TITLE: An example of uncountable set with zero Lebesgue measure
QUESTION [6 upvotes]: It is known that there is a bijection between $\mathbb{I}$ and $\mathbb{I}^2$ (let's denote this map as $f$). Consider the set $X_0 = \{ (0, x) \mid x \in [0, 1] \}$ (just a vertical unit interval in $\mathbb{I}^2$). Now consider the set $Y_0 = f(X_0)$.
If it is Lebesgue measurable, its measure $\lambda(Y_0)$ is zero. Proof by contradiction: suppose $\lambda(Y_0) = c > 0$, then we take countable number of unit intervals in $\mathbb{I}^2$, let them be $X_i = \{ (i, x) \mid x \in [0, 1] \}$ where $i$ indexes over rational numbers in $\mathbb{I}$ and map them. All $Y_i$ will be disjoint and in assumption of $Y_0$ being measurable, all $Y_i$ should be measurable and all $\lambda(Y_i)$ should be equal to $c$ (not sure how to prove it, but it's kind of intuitive. Or not?). Since $\bigcup\limits_{i \in \mathbb{Q} \cap \mathbb{I}} Y_i \subset \mathbb{I}$, then $\lambda(\bigcup\limits_{i \in \mathbb{Q} \cap \mathbb{I}} Y_i) = \sum\limits_{i \in \mathbb{Q} \cap \mathbb{I}} c \le \lambda(\mathbb{I}) = 1$, which is obviously contradiction.
However, I think it is very unlikely to such a set $Y_0$ to be Lebesgue measurable, but still, is there a way to prove its non-measurability?
P.S. Yeah, I know about Cantor set, just trying to construct something myself.

REPLY [2 votes]: For another approach, take $b\in\Bbb R$. Then $\{b\}\times\Bbb R$ is zero subset of $\Bbb R^2$. See 1. here to see why.