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.