Small derivative and the measure of a set.

Question

Suppose that $f:\mathbb{R}\to\mathbb{R}$ is a differentiable function, and that on some interval $(a,b)$, $|f'|\leq1$. Is it true that for all measurable sets $E\subset(a,b)$, $\lambda(f(E))\leq\lambda(E)$? (Here $\lambda$ denotes Lebesgue measure.) The intuition is that a small derivative means that $f$ is "shrinking sets" in some sense.

Answer 1

I don't remember much about neasure theory. This is a proof when $E$ is a closed interval. It should be helpful, together with Chival's solution, that assumes this one proved.

So, let $E=[c,d]\subset(a,b)$. Since $f$ is differentiable, it is also continuous. Weierstrass' extreme value theroem says that $f$ obtains its maximum $M$ and its minimum $m$ on $[c,d]$. Say, $f(r)=M$ and $f(s)=m$, where $r$ and $s$ are in $[c,d]$. Continuity also guarantees that $f$ reaches every point in $[m,M]$.

Then, $\lambda(f(E))=M-m$ and mean value theroem says that there exists some $t\in[c,d]$ such that $f'(t)(r-s)=M-m$. Therefore

$$\lambda(f(E))=|f'(t)|\cdot|r-s|\le1\cdot|d-c|=\lambda(E)$$

Answer 2

Yes, the (absolute value/ norm) Jacobian , here the derivative gives you a measure of the scaling, or change of volume of sets resulting from a transformation. Since ||J(f)||=|f'(x)|<1 , the sets are non-expanding.