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 $U$.
I asked about the relevance of openness of $U$ here. Now my question is about the proof which relies on Lebesgue's theorem which assumes a bounded set. We're not given that $U$ is bounded. However, we deduce that $\text{supp} f$ is bounded.
As with the previous question, I'm actually not sure if the $\tilde{f}$ is the one in Section 13 or the one in Subsection 23.1.
Case 1: $\tilde{f}$ is the one in Subsection 23.1:
Is $\tilde{f}$ actually meant to be $\tilde{g}$, the extension of $g$ by zero, where $g$ is the restriction of $f$ to its support, i.e. $g = f|_{\text{supp} f}$?
It seems the proof takes cases between $U$ and $U^c$ for the original $f$'s extension by zero to $\tilde{f}$. I think the proof should take cases between $\text{supp} f$ and $(\text{supp} f)^c$ for the restricted $f$'s (the restricted $f$ is $g = f|_{\text{supp} f}$), extension by zero to $\tilde{g}$. This is how I do it:
Method 1: For $\tilde{g}: \mathbb R^n \to \mathbb R$, the definition for $\tilde{g}$ to be continuous at a point $x \in \mathbb R^n$ is that for all neighborhoods $V_{\tilde{g}(x)}$ of $\tilde{g}(x)$ in $\mathbb R$, there is a neighborhood $W_x$ of $x$ in $\mathbb R^n$ such that $\tilde{g}(W_x) \subseteq V_{\tilde{g}(x)}$.
For $x \in {\text{supp} f}$, I don't think we need to use that definition. We have $\tilde{g}(x)=g(x)$, so $\tilde{g}$ is continuous at $x$ if $g$ is continuous at $x$, which is the case because restrictions of continuous functions are continuous. (I actually had a long answer which relied on openness of $U$ but then realized that I might be mistaken. Thus, the question of the relevance of the openness of $U$ remains.)
For $x \in (\text{supp} f)^c$, we have that $\tilde{g}(x) = 0$. For any neighborhood $V_{\tilde{g}(x)} = V_0$ of $\tilde{g}(x) = 0$ in $\mathbb R$, we choose $W_x = (\text{supp} f)^c$ to get $\tilde{g}(W_x) = \tilde{g}((\text{supp} f)^c) = \{0\} \subseteq V_0$.
Method 2: I think $\tilde{g}$ and $\tilde{f}$ are identical: They are both $0$ on $U^c$, both $f$ on $\text{supp} f$ and I think can be shown to be identical on $U$ and identical $(\text{supp} f)^c$. Therefore, we use Lebesgue's theorem on $A = \text{supp} f$ and show $\tilde{g}$ is continuous by showing $\tilde{f}$ is continuous.
Case 2: $\tilde{f}$ is the one in Section 13:
Well, kind of the same question. $U$ isn't given to be bounded, so how are we using Lebesgue's theorem?
- I'm thinking we extend the original $f$ from $U$ to all of $\mathbb R^n$ to get $\tilde{f}$ and then restrict $\tilde{f}$ to $\text{supp} f$ or something.
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.
