If $X = g(Y)$, then Is the sigma-algebra generated by $X$ a subset of the sigma-algebra generated by $Y$?

Question

Let $g$ be a continuous function. Let random variables $X$ and $Y$ satisfy $X = g(Y)$. Do we always have $\sigma(X) \subset \sigma(Y)$?

I want to disprove this. My thought is that there must be a set $S$ such that $g(Y)(S) = X^{-1}(B)$ where $B$ is a Borel set. $g$ might not have an inverse, but even if it does, it doesn't necessarily map to $B$ (that is, to be an element of $\sigma(Y)$.

Is this a correct approach?

Answer 1

Let $\mathcal B$ be the Borel $\sigma$-algebra of $\mathbb R$. The definition is that $\sigma(X)=\{X^{-1}(B):B\in\mathcal B\}$ and $\sigma(Y)=\{Y^{-1}(B):B\in\mathcal B\}$.

Notice that a continuous $g:\mathbb R\to\mathbb R$ must be Borel measurable, meaning $g^{-1}(B)\in\mathcal B$ for all $B\in\mathcal B$.

That is to say $\sigma(X)=\{(g\circ Y)^{-1}(B):B\in\mathcal B\}=\{Y^{-1}(g^{-1}(B)):B\in\mathcal B\}\subseteq\{Y^{-1}(C):C\in\mathcal B\}=\sigma(Y)$.