Show that if the projection of a set is negligible, then the set is negligible as well

Question

I'd like a hint in the right direction, im drawing a complete blank.

let $E \subset \mathbb R^2$. We'll define the projection of $E$ unto the $x$ axis as:

$P_x(E)=\{x| \exists y \in \mathbb R s.t (x,y) \in E\}$

Show that if $P_x(E)$ is a negligible set in $\mathbb R$, then $E$ is a negligible set in $\mathbb R^2$. Is the opposite direction true as well?

By negligible I mean that the jordan measure is zero. meaning for any $\epsilon >0$ I can cover the set with $\aleph_0$ many open intervals and the sum of the lengths of the intervals is less than $\epsilon$

Could anyone point me in the right direction?

Answer 1

If $E$ is negligible, it is not necessarily the case that $P_x(E)$ is negligible. Consider the line $y = 1$, which is negligible in $\mathbb{R}^2$ but its projection to the $x$ axis is the entire real line.

On the other hand, if $P_x(E)$ is negligible, you can prove $E$ is negligible. I'll prove the following statement: If $E_0 \subset \mathbb{R}$ is negligible, then the 'extension' $E_0 \times \mathbb{R} \subset \mathbb{R}^2$ is also negligible. Since for any $E \subset \mathbb{R}^2$ we have $E \subset P_x(E) \times \mathbb{R}$, this will imply the original statement.

So: assume $E_0 \subset \mathbb{R}$ is negligible. Pick $\varepsilon > 0$. The trick goes like this: We can cover $\mathbb{R}$ using countably many open intervals $I_1, I_2, ...$ each having length $1$. For example, we can take $I_{2n + 1} = (-\frac{n}{2},-\frac{n}{2} + 1)$ and $I_{2n} = (\frac{n}{2},\frac{n}{2} + 1)$, but it doesn't really matter. Now, for each integer $n$, we can cover $E_0$ with intervals $J_{n,1}, J_{n,2}, ...$ whose total length is less than $\varepsilon 2^{-n}$. Now my claim is that the boxes $J_{n,k} \times I_n$ cover all of $E_0 \times \mathbb{R}$, and their total area is less than $\varepsilon$. The fact that they cover $E_0 \times \mathbb{R}$ is easy: If $(x,y) \in E_0 \times \mathbb{R}$ there is some $I_n$ containing $y$, and for that $n$ there is some $k$ such that $J_{n,k}$ contains $x$, so $(x,y) \in J_{n,k} \times I_n$. As for the total area:

$$\sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty} A(J_{n,k} \times I_n) = \sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}L(J_{n,k}) < \sum_{n = 1}^{\infty}\varepsilon 2^{-n} = \varepsilon$$

Answer 2

I think there might be a problem with this question. I would very much appreciate someone to review this answer.

I gave it a good thought and I came up with this example, let's suppose that $E$ is the straight line $x=1$. It is not negligible. it's isometric to the entire real line.

But in this case, $P_x(E)=\{1\}$ a finite set, it is negligible. So I dont think what they ask to prove is correct.

Answer 3

I'll use your definition of negligible. If $\varepsilon>0$, then choose open covers $\{(a_k,b_k)\}_k$ and respectively $\{(c_k,d_k)\}_k$ for $P_i(A)$ $i=1,2$. That means A can be covered by $\{(a_k,b_k)\times (c_k,d_k)\}_k$. Therefore $$\mu(A)\leq\sum\mu((a_k,b_k))\mu((c_k,d_k))\leq \varepsilon\sum\mu((a_k,b_k))\leq \varepsilon^2$$ where $\varepsilon>0$ was arbitrarily and $\mu$ is your 'measure' (Lebesgue, Jordan?). Since this argument works both for finite as for countable covers, you may use Lebesgue or Jordan zero sets.