# Change of order of quantifiers "every$\text{ }t$" and "almost surely": what difference does it make?

Given a certain probability space $$(\Omega,\mathcal{A},\mathbb{P})$$ and a random variable $$X:t\mapsto X(t)$$ defined on it, which is the difference between the following statements: $$\color{blue}{\text{every }t}\text{ is }\color{red}{\text{almost surely}}\text{ a nondifferentiability point for }X(t)\tag{1}$$ $$\color{red}{\text{almost surely }} \color{blue}{\text{every }t}\text{ is a nondifferentiability point for }X(t)\tag{2}$$ ?

I would rewrite $$(1)$$ as: $$\tag{1.int}\forall t\text{, }\mathbb{P}(t\text{ is a nondifferentiability point for}X(t))=1$$ and $$(2)$$ as: $$\tag{2.int}\mathbb{P}(t\text{ is a nondifferentiability point for}X(t), \forall t)=1$$

First, I don't know whether $$(1.\text{int})$$ and $$(2.\text{int})$$ are correct "rewritings" of $$(1)$$ and $$(2)$$ (resp.).

In general, whichever the difference between $$(1)$$ and $$(2)$$, which is the gist of such a difference from a mathematical standpoint? I cannot grasp it.

Could you please give an example of a random variable for which $$(1)$$ holds true, but $$(2)$$ does NOT hold true? (or viceversa)

• You already know the correct interpretation! Oct 8, 2020 at 11:24
• Good, but I cannot understand what it does mean "practically": could you give me an example of a random variable for which $(1)$ holds true, but $(2)$ does NOT hold true? @KaviRamaMurthy Oct 8, 2020 at 11:29
• For any fixed $t_0$, note that the set in (2.int) is a subset of the set in (1.int) at $t=t_0$. So (2.int) implies (1.int) by subadditivity. Oct 8, 2020 at 11:33

On $$(0,1)$$ with Lebesgue measure let $$X_t(\omega)=|t-\omega|$$. Then $$P(X_t \, \text {is differentiable at } \, t)=1$$ for each $$t$$ and $$P(X_t \, \text {is differentiable at every point} \, t)=0$$.
• Why, for $X_t(\omega)=|t-\omega|$, does it hold true that $P(X_t \text{ is differentiable at }t)=1$ for each $t$? Oct 8, 2020 at 17:15
• @Strictly_increasing Because $P$ here is the Lebesgue measure and Lebesgue measure of $(0,1) \setminus \{t\}$ is $1$. Oct 8, 2020 at 23:12