequential Limits. Let $(s_{n}) \subset A$ be a Sequencethat converges to $t \in \mathbb{R}$. By definition of $A$ $$F(s_{n}) \geq x$$ Recall that Every Sequence on the Reals has a Monotone Subsequence. Let this Subsequence be $(s_{n_{k}})$. Every Subsequence of a Convergent Sequence converges to the same Limit, so $s_{n_{k}} \to t$. Now consider the following two cases:

   1. $(s_{n_{k}})$ is Non-Decreasing Function: Then, because Monotone Sequences converge to their Extremum, we know $s_{n_{k}} \to \sup{s_{n_{k}}} = t$. Thus $t \geq s_{n_{1}} \in A$ (by definition of Supremum). Since $F$ is Non-Decreasing Function, we have that $$F(t) \geq F(s_{n_{1}}) \geq x$$ and $t \in A$.
   2. $(s_{n_{k}})$ is Non-Increasing Function: Then we know $s_{n_{k}} \to \inf{s_{n_{k}}} = t$. Since $F$ is Right-Continuous and $(s_{n_{k}}) \subset [t, \infty)$, we know that $F(s_{n_{k}}) \to F(t)$. Since $F(s_{n_{k}}) \geq x$ $\forall k \in \mathbb{N}$, then by the Order Limit Theorem$$F(t) \geq x$$ and thus $t \in A$.

   Since $(s_{n})$ was arbitrary, we have that $A$ is Closed. $\blacksquare$

2. $F^{\leftarrow}$ is Non-Decreasing Function.

   Proof: Let $A_{x}= {s \in \mathbb{R} : F(s) \geq x}$. Suppose $y \leq x$. Then if $s \in A_x$, we have that $$F(s) \geq x \geq y$$ and thus $s \in A_{y}$. Therefore $A_{x} \subset A_{y}$. Since Infimums are Non-Increasing Set Functions we see that $F^{\leftarrow}(x) = \inf A_{x} \geq \inf A_{y} = F^{\leftarrow}(y)$, proving that $F^{\leftarrow}$ is Non-Decreasing Function. $\blacksquare$

3. $F^{\leftarrow}$ is Left-Continuous.

   Proof: Let $x \in \mathbb{R}$. Let $A_{x}$ be defined as in (2). First we claim that $$\bigcap_{\epsilon>0} A_{x - \epsilon} = A_{x}$$ To see this, first observe that since $A_{x} \subset A_{x-\epsilon}$ $\forall \epsilon > 0$ (as shown in (2)), we see that $$\bigcap_{\epsilon>0} A_{x - \epsilon} \supset A_{x}$$ On the other hand, suppose $s \in A_{x - \epsilon}$ for all $\epsilon > 0$. Then $F(s) \geq x - \epsilon$ for all $\epsilon > 0$. By the Epsilon Principle, we have that $F(s) \geq x$ and $s \in A_{x}$. Thus $$\bigcap_{\epsilon>0} A_{x - \epsilon} \subset A_{x}$$ Putting those two directions together we get that $$\bigcap_{\epsilon>0} A_{x - \epsilon} = A_{x}$$ so their Infimums are also the same. Now we claim that $$\inf\bigcap_{\epsilon>0} A_{x - \epsilon} = \lim\limits_{\epsilon \downarrow 0}\inf A_{x - \epsilon}$$ To see this, first recall from (2) that $\inf A_{x-\epsilon} \leq \inf A_{x}$ for all $\epsilon > 0$ . Thus, by the Order Limit Theorem, $$\lim\limits_{\epsilon \downarrow 0} \inf A_{x-\epsilon} \leq \inf A_{x} = \inf\bigcap_{\epsilon>0} A_{x - \epsilon}$$ Now on the other hand, if $$\lim\limits_{\epsilon \downarrow 0} \inf A_{x-\epsilon} < \inf\bigcap_{\epsilon>0} A_{x - \epsilon}$$ Then there exists $y \in \mathbb{R}$ s.t. $$\lim\limits_{\epsilon \downarrow 0} \inf A_{x-\epsilon} < y < \inf\bigcap_{\epsilon>0} A_{x - \epsilon}$$ Since Infimums are Non-Increasing Set Functions, we have that $y > \inf A_{x-\epsilon}$ for all $\epsilon > 0$ (otherwise the limit would exceed $y$ and never return to fall below $y$). However, since $y < \inf\bigcap_{\epsilon>0} A_{x - \epsilon}$, there exists $z \in \mathbb{R}$ s.t. $y < z < \inf\bigcap_{\epsilon>0} A_{x - \epsilon}$ and by definition of Infimum, $z < \bigcap_{\epsilon>0} A_{x - \epsilon}$. Thus there must be some $\epsilon > 0$ s.t. $z < A_{x - \epsilon}$ (otherwise, there is always an element at or below $z$; recall $A_{x-\epsilon}$ is a Non-Increasing Function Sequence of Sets). But then $y < z \leq \inf A_{x - \epsilon}$, contradicting above. Therefore $$F^{\leftarrow}(x) = \inf A_{x} = \inf\bigcap_{\epsilon>0} A_{x - \epsilon} = \lim\limits_{\epsilon \downarrow 0}\inf A_{x - \epsilon} = \lim\limits_{y \uparrow x} F^{\leftarrow}(y)$$ and $F^{\leftarrow}$ is Left-Continuous. $\blacksquare$

• Real Numbers
• Set Intersection

1. Resnick - A Probability Path - Ch unknown, pg unknown (I think ch 2)