Let $F:\mathbb{R}\to [0,1]$ be a right continuous non-decreasing function. Prove that $\sup\{x\in\mathbb{R}:F(x)

Question

Let $F:\mathbb{R}\to[0,1]$ be a right continuous non-decreasing function. Prove that $\sup\{x\in\mathbb{R}:F(x)<w\}\leq x\iff w\leq F(x)$.

The common trick involving supremum (show the supremum is always less than arbitrary dominant factor) doesn't work well for me and I am not sure how I should use right continuity in this case. This comes up in the proof that the random variable corresponding to any distribution function always exists. Could someone give any insight?

wait i closed as duplicate and then took it back but actually, i think my old question does answer your question? : Prove $X(\omega) = \sup{y \in \mathbb{R}: F(y) < \omega}$ is a random variable. in particular the '1' in the link i think doesn't assume the entirety of that the '$F$' there is a distribution function. based on answer of drhab, the assumptions used are and are only the right continuous and non-decreasing — BCLC, Jul 22 '21 at 09:01
@BCLC thanks for the link :) wow that was the post 7 years ago — jk001, Jul 22 '21 at 09:26

vmastr · Accepted Answer · 2021-07-22T08:10:36.107

3

First, note that if $$ x_0= \sup\{y\in\mathbb{R}: F(y)<w\}<+\infty, $$ then $F(x_0)\geq w$. To see why, assume $F(x_0)<w$, then by right continuity, for $\varepsilon= \frac{w- F(x_0)}{2}>0$ we may find $\delta>0$ such that for all $y\in [x_0, x_0+\delta)$

$$ F(y)\leq F(x_0)+ \varepsilon= \frac{w+ F(x_0)}{2}< w, $$ that is we may find $y>x_0$ with $F(y)<w$, which contradicts that $x_0$ is the supremum of that set.

As a result, since $F$ is non-decreasing we have that, $x_0\leq x$ if and only if $$ w\leq F(x_0)\leq F(x). $$

edited Jul 22 '21 at 08:10

answered Jul 22 '21 at 07:42

vmastr

880

Thank you. Shouldn't we still show that $F(x_0)>w$ doesn't hold? – jk001 Jul 22 '21 at 07:55
1

Good point. We don't actually need equality, let me edit my answer. – vmastr Jul 22 '21 at 08:08
@Tito Eliatron $t_n$ is approaching $x_0$ from the left, so I don't think we can use right-continuity. Or do I miss something? – jk001 Jul 22 '21 at 09:19
@jk001 right=derecha right=derecha. I would NEVER forget it. Sorry. You're right (or left)! – Tito Eliatron Jul 22 '21 at 09:29

score 2 · Answer 2 · answered Jul 22 '21 at 07:32

2

For simplicity, let's call $A_w:=\{t\in\Bbb R:\,F(t)<w\}$.

$\Leftarrow$ part is easy. If $t\in A_w$ then $F(t)<w$. but by hypothesis, $w\le F(x)$, so $F(t)<F(x)$.

By the way of contradiction, assuume that $x\le t$. Since $F$ is not decreasing, $F(x)\le F(t)$, but this yields the contradiction $F(x)\le F(t)<F(x)$. So $t\le x$.

That means that $x$ is an upperbound of $A_w$ and it follows that $\sup A_w\le x$.

answered Jul 22 '21 at 07:32

Tito Eliatron

6,678

1

Thanks so much! – jk001 Jul 22 '21 at 09:10

Let $F:\mathbb{R}\to [0,1]$ be a right continuous non-decreasing function. Prove that $\sup\{x\in\mathbb{R}:F(x)

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