Let ${\cal X} \subset {\mathbb R}^n$ be a compact set with $\mu_n({\cal X}) > 0$, where $\mu_n(\cdot)$ refers to $n$-dimensional Lebesgue measure. Let ${\cal Y} \subset {\mathbb R}^d$ be \begin{equation} {\cal Y} = \{y\colon y = I_{d\times n}x\}, \end{equation} where $I_{d\times n} = [I_d~0]$, and $I_d$ is a $d$-dimensional identity matrix. To be more specific, the relationship between $x$ and $y$ is that: if $x = (x_1,\ldots,x_d,\ldots,x_n)^T$, then $y = (x_1,\ldots,x_d)^T$.
Then, does $\mu_d({\cal Y}) > 0$ hold?
Thanks very much!
Yes, it does hold. Otherwise, we would have $$ X \subset Y\times \Bbb{R}^{n-d}, $$ where the right-hand side is easily seen to be a Lebesgue null-set (since $Y $ is). But this contradicts $\mu (X)>0$.