Exercise 1.6.1 (Differentiability and measurability)

Question

Let $F:[a,b]	o\mathbb{R}$ be everywhere differentiable function. Show that \begin{enumerate}[label=(oman*)]
\item $F$ is continuous
\item $F'$ is measurable.
\end{enumerate}    Show that if $F$ is almost everywhere differentiable (instead of everywhere differentiable) then
    \begin{enumerate}[label=(oman*)]
\item $F$ need not be continuous
\item $F'$ is measurable (equal to an almost everywhere defined measurable function).
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\roman*)]

\item By definition, we have
	$$\forall x \in [a,b]: \quad \lim_{y\to x, y\neq x} \frac{F(y)-F(x)}{y-x}$$
	exists. In other words, for an arbitrary $x \in [a,b]$ we have
	$$\forall \epsilon >0 \quad \exists \delta>0 \quad \forall y\in [a,b]/\{x\}: \quad |y-x|\leq \delta \implies \left|\frac{F(y)-F(x)}{y-x} - F'(x)\right| \leq \epsilon$$
	This implies (cf. \href{../analysis_I}{Analysis I}, Proposition 10.1.7):
	\begin{align*}
	&\left|\frac{F(y)-F(x)}{y-x} - F'(x)\right| \leq \epsilon\
	&\implies \left|F(y)-F(x) - F'(x)(y-x)\right| \leq \epsilon |y-x|\
	&\implies |F(y)-F(x)| - |F'(x)(y-x)| \leq \epsilon |y-x|\
	&\implies |F(y)-F(x)| \leq \epsilon |y-x| + |F'(x)(y-x)|\
	&\implies |F(y)-F(x)| \leq \epsilon \cdot \delta + |F'(x)|\delta\
	&\implies |F(y)-F(x)| \leq (\epsilon + |F'(x)|) \cdot \delta\
	\end{align*}
	Setting $\delta':=\min\{\delta_\epsilon, \frac{\epsilon}{\epsilon + |F'(x)|}\}$ we obtain
	$$\forall \epsilon >0 \quad \exists \delta=\delta' \quad \forall y\in [a,b]/\{x\}: \quad |y-x|\leq \delta' \implies \left|F(y)-F(x)\right| \leq \epsilon$$
	Since our choice of $x\in [a,b]$ was arbitrary, $F$ must be continuous everywhere.

\item By \href{./exercise_1-3-8}{Exercise 1.3.8}(iv) if $F'$ is a pointwise limit of Lebesgue measurable functions, then it must be Lebesgue measurable itself. By definition (and axiom of choice) in each $x\in [a,b]$, $F'$ is the limit of
	$$F'(x) = \lim_{n\to\infty} \frac{F(x_n)-F(x)}{x_n-x}$$
	where $(x_n)_{n\in\mathbb{N}}\to x$ with $x_n\in [a,b]/\{x\}$. By Analysis I, Proposition 9.4.9, $g_n(x) = (F(x_n)-F(x))/(x_n-x)$ must be continuous as a combination of continuous functions. Since continuous functions are automatically Lebesgue measurable by \href{./exercise_1-3-8}{Exercise 1.3.8}(i), we are done.

\item Consider a simple linear function that explodes in the middle of the domain:
	$$f:[0,1]\to [0,1], \quad f(x) = \left\{\begin{array}{ll} x &\text{ if } x\in \left[0,\frac{1}{2}\right)\cup \left(\frac{1}{2},1\right]\[5pt] 1 &\text{ if } x=\frac{1}{2} \end{array}\right.$$
	
	\begin{figure}
	\includegraphics[width=0.50\textwidth]{tao_measure_exercise_1-6-1.png}
	\caption{Graph of a function exploding at $x=1/2$.}
	\end{figure}
	
	Then this function is not continuous. It is differentiable at any $x\in [0,0.5)\cup (0.5,1]$ with $f'(x) = 1$. At the point $x=0.5$, however, we have
	$$\lim_{x\to 0.5 -} \frac{x-1}{x-0.5} = +\infty \quad \lim_{x\to 0.5 +} \frac{x-1}{x-0.5} = -\infty $$
	(easy to verify that there is no upper bound, or see: \url{https://www.wolframalpha.com/input/?i=lim_{x \to 0.5} \frac{x-1}{x-0.5}}). Thus, $f'$ is strictly almost everywhere differentiable.

\item Follows directly from the definition.
	
\end{enumerate}

Exercise 1.6.1 (Differentiability and measurability)

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Exercise 1.6.1 (Differentiability and measurability)

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