What is the difference between "almost surely" property of a measurable function and the property which holds "on the support of a random variable"?

Question

Consider a random variable $X: \Omega \to \mathbb{R}$ on $(\Omega, \mathcal{F}, P)$ and a measurable function $g:\mathbb{R} \to \mathbb{R}$.

And let's choose some function property, for example continuity.

Does the sentence "function $g$ is continuous almost surely" (i.e. $g$ is continuous at every point of $B \subset \mathbb{R}, \, P(X \in B) = 1$) mean the same as "function $g$ is continuous on the support of $X$"?

This looks like to be true when $X$ is a discrete random variable but what about other possible cases (continuous r.v.)? And what if we choose another function property (monotonicity, differentiability, etc.)?

Answer 1

No, these mean different things. If the support of $X$ is $B$ then $B$ must be closed, but this is not the case if we merely assume $P(X\in B)=1$.

For an example, let $\{q_k\}$ be an enumeration of the rationals and suppose $P(X=q_k)=2^{-k}$, $k\ge1$. Then the support of $X$ is $\mathbb R$, since $P(X\in U)>0$ for all non-empty open sets $U$, but $X$ only takes values in $\mathbb Q$. It is not difficult to give an example of a function $g$ which is continuous at $x$ for every $x\in\mathbb Q$, but is not continuous on all of $\mathbb R$. In fact, we could even construct such a $g$ for which the set on which is continuous is "small" in some sense. For instance, let $I=\bigcup_k(q_k-2^{-k}\varepsilon,q_k+2^{-k}\varepsilon)$ where $\varepsilon>0$ is small, and let $g(x)=\mathbf1_{I}(x)$. Then $g$ is continuous at $x$ if and only if $x\in I$; in particular $g$ is continuous on $\mathbb Q$ and so is continuous almost surely with respect to the law of $X$, but the Lebesgue measure of $I$ is no larger than $2\varepsilon$, which is small.

On the other hand, if a property holds on the support of a probability measure $\mu$, then it also holds $\mu$-almost surely. This is because $\mu(\operatorname{supp}\mu)=1$.