1. ## Zero content subset

I'm trying to find the proof of the following proposition.

Let $S^n$ be the unit sphere in $R^n$. If $K \subset S^n$ is $n-1$ dimensional, then K has Lebesgue measure zero.

I'd b very thankful if any of you could give me the general idea behind the proof or the link to a textbook or internet source where the proof of this is explained.

I'm trying to find the proof of the following proposition.

Let $S^n$ be the unit sphere in $R^n$. If $K \subset S^n$ is $n-1$ dimensional, then K has Lebesgue measure zero.

I'd b very thankful if any of you could give me the general idea behind the proof or the link to a textbook or internet source where the proof of this is explained.
Are you aware of the theorem that states that if $M$ is an open submanifold of $\mathbb{R}^m$ with $m and $F:M\to\mathbb{R}^n$ is smooth then $F(M)$ has measure zero?

3. What's the definition of "n-1 dimensional"?

4. Hm, now that I think about it, the formulation I have written up there may be a little sloppy. The whole situation is the following;

I'm trying to prove that a random matrix is always full rank, or, in other words, that if we are in $R^n$, a set of vectors with random coefficients will always be linearly independent, so long as we don't choose more than n vectors.

Now I can easily show this in 2 dimensions, and the idea is that, if we choose one vector at random, $V_1=(r,\theta_1)$, we can calculate the probability that a second vector chosen at random will linearly dependent with $V_1$ by calculating the probability that the vector will be on the subspace of 1 dimension spanned by $V_1$. You can then show that this is going to happen if the second vector has an angle of $\theta_1$ or $\theta_1 + \pi$, and this set on the unit circle has measure zero, and thus an integral of any probability distribution over this set is zero.

I'm now trying to generalize this idea on $R^n$.

5. I'd first show that, generically, you won't ever have the zero vector as a column in your random matrix.

Then go by induction: change the basis so that the first column becomes the first standard basis vector (1,0,...,0). Now unless the second column is all zero except for possibly the first entry, you can change the basis again so that it becomes the second standard basis vector (0,1,0,...,0). And so forth.

You're just using over and over again the fact that a point chosen "at random" from R^n won't be the origin.