## Problem #6.5, MA598R, Summer 2009, Weigel

$\text{Let } f: [0,1] \rightarrow R \text{ be } C^1 \text{, and let}$

$Z = \{x\in [0,1] : f'(x) = 0\}$

$\text{Show that } m(f(Z)) = 0$

Note: In the original problem statement, it uses $\mu$, but I didn't want to have to keep typing it, so I used m to denote the measure.

First, we show a lemma:

$\text{Let } f\in L^1([0,1])\text{ and let } F(x) = \int_{0}^{x} f(t) dt\text{. If } E \subset [0,1]\text{ is measurable, then:}$

$\text{(a) } F(E) \text{ is measurable.}$

$\text{(b) } m(F(E)) \leq \int_{E} |f(t)| dt$

Proof: (a) F is continuous, since $f \in L^1([0,1])$, and F is absolutely continuous, so for some compact interval $I\subset [0,1]$, there is a compact interval $I' \subset R$ such that $F(I) = I'$. So F(I) is measurable. Hence F(G) is measurable for $G\in F_{\sigma}$. But for E measurable,

$\exist K\text{, } m(K) = 0 \text{, and } G \in F_{\sigma} \text{ such that } E = K \bigcup G$

$\text{But F takes null sets to null sets, so }$

$F(G\bigcup K) = F(G) \bigcup F(K)$

$\text{Since } F(K) \text{ is of measure 0, hence measurable, and } F(G) \text{ is measurable, so too is the union.}$

Proof: (b) Since F is continuous, for $I\subset [0,1]$ compact, there exists $J \subset R$, $F(I)) = J$

$\text{Hence } m(F(I)) = m([\alpha, \beta]) = \beta - \alpha$

$\exist c \in [0,1] \text{, } \exist d \in [0,1] \text{, such that}$ $F(c) = \alpha \text{ and } F(d) = \beta$

$m(F(I)) = | \left(\int_{c}^{d} f(x) dx\right) | \leq \int_{c}^{d} |f(x)| dx \leq \int_{a}^{b} |f(x)| dx$

So it's true for intervals. But any measurable set can be represented as the union of an $F_\sigma$ set and a zero measure set.

$\text{Let } G\in F_\sigma \text{ and let F be such that } m(F) = 0 \text{, and further let } E = F \bigcup G$

Hence, using the fact that absolutely continuous functions take null sets to null sets:

$m(F(E)) = m(F(F\bigcup G)) = m(F(G)) = \int_{G} |f(x)| dx \leq \int_{a}^{b} |f(x)| dx$

Now, another lemma:

$\text{Lemma: } f\in C^1([0,1]) \Rightarrow f \in AC([0,1])$

Proof:

Given $\epsilon > 0$

Consider: $\sum_{i=1}^{N} |f(x_{i}) - f(x_{i-1})| = \sum_{i=1}^{N} f'(\zeta_{i}) (x_{i} - x_{i-1}) \leq M \sum_{i = 1}^{N} (x_{i} - x_{i-1}) \leq M\delta$

This works, because f' is continuous on a compact set, hence f' is bounded, and the first equality comes about from the Mean Value Theorem.

So choosing $\delta < \frac{\epsilon}{M}$, we get the desired result, namely that $f\in AC([0,1])$.

Now, we just use the lemma proved above (namely part (b)):

$0 \leq m(f(Z)) \leq \int_{Z} f' = \int_{Z} 0 = 0$

Written by Nicholas Stull

## Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy Francisco Blanco-Silva