Measurability of the Derivative

Theorem (Measurability of the Derivative).

Let $f:(a,b)\to \mathbb{R}$ be a differentiable function on an open interval $(a,b)$. Then its derivative, $f^{\prime }:(a,b)\to \mathbb{R}$, is a measurable function with respect to the Borel $\sigma$-algebra.

Proof. Let $x\in (a,b)$. By the definition of the derivative,

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

We can express this limit using a sequence. For each integer $n\ge 1$, define a sequence of real numbers $h_n=\frac{1}{n}$. For any fixed $x\in (a,b)$, there exists an integer $N_x$ such that for all $n>N_x$, both $x+h_n$ and $x-h_n$ are in $(a,b)$. To avoid issues with the domain boundary for all $x$ simultaneously, we define a sequence of functions $f_n:(a,b)\to \mathbb{R}$ as follows. For each $n\in \mathbb{N}$, choose a sequence of non-zero numbers $\{h_k{\}}_{k=1}^{\infty }$ such that $h_k\to 0$ as $k\to \infty$. For instance, let $h_k=1/k$. For each $k$, define the function $g_k:(a,b)\to \mathbb{R}$ by:

$$g_k(x)=\frac{f(x+h_k)-f(x)}{h_k}$$

This function is well-defined for all $x\in (a,b-h_k)$ if $h_k>0$ or $x\in (a-h_k,b)$ if $h_k<0$. To simplify the domain, we can extend $f$ to be 0 outside of $(a,b)$ and define for each $n\in \mathbb{N}$ the function $f_n:\mathbb{R}\to \mathbb{R}$ by

$$f_n(x)=n\left(f\left(x+\frac{1}{n}\right)-f(x)\right)$$

where we consider $f$ extended to all of $\mathbb{R}$ by setting it to zero outside $(a,b)$.

A differentiable function is continuous. Since $f$ is differentiable on $(a,b)$, it is continuous on $(a,b)$. The extended function may have discontinuities at $a$ and $b$, but since these form a set of measure zero, its measurability is not affected. As a continuous function, $f$ is Borel measurable.

The function ${\varphi }_n(x)=x+1/n$ is continuous, hence measurable. The composition $f({\varphi }_n(x))=f(x+1/n)$ is a composition of measurable functions. Since $f$ is continuous, $f(x+1/n)$ is also measurable.

The function $f_n(x)$ is constructed from the measurable functions $f(x+1/n)$ and $f(x)$ through subtraction and scalar multiplication. The set of measurable functions is closed under linear combinations. Therefore, each function $f_n$ in the sequence is measurable.

By the definition of the derivative, for every $x\in (a,b)$, the sequence of values $\{f_n(x)\}$ converges to $f^{\prime }(x)$:

$$\underset{n\to \infty }{\lim }{f}_n(x)=f^{\prime }(x)$$

The function $f^{\prime }$ is the pointwise limit of a sequence of measurable functions $\{f_n\}$. A standard theorem of measure theory states that the pointwise limit of a sequence of measurable functions is itself measurable.

Therefore, $f^{\prime }$ is a measurable function. $\square$