# Prove that $-X$ is measurable with respect to some sigma field.

I am reading A Second Course in Probability by Ross and Peköz. I came across the following question:

1.10.4. Show that if $$X$$ and $$Y$$ are real-valued random variables measurable with respect to some given sigma field, then so is $$XY$$ with respect to the same sigma field.

My attempt was to first show that:

(i) $$X+Y$$ is measurable

(ii) $$cX$$ is measurable for any $$c\in\mathbb R$$

(iii) $$X^2$$ is measurable.

Then, since $$XY = \frac14[(X+Y)^2-(X-Y)^2]$$, we just use the above properties.

Property (i) was proven in Chapter 1. However, I am having a bit of trouble showing property (ii) in the case that $$c < 0$$. From the problem setup, we know that for any $$x\in\mathbb R$$, $$\{X\leq x\}\equiv\{\omega\in\Omega:X(\omega)\leq x\}\in \mathcal F$$.

So $$\{-X\leq x\}=\{X\geq -x\} = \{X<-x\}^c$$, but I don't know if $$\{X<-x\}\in\mathcal F$$. If I expand a little more, it seems I essentially need to show that $$\{X=x\}\in\mathcal F$$, but I don't see how that is necessarily true.

I have seen the question Product of two random variables, though in the answers, they assume knowledge of the fact that $$cX$$ is measurable, and they also seem to be working with the Borel $$\sigma$$-algebra.

I have also seen these notes, but they omit the case where $$c < 0$$.

My Question: How do we show that $$\{-X\leq x\}\in\mathcal F$$ if we already know that $$\{X\leq x\}\in\mathcal F$$? Do we need to assume something about the sigma field in order to show this for real-valued random variables (e.g. Borel sigma field)?

• Do you mean that you want to prove this for an arbitrary sigma field on $\mathbb{R}$? Rather than simply the Borel sigma field? Jun 7, 2020 at 21:55
• Hint: write $\lbrace X <-x \rbrace = \bigcup_{n} \lbrace X \leq -x-1/n\rbrace$ Jun 7, 2020 at 21:56
• @Dasherman Maybe that is what I am failing to recognize. The question never states anything about the Borel sigma field, so I assume that I should prove it for an arbitrary sigma field. Jun 7, 2020 at 22:00

Note that $$\{X. Since a Sigma field is closed under countable unions this implies $$\{X for every $$x\in\mathbb{R}$$.
A general note: If $$X:\Omega\to\mathbb{R}$$ is a random variable then we have $$X^{-1}(B)\in\mathcal{F}$$ for every Borel set $$B\subseteq\mathbb{R}$$. In many books this is actually the definition of a random variable. The definition you are using, that is $$\{X\leq x\}\in\mathcal{F}$$ for every $$x\in\mathbb{R}$$, is an equivalent definition. It is good to know both definitions. To prove the equivalence note that the collection of sets $$\{B\subseteq\mathbb{R}: X^{-1}(B)\in\mathcal{F}\}$$ is a Sigma field on $$\mathbb{R}$$, and from your definition it is not too hard to show it contains all open subsets of $$\mathbb{R}$$, hence must contain the whole Borel sigma field.