Prove that if the derivative $f'(x)$ of a function exists on the measurable set $E$, then $f'(x)$ is measurable on $E$.

Question

Prove that if the derivative $f'(x)$ of a function exists on the measurable set $E$, then $f'(x)$ is measurable on $E$.

We are told to only consider 1 dimensional spaces,that f is a measurable function in one variable.that is, f is a measurable function in one variable.

Following is my solution, I am not sure whether I am on the right track.Can someone have a look? Many thanks, I can explain further if I should. Thanks in advance!

Answer 1

Sorry for the late answer

It seems to me there are two flaws in your proof.

Actually, for $x \in E$, $h_n(x)$ needs not tend towards $f'(x)$. Consider for instance$$f : x \longmapsto\begin{cases} 1&\text{if $x \le 0$ or $x \notin \mathbb{Q}$}\\ 1+x^2&\text{otherwise} \end{cases}$$

Then you can check that $f$ is differentiable on $]-\infty,0]$ with derivative $0$, but $f$ is discontinuous on $]0,+\infty[$. Thus $g : x \mapsto\begin{cases} 1&\text{if $x \le 0$}\\ 0&\text{if $x>0$} \end{cases} \ \ $ so $h_n(0) \underset{n \to + \infty}{\longrightarrow} + \infty$, and so $\big(h_n(x)\big)_{n \ge 1}$ does not converge to $f'(0) = 0$. Note that, using a Cantor set, you can adapt this so that $h_n$ does not converge to $f'(x)$ on a subset of $E$ with positive measure.

Moreover, your last claim at the end of the proof (with some set being countable, or having zero measure) also seems false to me.

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More constructively, I think that you can follow the standard reasoning to prove your result.

For $n \in \mathbb{N}^*$, denote $g_n : x \mapsto n \left ( f\big( x + \frac{1}{n} \big) - f(x) \right)$. We consider that $f'$ is defined on $E$. Then $(g_n)$ converges pointwise to $f'$ on $E$.

We take an closed set $C \subset \mathbb{R}$, and we want to prove that $f'^{-1} (C)$ is measurable.

For every $x \in f'^{-1}(C)$, for $k \in \mathbb{N}^*$, $d \big( g_k(x), C \big) \le d \big( g_k(x), f'(x) \big) \underset{k \to + \infty}{\longrightarrow} 0$. Now for $n \in \mathbb{N}^*$, we denote the open set $C_n = \left \{ y \in \mathbb{R},\ d(y, C) < \frac{1}{n} \right \}$. Using the previous inequality, we get that for all $n \in \mathbb{N}^*$, there exists $m>n$ such that $g_m(x) \in C_n$, so $x \in g_m^{-1} (C_n)$. Moreover, $x \in E$. Hence $$ f'^{-1}(C) \subset \bigcap \limits_{n \in \mathbb{N}^*} \bigcup \limits_{m \ge n} E \cap g_m^{-1} (C_n)$$

Now for the other inclusion, with $x$ in the RHS set, you have a sequence of integers $(m_n)$ such that for all $n > 1$, $m_n \ge n\ $ and $\ d \big ( g_m(x), C \big) \le \frac{1}{n}$, so using pointwise convergence, as $x \in E$, $d \big( f'(x), C \big) = 0$, and $C$ is closed, so $f'(x) \in C$, and thus $x \in f'^{-1}(C)$.

Hence $f'^{-1}(C) = \bigcap \limits_{n \in \mathbb{N}^*} \bigcup \limits_{m \ge n} E \cap g_m^{-1} (C_n)$ and every set $g_m^{-1}(C_n)$ is measurable because $C_n$ is Borel and $g_m$ is measurable (because $f$ is measurable). As $E$ is also measurable, we get an intersection of unions of measurable sets, so $f'^{-1}(C)$ is measurable.

Finally, we get that $f'$ is measurable on $E$.

Answer 2

If $f:[a, b]\to\mathbb{R}$ arbitrary function, then the $$\text{upper derivative }\overline{D}f(x) := \limsup_{y\to x} \frac{f(x)-f(y)}{x-y}$$ and $$\text{lower derivative }\underline{D}f(x) := \liminf_{y\to x} \frac{f(x)-f(y)}{x-y}$$ are always measurable. Thus the set $E = \{x\in [a, b] : \overline{D}f(x) = \underline{D}f(x)\in\mathbb{R}\}$ is measurable, and so is $f':E\to\mathbb{R}$ since $f'\restriction_E = \overline{D}f\restriction_E$.

The proof that the upper and lower derivatives are measurable involves the following lemma:

Lemma. If $\mathcal{A}$ is an arbitrary family of closed non-degenerate intervals, then $\bigcup \mathcal{A}$ is measurable.

Proof: Let $\mathcal{K}$ the family of all closed non-degenerate intervals which are subsets of some $I\in\mathcal{A}$. Then $\mathcal{K}$ is a Vitali covering of $\bigcup \mathcal{A}$, so by Vitali's covering theorem there are intervals $I_1, I_2, ...\in \mathcal{K}$ such that $Z = \bigcup \mathcal{A}\setminus \bigcup_{n=1}^\infty I_n$ is of measure zero, so $\bigcup\mathcal{A} = Z\cup \bigcup_{n=1}^\infty I_n$ is measurable $\square$

Theorem. Upper and lower derivatives are measurable.

Sketch of proof: Consider some $r$ and sets $$E_n^k = \bigcup\left\{[c, d]\subseteq [a, b] : 0 < d-c\leq \frac{1}{k},\ \frac{f(d)-f(c)}{d-c}\geq r+\frac{1}{n}\right\},\ k, n\in\mathbb{N}.$$

From above lemma, the sets $E_n^k$ are measurable. Let $E = \{x\in [a, b] : \overline{D}f(x) > r\}.$ Its not hard to show that the equality $E = \bigcup_{n=1}^\infty \bigcap_{k=1}^\infty E_n^k$ is true (see comment after proof). Thus $E$ is measurable. Since $r$ was arbitrary, $\overline{D}f$ is measurable. Since $\underline{D}f = -\overline{D}(-f)$, the lower derivative is also measurable. $\square$

The inclusion $\bigcup_{n=1}^\infty\bigcap_{k=1}^\infty E_n^k\subseteq E$ in above proof might be trickier to see without some argument. To show it, note that if $x\in (c, d)$ then we can write $$\frac{f(d)-f(c)}{d-c} = \frac{d-x}{d-c}\cdot \frac{f(d)-f(x)}{d-x} + \frac{x-c}{d-c}\cdot\frac{f(x)-f(c)}{x-c}$$ where $\frac{d-x}{d-c}, \frac{x-c}{d-c}\geq 0$ and $\frac{d-x}{d-c}+\frac{x-c}{d-c} = 1$, which makes the above quotient a convex combination of quotients involving $x$.