Bogachev change of variables theorem.

Question

I am reading the Bogachev's change of variable theorem proof. And I am stuck in an argument contained in it:

Let $F:U\subset \mathbb{R}^n\to \mathbb{R}^n$ an injective $C¹$ map. Given $\epsilon>$ there exist $\delta>0$ s.t $\forall x,y \in U,~\|x-y\|<\delta$ implies $$ F(x)-F(y)=F^{'}(x)(y-x)+r(x,y) $$ where $\|r(x,y)\|\leq \epsilon \|x-y\|$.

Here is where I am stucked:

Claim: Let $Q$ be a cube with center $x_0$ and diameter less than $\delta.$ If $|\det F'(x_0)|\leq \sqrt{\epsilon}$ then we have that $$ \lambda (L(Q))-\lambda(F(Q))\leq \sqrt{\epsilon}\lambda (Q) $$ where $L(x)=F^{'}(x_0)(x-x_0)+F(x_0)$ and $\lambda$ is the $n$-dimensional lebesgue measure.

Can someone give me some help?

Comment: I think the argument pass over to give some inferior bound for $\text{diam}( F(U))$ by using in some way the inequality $\|F(x)-F(y)\|\geq \|F^{'}(x_0)(x-y)\|-\|r(x,y\|$.

Answer 1

I don't know enough measure theory but since $\lambda(L(Q))=\det L \cdot \lambda(Q)$, you could try to prove that $\det L \le \frac{\lambda(F(Q))}{\lambda(Q)} + |\det F'(x_0)|$.

Answer 2

$\lambda(L(Q)) - \lambda(F(Q) \leq \lambda(L(Q)) = \lambda(F'(x_0)Q + (F(x_0) - F'(x_0)x_0)) = \lambda(F'(x_0)Q) = |\det F'(x_0)| \lambda(Q) \leq \sqrt{\epsilon}\lambda(Q).$

Where we use shift invariance as well as the scaling property under linear maps of Lebesgue measure, proofs of which can be seen in Folland, Real Analysis, Theorem 2.42 and 2.44 resp.