A compactly supported continuous function on an open subset of $\mathbb R^n$ is Riemann integrable. What is the relevance of openness in the proof?

1. In the proof of Prop. 23.4, by the extended function, I mean the extension by zero defined in Section 23.1. If $U$ is not open, the extended function $\overline{f}$ need not be continuous at a boundary point of $U$. For example, if $U$ is the closed interval $[-1,1]$ in the real line and $f(x) = x^2$ for $x$ in $U$, then $\overline{f}$ agrees with $f$ on $U$, but $\overline{f}$ is not continuous at $+1$ or $-1$.

Since $U$ is open in $\mathbb{R}^n$, an open subset of $U$ is also an open subset of $\mathbb{R}^n$. Let $p$ be a point of $U$. Then $\overline{f}$ agrees with $f$ on an open neighborhood of $p$ in $\mathbb{R}^n$, so $\overline{f}$ is continuous at $p$.

2. To Apply Lebesgue's theorem, which requires the domain of $f$ to be bounded, let $K$ be the support of $f$, a bounded set. Note that $\overline{f}$ is also the zero extension of the restriction of $f$ to $K$. Since $\overline{f}$ is continuous, $f$ is Riemann integrable on $K$. This implies that $f$ is Riemann integrable on $U$, since $f$ is zero on $U - K$.

Because on ugly enough domains, even constant functions are non-integrable (see, for instance, this question. Assuming $U$ to be open stops that from happening.