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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.

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sorry, mistyped in the title, I wanted not "non-measurable uncountable" but "uncountable with zero measure", fixed it –  karlicoss Jan 27 '13 at 8:48
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2 Answers

up vote 2 down vote accepted

What if $f$ maps $X_0$ homeomorphically to $\left[0,\frac12\right]$ and $\Bbb I^2\setminus X_0$ bijectively to $\left(\frac12,1\right]$?

Added: It’s quite possible that $f$ maps some of the verticals $\{x\}\times\Bbb I$ to measurable sets of different measures and others to non-measurable sets. For example, $f$ might map $X_0$ homeomorphically onto $\left[0,\frac12\right]$ (of measure $1/2$), $\left\{\frac12\right\}\times\Bbb I$ homeomorphically onto $\left[\frac34,1\right]$ (of measure $1/4$), and (assuming the axiom of choice) the rest of $\Bbb I^2$ bijectively to $\left(\frac12,\frac34\right)$ in such a way that no other vertical $\{x\}\times\Bbb I$ maps to a measurable set at all.

The behavior of $f$ on one vertical says nothing about its behavior on another (save that the images have to be disjoint, of course, and the measures of any measurable images cannot sum to more than $1$).

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You may want to edit now. –  Asaf Karagila Jan 27 '13 at 8:45
    
@Asaf: Thanks. It was still relevant, but I decided to expand it anyway. –  Brian M. Scott Jan 27 '13 at 8:52
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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.

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