Sangchul Lee's answer does a fantastic job of answering my question. But I wanted to share my own answer that I have come up with since, to provide an even more extreme example. A function $f : \mathbb{R} \rightarrow \mathbb{R}$ with the following properties:
- $f$ is continuous.
- $f$ is differentiable almost everywhere, with $f' \geq 0$ almost everywhere.
- $\int_a^b f' = +\infty$ for all $a, b \in \mathbb{R}$ with $a < b$.
Preliminary Constructions
Start by letting $C$ denote the Cantor function. Then define $D : \mathbb{R} \rightarrow \mathbb{R}$ as:
$$D(x) := \begin{cases}
x & \text{if $0 \leq x < \frac{1}{2}$} \\
x - C(2x-1) & \text{if $\frac{1}{2} \leq x \leq 1$} \\
0 & \text{otherwise}
\end{cases} $$
This gives an example of a continuous non-negative function, with a maximum output of $\frac{1}{2}$, which is almost everywhere differentiable with $D'=1$ almost everywhere in $[0,1]$, and $D(0) = D(1) = 0$.
So given any interval $[a,b]$, we can lay down multiple copies of rescaled versions of $D$ back to back to construct a function $D_{[a,b]} : \mathbb{R} \rightarrow \mathbb{R}$ satisfying:
- $D_{[a,b]}$ is non-negative, with $D_{[a,b]}(x) \leq b-a$ everywhere.
- $D_{[a,b]}$ is continuous, with $D_{[a,b]}(x) = 0$ whenever $x \not\in (a,b)$.
- $D_{[a,b]}$ is differentiable almost everywhere, with $D_{[a,b]}'(x) \geq \frac{1}{b-a}$ almost everywhere in $[a,b]$.
Main Construction
Let $\{ \frac{p_n}{q_n} : n \in \mathbb{N^{\geq 1}}\}$ be an enumeration of the rational numbers in their simplest form, with $q_n > 0$. Then for each $n \in \mathbb{N^{\geq1}}$ define the interval $I_n := [\frac{p_n}{q_n} - \frac{1}{2^{n+1}}\frac{1}{q_n^3}, \frac{p_n}{q_n} + \frac{1}{2^{n+1}}\frac{1}{q_n^3}]$.
Then finally define $f(x) := \sum_{n=1}^{\infty} D_{I_n}(x)$.
I claim that this $f$ satisfies the criteria stated at the beginning. Firstly, the convergence and continuity of $f$ is immediate by using the Weierstrass M-test with $\frac{1}{2^n}$.
Preliminaries for Differentiability
The set $A_1 := \{x \in \mathbb{R} : x \text{ does not have a Liouville-Roth irrationality measure of } 2\}$ is known to be a null set.
The set $A_2 := \bigcup_{n=1}^\infty \{x \in \mathbb{R} : D_{I_n} \text{ is not differentiable at } x\}$ is a countable union of null sets, and is therefore also null.
Finally, the set $B := A_1 \cup A_2$ is a null set.
Differentiability
Given that $r \not\in B$.
Since $r$ has an irrationality measure of $2$, there exists $N \in \mathbb{N}^{\geq 1}$ such that $|r - \frac{p_n}{q_n}| \geq \frac{1}{q_n^3}$ whenever $n \geq N$.
So if $n \geq N$, this means $D_{I_n}(r+h) = 0$ whenever $|h| \leq \frac{1}{q_n^3} - \frac{1}{2^{n+1}}\frac{1}{q_n^3}$, and hence:
$$\small{\frac{|D_{I_n}(r+h)|}{|h|} \leq \frac{\frac{1}{2^n} \frac{1}{q_n^3}}{\frac{1}{q_n^3} - \frac{1}{2^{n+1}} \frac{1}{q_n^3}}}$$
$$\small{\frac{|D_{I_n}(r+h)|}{|h|} \leq \frac{2}{2^{n+1} - 1}}$$
$$\small{\frac{|D_{I_n}(r+h)|}{|h|} \leq \frac{2}{2^n}}$$
Also clearly $r \not\in I_n$, meaning that $D_{I_n}(r) = 0$ and $D_{I_n}'(r) = 0$. Therefore:
$$\small{- \sum_{k=n+1}^{\infty} \frac{2}{2^k} \leq \frac{f(r+h) - f(r)}{h} - \frac{\sum_{k=1}^n D_{I_k}(r+h) - \sum_{k=1}^n D_{I_k}(r)}{h} \leq \sum_{k=n+1}^{\infty} \frac{2}{2^k}}$$
$$\small{-\frac{2}{2^{n}} \leq \frac{f(r+h) - f(r)}{h} - \sum_{k=1}^n \frac{D_{I_k}(r+h) - D_{I_k}(r)}{h} \leq \frac{2}{2^{n}}}$$
$$\small{- \frac{2}{2^{n}} \leq \liminf_{h \rightarrow 0}\frac{f(r+h) - f(r)}{h} - \sum_{k=1}^n D_{I_k}'(r) \leq \limsup_{h \rightarrow 0} \frac{f(r+h) - f(r)}{h} - \sum_{k=1}^n D_{I_k}'(r) \leq \frac{2}{2^{n}}}$$
$$\small{- \frac{2}{2^{n}} \leq \liminf_{h \rightarrow 0}\frac{f(r+h) - f(r)}{h} - \sum_{k=1}^{\infty} D_{I_k}'(r) \leq \limsup_{h \rightarrow 0} \frac{f(r+h) - f(r)}{h} - \sum_{k=1}^{\infty} D_{I_k}'(r) \leq \frac{2}{2^{n}}}$$
As $n$ is arbitrary, we conclude that $\lim_{h \rightarrow 0} \frac{f(r+h) - f(r)}{h} = \sum_{k=1}^{\infty} D_{I_k}'(r)$.
Hence $f$ is differentiable at $r$ with $f'(r) = \sum_{k=1}^{\infty} D_{I_k}'(r)$.
Integrability
Due to the above, clearly $\int_{I_n} f' \geq \int_{I_n} D_{I_n}' \geq 1$ for all $n \in \mathbb{N^{\geq 1}}$. Since any non-empty open interval contains infinitely many disjoint $I_{n_1}, I_{n_2}, ...$ we must have $\int_a^b f' = +\infty$ for all $a,b \in \mathbb{R}$ with $a < b$.
