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? My book is An Introduction to Manifolds by Loring W. Tu. A proposition in Subsection 23.2 (Proposition 23.4) is

If a continuous function $f: U \to \mathbb R$ defined on an open subset $U$ of $\mathbb R^n$ has compact support, then $f$ is Riemann integrable on $U$.

What is the relevance of openness of $U$, specifically the proof given?


*

*One might argue it's about smooth extensions of $f$ like in Proposition 13.2 for $M = \mathbb R^n$ (which relies on openness of $U$ specifically through Exercise 13.1 although the openness of $U$ is not explicitly used based on the solution provided, but assuming I understand right, I think I know where the openness of $U$).

*But I think the $\tilde{f}$ is not the one in Proposition 13.2 but rather the one in Subsection 23.1 (see (1) below), which does not make use of any bumps and has no claims of smoothness or continuity, though the $\tilde{f}$ in Proposition 23.4 is of course proven continuous on $\mathbb R^n$ assuming $f$ is continuous on $U$, open in $\mathbb R^n$ in the same way that the $\tilde{f}$ in Proposition 13.2 is proven smooth on $M$ assuming $f$ is smooth on $U$, open in $M$.


*

*Part of why I think so is that it says in the proof that $\tilde{f}$ agrees with $f$ on $U$, but that's not even what the $\tilde{f}$ of Proposition 13.2 does: $\tilde{f}$ agrees with $f$ on an equal or a smaller neighborhood: This is not only explicit in Proposition 13.2 but also emphasized in a remark just before Proposition 13.2.


*Additionally, I think $\tilde{f}$ is meant to be an extension not of $f$ but rather of $f|_{\text{supp} f}$. I ask about this here. Perhaps this should be resolved first.

*One thing I have in mind about how $U$'s openness is relevant is some kind of rule like $f$ Riemann integrable on $U$ if $f$ is Riemann integrable on $\text{supp} f (\subseteq U)$.

*Another thing I had in mind is something like $\overline \int_U f = \overline \int_U f 1_{\text{supp} f} = \overline \int_{\text{supp} f} f$ and $\underline \int_U f = \underline \int_U f 1_{\text{supp} f} = \underline \int_{\text{supp} f} f$ by the overline and underline versions of Remark 10.7(i) in another book From Calculus to Cohomology by Ib Madsen and Jørgen Tornehave because $U$ is a manifold, because $U$ is open and because $\mathbb R^n$ is manifold...but I think this is overthinking not because this is from another book but because this uses geometry for an elementary analysis proposition.

Update: I think I know it now:


*

*My mistake in Method 1 is deducing $\tilde{g}$ continuous at $x \in \text{supp} f$ simply because $g$ is continuous at $x$. This is not necessarily true, but $\tilde{f}$ continuous at $x \in U$ from $f$'s continuity at $x$ BECAUSE $U$ is open.

*However, we must also prove $\tilde{f}$ and $\tilde{g}$ are identical. Much like with $T_eG$ and $L(G)$ in Lie algebras, each hand washes the other: $\tilde{g}$ is an extension from a bounded but closed set and therefore while we can use Lebesgue, we can't deduce continuity. $\tilde{f}$ is an extension from an open set but possibly unbounded set and therefore while we can't use Lebesgue, we can deduce continuity.

*Therefore, the proof of Proposition 23.4 is as follows:


*

*For continuous $f: U \to \mathbb R$ with $U$ open in $\mathbb R^n$ and with compact support, there exists a extension of $f$ by zero $\tilde{f}: \mathbb R^n \to \mathbb R$, i.e. $\tilde{f}(x) = f(x) \cdot 1_U(x) + 0 \cdot 1_{U^c}(x)$ which is a continuous extension because $f$ is continuous AND because $U$ is open (and compact support is not used here, I think). Lebesgue's theorem does not directly apply here because $U$ is not given to be bounded.

*For $g=f|_{\text{supp} f}$, the restriction of $f$ to its support, we have $\tilde{g}: \mathbb R^n \to \mathbb R$, $\tilde{g}(x) = g(x) \cdot 1_{\text{supp} f}(x) + 0 \cdot 1_{(\text{supp} f)^c}(x))$, the extension of $g$ by zero.

*Underemphasized in my opinion: Observe $\tilde{g}$ is not only an extension from a bounded set but also identical to $\tilde{f}$.

*In Lebesgue's theorem choose the bounded set $A = \text{supp} f$, and the bounded function to be $g$ which we can do because $\text{supp} f$ is bounded because $f$ has compact support and say that $\tilde{g}$ is continuous, not because $\tilde{g}$ is an extension of a continuous function, namely $g$, but because $\tilde{g}$ is identical to a continuous function, namely $\tilde{f}$.

*Oh, there's something missing here: We have shown $g=f|_{\text{supp} f}$ is Riemann integrable. How exactly do we get that the original $f$ is Riemann integrable? Intuitively, I guess this has something to do with $(\text{supp} f)^c \subseteq \{f=0\}$. At this point, I think geometry's/topology's role is done, and analysis has to take over. I guess $$\int_U f = \int_{\text{supp} f} f + \int_{U \cap (\text{supp} f)^c} f = \int_{\text{supp} f} f + 0 = \int_{\text{supp} f} f = \int_{U} f|_{\text{supp} f},$$ but that relies on $\int_U f$ being well-defined in the first place. Hence I suspect some equivalent definition of Riemann integrability or at least some property of Riemann integrability is that $f$ is Riemann integrable if $f_{\text{supp} f}$ is Riemann integrable.

(1) Actually, the solution to Exercise 13.1 makes use of something called "extended by zero", but I don't think that's defined until Section 23.
 A: *

*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 $p$.


*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 $U - K$.

A: 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.
