Is the derivative of differentiable function $f:\mathbb{R}\to\mathbb{R}$ measurable on $\mathbb{R}$?

Question

Suppose we have a bounded differentiable function $f:\mathbb{R}\to[a,b]\subset\mathbb{R}$. Hence $f$ is continuous and measurable (in terms of standart Lebesgue measure) on $\mathbb{R}$.

I want $f$ to have bounded measurable derivative $f'$. Derivative $f'$ doesn't have to be bounded by default, so I add additional constrain on $f$, saying that $f$ is bounded differentiable function with bounded derivative. Now do I have to explicitly postulate that $f'$ is measurable, or do I have it automatically from already mentioned conditions?

Answer 1

Assuming merely that $f$ is differentiable, $f'$ is measurable because $\{x\mid f'(x)>L\}$ is Borel for every $L$:

Since we know $f$ to be differentiable everywhere, the $x$ for which $f'(x)>L$ are those where $\dfrac{f(x+1/n)-f(x)}{1/n}> L$ for all sufficiently large $n$. And once we choose an $n$, the set $$ A_n = \left\{ x \;\Big|\; \frac{f(x+1/n)-f(x)}{1/n} > L \right\} $$ is open, so $$\{x\mid f'(x)>a \} = \bigcup_{k=1}^\infty \bigcap_{n=k}^\infty A_n $$ is Borel.

Answer 2

I have read the only answer posted to this question, but I am afraid it is not right:

Of course, if the derivative of $f$ at a point $x$ is greater than a real number $L$, there will be a natural number $N$ such that $$\forall n \in \mathbb{N}, n \geq N \implies \frac{f(x + \frac{1}{n}) – f(x)}{\frac{1}{n}} > L$$

(it follows from the fact that the sequence $\{\frac{f(x + \frac{1}{n}) – f(x)}{\frac{1}{n}}\}_{n = 1}^{\infty}$ is convergent and its limit is $f^{\prime}(x)$, which is such that $f^{\prime}(x) – L$ is positive). However, the converse is not necessarily true. If it is assumed that there exists a natural number $N$ such that $$\forall n \in \mathbb{N}, n \geq N \implies \frac{f(x + \frac{1}{n}) – f(x)}{\frac{1}{n}} > L,$$

then the limit of the (convergent) sequence $\{\frac{f(x + \frac{1}{n}) – f(x)}{\frac{1}{n}}\}_{n = 1}^{\infty}$ is greater or equal than $L$, and, as far as I am concerned, not enough information is given to state that such limit cannot be equal to $L$.

This idea, however, can be useful still:

Let $L$ be a real number. If $f$ is differentiable at $x$, then $$f'(x) > L \iff \exists N \in \mathbb{N}: \inf_{k \geq N}\{\frac{f(x + \frac{1}{n}) – f(x)}{\frac{1}{n}}\} > L.$$

Indeed. For any natural number $k$, let $f_{k}$ be the function defined as $$x \mapsto f_{k}(x) = \frac{f(x + \frac{1}{k}) – f(x)}{\frac{1}{k}}.$$

If the derivative of $f$ at $x$ is greater than $L$, then the limit of the (convergent) sequence $\{f_{k}(x)\}_{k = 1}^{\infty}$ will be greater than than $L$. But, since such sequence is convergent, its limit equals its lower limit (also known as inferior limit), so $$\lim_{n \to \infty} \inf_{k \geq n}\{f_{k}(x)\} > L.$$

Consequently, for the positive number $f^{\prime}(x) – L$, there exists a natural number $N$ such that $$\forall n \in \mathbb{N}, n \geq N \implies f^{\prime}(x) – (f^{\prime}(x) – L) < \inf_{k \geq n}\{f_{k}(x)\} < f^{\prime}(x) + (f^{\prime}(x) – L)\,$$

and it follows that $N$ is such that $$\inf_{k \geq N}\{f_{k}(x)\} > L.$$

The converse is also true: if there exists a natural number $N$ such that the infimum of the set $\{f_{k}(x) \mid k \geq N\}$ is greater than $L$, then $$\forall n \in \mathbb{N}, n \geq N \implies f_{n}(x) \geq \inf_{k \geq N}\{f_{k}(x)\},$$

so $$\lim_{n \to \infty}f_{n}(x) \geq \inf_{k \geq N}\{f_{k}(x)\},$$

and, since it is assumed that such infimum is greater than $L$, the result follows.

Thus, $$\{x \in \mathbb{R} \mid f^{\prime}(x) > L\} = \bigcup_{N \in \mathbb{N}} \{x \in \mathbb{R} \mid \inf_{k \geq N}\{f_{k}(x)\} > L\}.$$

Now, since $\{f_{k}\}_{k = 1}^{\infty}$ is a sequence of measurable functions (it follows from the fact that any linear combination of measurable functions is measurable and the fact that $f$ is continuous), the function $\inf_{k \geq N}\{f_{k}\}$ is also measurable. It implies that all the sets of the (countable) family $\{\{x \in \mathbb{R} \mid \inf_{k \geq N}\{f_{k}(x)\} > L\} \mid N \in \mathbb{N}\}$ are measurable. The last equality of sets then shows that $\{x \in \mathbb{R} \mid f^{\prime}(x) > L\}$ is a countable union of measurable sets, thus measurable.