Rigorous proof of CDF $F_X(X)$ is uniform

Question

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be given. It is well known that, for continuous random variable $X(\omega)$, with CDF $F(x)$, then $F(X(\omega))$ is a uniform random variable. I want to look up a proof of this, and found this post. I believe the highest voted answer by JimmyK4542 to be flawed, in that, the answerer uses $F^{-1}$. Let $$ A =\{ \omega: F(X) \leq y \} \\ B =\{ \omega: X \leq F^{-1} (y) \} $$ Then it would seem that $$ \mathbb{P} (A) =\mathbb{P} (B) $$ but if $F'(\xi)=0$ for some $\xi$, $F^{-1}$ must not exist at $\xi$, by inverse function theorem, and $B$ is undefined.

So I tried the following proof, please check for me, and do tell me whether there are more concise ones. As someone suggests in the comment, I try $$ \tilde{A} =\{\omega: X \leq x_0\} \\ x_0 :=\inf\; \{\tilde{x}: F(\tilde{x}) \geq y\} $$ I say that $$ A =\tilde{A} $$ Indeed, fix any $x \in A$, then $F(x) \leq y$. For any $\tilde{x}$ so that $F(\tilde{x}) \geq y$, we have $F(\tilde{x}) \geq F(x)$, and, since $F$ is nondecreasing, $\tilde{x} \geq x$, and $x \in \tilde{A}$.

Conversely, fix any $x \in \tilde{A}$, so for any $F(\tilde{x}) \geq y$, we have $\tilde{x} \geq x$, and $F(\tilde{x}) \geq F(x)$ by monotonicity. Prove by contradiction, assuming $F(x) >y$. Let $\tilde{x}$ be chosen, such that $x-\tilde{x}>0$ be so small that $F(x) > F(\tilde{x}) >y$, which is possible by continuity. This simply contradicts the previous statement.

Thus $A =\tilde{A}$ is true, and $\mathbb{P} (A) =\mathbb{P} (\tilde{A})$.

Next, I claim that $F (x_0) =y$. Choose any $x >x_0$, but $x -x_0$ is so small that $F(x) \leq F(x_0) +\epsilon$, due to continuity. Then there is a $z$ so that $x > z > x_0$, and $F(z) \geq y$, by the nature of inf. So $F(x_0) +\epsilon \geq F(x) \geq F(z) \geq y$.

It remains to show $F(x_0) \leq y$. Since any single point $x$ has probability measure 0, we can establish that $F$ is onto $(0,1)$. Thus, there is some $z$ so that $F(z) =y$. Now, $F(z) \geq y$ and so $x_0 \leq z$ by the definition of inf. We have $F(x_0) \leq F(z) =y$.

The intended result is immediate by invoking the definition of CDF and expression of $\mathbb {P} (A)$. Am I overthinking? Are there more straightforward proofs?

Answer 1

So, I am not sure what you are exactly trying to prove, but I think that I can write something that might answer your question.

In general, we know that CDFS are càdlàg functions, so that if $F$ is a CDF, then

• the left limits $\lim_{x\to t^-} F(x)$ always exist
• the right limits $\lim_{x\to t^+} F(x)$ always exist and equal $F(t)$

Since $F$ is (not strictly) increasing, this means that the function can have jump discontinuities at worst, and that if there is a jump at $x=t$, then $F(t)$ is equal to the limit from the right.

This means that we can uniquely define the "quantile function", which is written as $F^{-1}$, but it's not the usual inverse. It is the same as the inverse for strictly increasing $F$, but otherwise it is defined (as you have written above) as $$F^{-1}(y)=\inf\{x\in\mathbb{R}:F(x)\geq y\}.$$ The "continuous from the right" property ensures that this infimum works nicely.

Using this definition of $F^{-1}$, it holds that $A=B$.