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First post! I'm currently going through Grimmett & Stirzaker's Probability and Random Processes. Exercise 2.3.2 says:

Let $X$ be a random variable and let $g:\mathbb{R}\rightarrow \mathbb{R}$ be continuous and strictly increasing. Show that $y=g(X)$ is a random variable

If we take $(\Omega, \mathcal{F}, \mathbb{P})$ as our probability space, the definition of a random variable on this space says that $X: \Omega \rightarrow \mathbb{R}$ is a random variable if $\{\omega \in \Omega : X(\omega) \leq x\} \in \mathcal{F}$ for each $x\in \mathbb{R}$.

The answer to the exercise is:

For $y$ lying in the range of $g$, $\{Y\leq y\} = \{X\leq g^{-1}(y)\}\in \mathcal{F}$

I don't understand the necessity of the continuity of $g$. My understanding is that we have the requirement that $g$ is invertible, so that $g^{-1}(y)$ is defined for each $y\in g(\mathbb{R})$, and then since $X$ is a random variable for each $y \in \mathbb{R}-g(\mathbb{R})$ we already have $\{\omega \in \Omega : X(w) \leq y\} \in \mathcal{F}$.

And indeed, strictly monotone continuous functions are bijective (when defined as mapping domain to image). But $g$ is a strictly monotone total function defined on the whole of $\mathbb{R}$ -- doesn't it suffice that it's injective? I can't come up with a counterexample where this would not be the case.

So my question is, after all, if $g:\mathbb{R}\rightarrow \mathbb{R}$ is total and strictly monotone (not necessarily continuous), does it always have an inverse? My best guess is yes because $|\mathbb{R}|$ and $|g(\mathbb{R})|$ should be the same, and so such a function should be bijective. But that's not a very formal statement.

I'm self-studying this and analysis and a bunch of other stuff, so apologies for any flaws in my reasoning.

geoant
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    Consider the function that sends $x$ to itself, if $x < 0$ and to $x+1$ if $x \ge 0$. This function is injective and not continuous at $0$. It is not surjective as $\frac12$ is not in the image. – Tzimmo Sep 11 '24 at 12:59
  • @Tzimmo Ah I see, a very simple one! So the continuity in this case is necessary for $g$ to be surjective? I think what confused me was "For $y$ lying in the range of $g$" in the answer. – geoant Sep 11 '24 at 13:12
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    Yes, you need continuity for $g$ to be surjective. Or you can corestrict to the image of $g$, but that might bring some other problems with it. – Tzimmo Sep 11 '24 at 13:27

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Continuity is not strictly required, but may complicate the proof. If $g$ is strictly increasing, then $g$ is a Borel measurable function. See here for a discussion of that proof. Thus if we just work functionally, the preimage of Borel sets of the composition $g\circ X$ will be in $\mathcal F$, and so $Y = g\circ X$ will be a random variable.

Without continuity though, the range of $g$ might not be connected or simple to describe, so using $g^{-1}$ to describe the preimages would be messier. Continuity is likely there to help simplify the exercise or keep the arguments aligned with the material presented in the text.

Trevor Norton
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