Suppose $\mu$ be a probability measure on $\mathbb{R}^d$. Let $V \subset \mathbb{R}^d$ be a linear subspace such that $\dim V = r < d$.
Then can $\mu(V) \approx 1 = \mu(\mathbb{R}^d)$?
I believe it can't be true. For example, let $\mu$ be the weighted Lebesgue measure (Gaussian distribution) in $\mathbb{R}^2$ where $$d\mu(x,y) = \frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}dxdy.$$ Then the measure of x-axis $V = \mathbb{R} \times \{0\}$ is $$ \mu(V) = 0, \qquad \text{and} \qquad \dim V = 1.$$
In general, let $V$ be a $r$-dimensional subspace of $\mathbb{R}^d$. Then by applying appropriate rotations, we may assume $x_1,\cdots,x_r$ axis generates $V$. Thus $\mu(V) = 0$ as $x_n$ axis integration is 0.
Am I missing something? Or my derivation is good enough to justify $\mu(V) = 0$?
Any comments or suggestions would be appreciated.
Of course this is possible. For instance, if you fix any point $a\in \mathbb{R}^d$, then you could define a probability measure $\mu$ by $\mu(A)=1$ if $a\in A$ and $\mu(A)=0$ if $a\not\in A$. This is a perfectly good probability measure. More generally, given your favorite probability measure $\nu$ on some subspace $V\subset\mathbb{R}^n$, you can get a probability measure $\mu$ on $\mathbb{R}^n$ by defining $\mu(A)=\nu(A\cap V$).
Your error seems to be assuming the measure must be given by integrating some distribution function with respect to Lebesgue measure. This is certainly not true of arbitrary measures; measures of this form are called absolutely continuous with respect to Lebesgue measure.