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
