sigma-algebra on the image of a random variable

Question

Usually a random variable is considered to be a function $X: (\Omega, \sigma, P) \to (\mathbb{R}, B(\mathbb{R}))$ but I wonder what advantage choosing $B(\mathbb{R})$ as a sigma algebra on $\mathbb{R}$ has over using the pushforward of $\sigma$ under $X$.

Answer 1

Here’s one immediate problem. Let $(\Omega, \sigma, \mathbb{P})$ be a probability space. Suppose we have an arbitrary function $X: \Omega \rightarrow \mathbb{R}$. Perhaps the most basic question we could ask is “what is the probability that $X \leq r$ for fixed $r \in \mathbb{R}$?”

If we equip $\mathbb{R}$ with the Borel sigma-algebra $\mathcal{B}(\mathbb{R})$, then this probability $\mathbb{P}[X \leq r]$ is well-defined for all $r$ if and only if $X$ is a measurable function. Since $X$ was an arbitrary function, this may not be the case! But “measurability” with respect to the Borel $\sigma$-algebra exactly detects whether or not these probabilities are always defined.

On the other hand, suppose we push the $\sigma$-algebra $\sigma$ forward by $X$. That is, we equip $\mathbb{R}$ with the $\sigma$-algebra $\mathcal{F} := \{ A \subseteq X : X^{-1}(A) \in \sigma \}$. It is clear that $X$ will always be measurable with respect to this $\sigma$-algebra, but it is no longer the case that $\mathcal{F}$ necessarily contains all the intervals. In other words, this all “makes sense” mathematically, but you may find you are no longer able to answer all the questions you’re interested in asking.

Further, suppose it is the case that every set of the form $(- \infty, r]$ is contained in the pushforward $\mathcal{F}$. Then it follows that $\mathcal{B} \subseteq \mathcal{F}$. In this case, we lose nothing by simply considering $X$ As a measurable function with respect to the Borel sets on $\mathbb{R}$.