Let $X$ be a random variable and $F_X(x) = P(X \le x)$ its cumulative distribution function (cdf). $P_X$ is the probability measure induced by $X$, which is defined by $P_X((a,b)) = P(X^{-1}((a,b))$ for $a,b \in \mathbb{R}$ .

  1. Is it correct to say that $F_X$ is strictly increasing $P_X$ almost everywhere?
  2. If so, what would be the correct way to express that formally?
  3. Does it imply $ F(x) = P(X \le x) = P(F(X) \le F(x))$ ?

If have seen that the equality in the 3. question is used to prove the probability integral transform. I think that it is not as straight forward to show 3. if $F_X$ is not strictly increasing.

I attempted to answer these questions, considering some very helpful comments from whuber and Zhanxiong (who provided detailed explanations of the probability transform here and here).

My attempt goes so far:

Is it correct to say that $F_X$ is strictly increasing $P_X$ almost everywhere?

Not really, since almost everywhere statements refer to properties of individual points. Strict monotonicity is not a property of a single point.

What would be the correct way to express that formally?

It holds that for any $a \le b$ with $F_X(a) = F_X(b)$ that $$P_X((a,b)) \le P_X((a,b])) = F_X(b) - F_X(a) = 0$$ So it is better to say that "Strict monotonicity fails to hold only on sets of measure 0" (by whuber)

Does this imply $F_X(x) = P(X\le x) = P(F(X) \le F(x))$ ?

Yes. See: Zhanxiong's answer for a shorter version. Edit: I have adapted this attempt of a proof a couple of times following recommendations from the comments.

Consider a fixed $x \in \mathbb{R}$. Because $F_X$ is monotonically increasing, $$ \{ X > x, F_X(X) \le F_X(x) \} = \{ X > x, F_X(X) = F_X(x) \}$$ It holds that $$P(\{ X> x, F_X(X) = F_X(x) \}) = 0 $$ To see this consider $x_0 $ s.t. $$x_0 = \text{sup}\{y\mid y> x, F_X(y) = F_X(x) \}$$ Then $F_X(y) = F_X(x)$ for all $x_0> y > x$ and thus $$P(\{X> x, F_X(X) = F_X(x) \}) = P_X((x, x_0)) =0$$

We have by monotonicity $$\{X \le x \} \subseteq \{F_X(X) \le F_X(x) \} $$ It follows that $$P(\{ F_X(X) \le F_X(x) \})= P(\{X\le x \})$$

Is there an easier way to show 3.? I expected it to straightforward.

Another question I have is,

  1. Can we say that $F_X$ is strictly increasing on the image of $X$ ?

I am not sure about this, but I guess we can remove all values from the image of $X$ that occur with probability 0 and then it would be true.

  • $\begingroup$ After $F$ is defined as $P[X \leq x]$, it is a deterministic function on $\mathbb{R}^1$. So be careful on using the qualifier "almost surely", it does not have anything to do with the original probability space anymore. One (unnatural) way of talking "almost surely" is with respect to the Lebesgue measure on $\mathbb{R}^1$, but if so, your statement is obviously wrong -- just consider the CDF of any discrete r.v. $\endgroup$
    – Zhanxiong
    Nov 24, 2023 at 12:59
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    $\begingroup$ OK, I see what you meant now. That seems to be a valid question but I am not sure if is answerable (as yourself have felt the difficulty) because "strict monotonicity" is an overall property of a function while "almost everywhere" usually needs to be accompanied by some point-wise property. For example, we can say a function $f$ is non-negative almost everywhere, but we don't say $f$ is integrable almost everywhere because the former is a pointwise property of $f$ while the latter is an overall property of $f$. $\endgroup$
    – Zhanxiong
    Nov 24, 2023 at 14:02
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    $\begingroup$ I see that I have should clarify which measure on $\mathbb{R}$ I intend. It is $P_X((a,b)) = P(X^{-1}((a,b))$ $\endgroup$
    – ChrisL
    Nov 24, 2023 at 16:32
  • 2
    $\begingroup$ But then the question is trivial: let $b\ge a$ with $F_X(b)=F_X(a)$ and immediately conclude $P_X((a,b))\le P_X((a,b])=F_X(b)-F_X(a)=0.$ In words: "strict monotonicity fails to hold only on sets of measure zero." $\endgroup$
    – whuber
    Nov 24, 2023 at 19:51
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    $\begingroup$ Perfect, thank you $\endgroup$
    – ChrisL
    Nov 25, 2023 at 8:52

1 Answer 1


To your specific question "Is there an easier way to show 3.?", the answer is "Yes". The problem of your "proof" is not only its verbosity (for example, I don't think it is necessary at all to introduce "$\omega$"s into the proof), but more importantly, it is probably incorrect (in my opinion many steps are hard to justify).

Here is my proof by discussing the range of $F(x)$.

Case 1: $F(x) = 1$. In this case $P[X \leq x] = F(x) = 1$. On the other hand, clearly we have $P[F(X) \leq 1] = 1$. Thus $P[X \leq x] = P[F(X) \leq F(x)]$.

Case 2: $0 \leq F(x) < 1$. For simplicity, denote by $C$ the set of all continuous points of $F$ and define $x^* := \sup\{y \geq x: F(y) = F(x)\}$. Because $F(x) < 1$, we have $x^* < \infty$. There are the following two sub-cases:

  • $x^* \in C$. In this case \begin{align*} P[F(X) \leq F(x)] = P[X \leq x^*] = F(x^*) = F(x). \end{align*} To help you understand the geometry of this case, see the graph of $F(\cdot)$ below:

enter image description here

  • $x^* \in C^c$. In this case \begin{align*} P[F(X) \leq F(x)] = P[X < x^*] = F(x^*-) = F(x). \end{align*} As in sub-case 1, below is the plot of illustrating the above identity.

enter image description here

This completes the proof.

  • $\begingroup$ Thank you for your answer. I had included $\omega$ to help me distinguish between Probability measures. Now I see it is not helpful. $\endgroup$
    – ChrisL
    Nov 27, 2023 at 6:36
  • $\begingroup$ I have adapted my attempt of a proof. It was indeed incorrect. I hope, it is correct now. Can you provide more detailes on the last part of your proof, please? The case where $x^*\in C^c$. $\endgroup$
    – ChrisL
    Nov 27, 2023 at 8:10
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    $\begingroup$ @ChrisL Try to draw a picture. Note that for all $x \leq y < x^*$, we have $F(y) = F(x)$, hence $F(x^*-) = \lim_{y \uparrow x^*} F(y) = F(x)$. $\endgroup$
    – Zhanxiong
    Nov 27, 2023 at 13:20
  • $\begingroup$ @ChrisL I added some visualizations. Hope it helps. $\endgroup$
    – Zhanxiong
    Nov 27, 2023 at 16:19
  • $\begingroup$ thank you! In particular explaining the notation $x^*-$ was helpful. $\endgroup$
    – ChrisL
    Nov 27, 2023 at 18:51

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