A ``continuous'' family of measurable subsets of [0,1], each with a positive measure, admits a ``continuous'' subfamily with non-empty intersection? Let $\{Y_x\}$ be a family of measurable subsets of [0,1] indexed by $x \in [0,1]$, and such that each set $Y_x$ has a positive (Lebesgue) measure. Is there always a subset $X \subseteq [0,1]$ with positive measure and such that $\bigcap\limits_{x \in X} Y_x$ is nonempty?
Intuitively it seems to be true, possibly under some mild extra assumptions (say compactness of $Y_x$'s).
 A: The comments contain essentially a full answer, and CH is not needed, though perhaps there is a more elementary counterexample.
Let $c$ be the cardinality of the reals.
Lemma: Every subset of $[0, 1]$ of positive measure has cardinality $c$.
This is well-known. Here is one proof. Let $X \subset [0, 1]$ be a set with $\mu(X) > 0$. Then the convolution $f(x) = 1_X * 1_{-X}$ is continuous and $f(0) = \mu(X) > 0$, so there is some neighbourhood of zero on which $f > 0$. This implies that $X - X$ contains an interval, so $|X| = |X - X| = c$.
Now by choosing an bijection between $[0, 1]$ and $c$, there is an ordering $\prec$ on $[0,1]$ for which every $x$ has $< c$ predecessors. This implies that $\{y \in [0, 1] : y \prec x\}$ has measure zero by the lemma, so $\{y \in [0, 1] : y \succ x\}$ has full measure. Let $Y_x$ be a compact subset of $\{y \in [0, 1] : y \succ x\}$ of positive measure. Now the intersection of any collection of $c$-many $Y_x$'s is empty.
EDIT: Alessandro Codenotti points out that $\{y \in [0, 1] : y \prec x\}$ can be nonmeasurable if CH does not hold. The lemma only shows the inner measure is zero. So I'm not adding anything to bof's solution after all.
EDIT 2: Here is a different solution. By inner regularity (and the lemma) it suffices to define $Y_x$ ($x \in [0, 1]$) such that $\bigcap_{x \in C} Y_x = \emptyset$ for every compact set $C$ of cardinality $c$. Note that there are only $c$ such sets (because there are only $c$ open sets). Let $\{C_\alpha : \alpha < c\}$ be all such sets. For each $\alpha < c$ in turn do the following. Choose $x_\alpha, y_\alpha \in C_\alpha \setminus \{x_\beta, y_\beta : \beta < \alpha\}$ and define $Y_{x_\alpha} = [0, 1/3]$ and $Y_{y_\alpha} = [2/3, 1]$. Define also $Y_x = [0, 1]$ for all $x \in [0, 1] \setminus \{x_\alpha, y_\alpha : \alpha < c\}$.
A reasonable hypothesis to add would be something about the measurability of the map $x \mapsto Y_x$, maybe the graph $\Gamma = \{(x, y) : y \in Y_x \}$ should be measurable. In this case I think the claim follows from Fubini's theorem:
$$\int \mu(\{x : y \in Y_x\}) \, dy = \mu(\Gamma) = \int \mu(Y_x) \, dx > 0,$$
so there is some $y \in [0, 1]$ such that
$$\mu(\{x : y \in Y_x\}) > 0.$$
