Let $\mathcal{R}^*(I)$ be the set of gauge integrable (generalized Riemann integrable) functions $f:I\to \mathbb{R}$ where $I = [a, b]$.
Say that a point $c\in [a, b]$ is a singular for $f\in \mathcal{R}^*(I)$ if for any neighbourhood $[c_0, c_1]$ of $c$ in $[a, b]$ we have $\int_{c_0}^{c_1} |f| = \infty$ interpreted as a Lebesgue integral (this makes sense as $f$ is Lebesgue measurable).
Call the set of all singular points or singularities of $f$ by $S_f$. This set is closed by definition.
Is it always the case that $|S_f| = 0$? As for example, one can construct $f$ such that $S_f = \{0\}\cup \{\frac{1}{n} : n\in\mathbb{N}_+\}$.
Answer: There exists a gauge integrable function $f:[a, b]\to\mathbb{R}$ with set of singularities of Lebesgue measure $\lambda$ iff $\lambda\in [0, b-a)$.
Note that topology-wise, the set of singularities of a gauge integrable function is nowhere dense. Hence small from perspective of topology.
A well-known theorem says that for every gauge integrable function $f:[a, b]\to\mathbb{R}$ there is a non-degenerate subinterval $J\subseteq [a, b]$ such that $f$ is Lebesgue integrable on $J$.
In the paper Henstock Integrable Functions are Lebesgue Integrable on a Portion by Zoltán Buczolich it was proven that any gauge integrable function $f:[a_1, b_1]\times\dots\times [a_n, b_n]\to\mathbb{R}$ contains a non-degenerate subrectangle $J$ on which $f$ is Lebesgue measurable.
In the article On Singularity of Henstock Integrable Functions by Peng-Yee Lee and Jitan Lu, they provide an example of a function $f:[0, 1]\to\mathbb{R}$ that has anti-derivative, thus is gauge integrable. Moreover, the set of singularities of $f$ is a generalized Cantor set of Lebesgue measure $1/2$. This can be generalized to generalized Cantor set of size $\lambda$ for any $0 < \lambda < 1$.