Number of random variables in a stochastic process taking a value Suppose I have a stochastic process $\{X_t : t \in [0,1]\}$ where the $X_t$ are iid continuous RVs supported on $\mathbb R$. For any $x \in \mathbb R$ I know $P(X_t = x)=0$, but I have uncountably many RVs. What is the (likely?) cardinality of $\{t \in [0,1] : X_t = 0\}$? Is this some kind of possibly-infinite valued random variable now? If my index set was countable I’d guess that I’d expect zero or maybe at most finitely many, but I’m not sure how to reason about this with an uncountable index. Do I need to have some distribution on $[0,1]$? 
Alternatively, what about the Lebesgue measure of $\{t : X_t = x\}$? I'm not sure if there's any reason for this set to be measurable unless it's at most countable, so I'm guessing if it is measurable then the measure will be zero but I'm not sure how to try showing that.
 A: This is a very nice question! But not so innocent as it seems.
Let us first look at the Lebesgue measure. It is tempting to say that it is zero almost surely since 
$$
\mathbb{E} [\lambda(\{t\in [0,1]: X_t = x\})] = \mathbb{E} \left [\int_0^1 \mathbf{1}_{X_t = x}\,dt \right ]\\ = \int_0^1 \mathbb{E} \left [\mathbf{1}_{X_t = x} \right ]dt = \int_0^1 \mathbb{P} (X_t = x)dt = 0.
$$
But hold on. This is incorrect, since the process $X$ is not measurable. Indeed, assume the contrary and note that wlog $X$ is centered and bounded (we can consider $X'_t = h(X_t)$ with appropriate $h$). Then, 
$$
\mathbb E\left[ \left(\int_0^t X_s ds \right)^2\right] = \mathbb E\left[ \int_0^t X_s ds \cdot \int_0^t X_u du\right] = \int_0^t \int_0^t \mathbb{E}[X_sX_u]ds\, du = 0.
$$
Therefore, for each $t\in [0,1]$, $f(t):=\int_0^t X_s ds$ is constant almost surely. This already sounds absurd, and, in view of continuity, we can claim that there is a set $\Omega'\subset \Omega$ of probability $1$ such that $\int_0^t X_s ds = f(t)$ for all $t\in [0,1]$ and $\omega\in \Omega'$. This, however, possible only if $X_t$ is a constant for each $t$, which contradicts the fact that the distribution is non-degenerate. 
This not only shows that we cannot speak of the measure of level sets, but also that the question is very delicate, as we cannot define $X$ on any space of "well-behaved" functions. Actually, we cannot go much further that defining it as an element of $\mathbb R^{[0,1]}$ equipped with cylindrical sigma-algebra $\mathcal B(\mathbb R^{[0,1]})$. Unfortunately, this sigma-algebra is very poor, it contains only sets "supported by countable set of coordinates": for any $B \in \mathcal B(\mathbb R^{[0,1]})$, there are some $(t_n,n\ge 1)\subset [0,1]$ and $A \in \mathcal B(\mathbb R^{\mathbb N})$ such that $B = \{f\in \mathbb R^{[0,1]}: \big(f(t_1),f(t_2),\dots)\in A\}$. In particular, from the probabilistic point of view, the events like $\{\exists t\in [0,1]: X_t = x\}$ or $\big\{\{t\in [0,1]: X_t = x\}\text{ is countable}\big\}$ are quite meaningless. 
But why can we speak of the set of zeroes of a Brownian motion, say? This is because the Brownian motion $W$ can be defined as an element of $C[0,1]$, and the events like $\{\exists t\in [0,1]: W_t = 0\}$ can be defined countably.
See also a related discussion here. 
