# An estimate of measure of a set

I have a question about measure theory.

Let $B(r)$ denotes open ball of $\mathbb{R}^{d}$ centered at origin with radius $r>0$. We assume $N \subset \mathbb{R}^{d}$ satisfies the follwing: $$N\subset B(r) \text{ and } m(N) \ge \varepsilon m\bigl(B(r)\bigr),$$ where $m$ is the Lebesgue measure on $\mathbb{R}^{d}$ and $\varepsilon \in (0,1]$.

My question

Let $\varphi$ denotes one to one mapping from $B(r)$ to $\varphi\bigl(B(r)\bigr)$ such that $\text{Lip}(\varphi) \le M$, $\text{Lip}(\varphi^{-1})\le M$.

Can we have lower estimate of $m\bigl(\varphi(N)\bigr)$ ?

My attempt

By a result of geometric measure theory, \begin{align*} m\bigl(\varphi(N)\bigr)=\int_{N}|J_{\varphi}(x)|\,dx, \end{align*} where $J_{\varphi}(x)=\text{det}\left(\frac{\partial_{j} \varphi}{\partial x_k}\right)_{j,k=1}^{d}$. But I couldn't know how to estimate $|J_{\varphi}(x)|$...

• You have that each partial derivative has absolute value less than $M$. The determinant is homogeneous polynomial of degree $n$ in the partial derivatives with $n!$ terms. So a (rough) bound of the Jacobian is $M^n n!$ – guestDiego Oct 7 '16 at 13:46
• Thank you for your comment. I know upper bounds of $|J_{\varphi}|$. But what I want to know is lower bounds. – sharpe Oct 7 '16 at 13:49
• Since $\varphi^{-1}$ has the same properties, $|J^{-1}_{\varphi}|=|J_{\varphi^{-1}}|\leq M^n n!$. But $|J^{-1}_{\varphi}|=|J_{\varphi}|^{-1}$, so $|J_{\varphi}|^{-1}\leq M^n n!$ and $|J_{\varphi}|\geq (M^n n!)^{-1}$ – guestDiego Oct 7 '16 at 14:14
• Thanks for your kind explanation. I understood. – sharpe Oct 7 '16 at 14:17
• The singular values of $J_\varphi$ are also bounded by the Lipschitz constant (and their product is the determinant), so you don't need the $n!$. – Anthony Carapetis Oct 7 '16 at 14:40

Let $L = \min\{M,M^{-1}\}$. Your conditions allow $\varphi$ to send $B(r \cdot L)$ to an arbitrarily small neighborhood of the origin. (The origin is not special in this argument and can be replaced with some $x_0$ throughout.) If $N$ is in this small ball, $m(\varphi(N))$ may be made as close to zero as you like, leading to a lower bound of zero.
If $m(N)$ is too large to fit in this small ball, approach a lower bound via $m\left(N \cap (B(r) \setminus B(r \cdot L)\right)$. If $N$ were all in the spherical shell $B(r) \setminus B(r \cdot L)$, concentrated at the smallest possible radii, its image could be in a new small ball.
Your lower bound would then look like (possibly with detail errors, because I'm too lazy to actually find a pen right now): $$m(\varphi(N)) \geq \begin{cases} 0 ,& m(N) \leq m\left(B(r \cdot L)\right) \\ m(B(r_0)) ,& m(N) - m(B(r\cdot L)) \leq m\left(B(r\cdot L + r_0) \setminus B(r \cdot L) \right) \end{cases}$$
In the first case, $N$ is all sent to an arbitrarily small neighborhood of the origin because it is all close enough to the origin to be so. In the second case, as much of $N$ as possible is sent to an arbitrarily small neighborhood of the origin and the remainder is assumed concentrated on the inner surface of the spherical shell. That inner surface is too far away from the origin to be sent there, so the thin shell of $N$ is radially contracted to the smallest possible ball. (Since your function is injective, this ball must be adjusted to avoid the neighborhood of the origin containing the ball of points that could be sent to the arbitrarily small neighborhood of the origin. So in both cases, the inequality in the above display is actually strict.)