Let $U\subseteq \mathbb{R}^n$ and $K\subseteq U$ be compact. Suppose $f:U\to \mathbb{R}$ is continuous and $f(x)=0$ if $x\not\in K$. Show that $\int_K f=\not\int_U f$
I don't know how to make an extended integral sign in latex but this $\int_U f$ should be an extended integral.
I believe I want to make use of this theorem, If $S\subseteq \mathbb{R}^n$ is jordan measureable then $\mu(S)=\int_S \chi_S$. With $S=\{ f(x): x\in U\setminus K\}$ in this case.
My attempt at a solution was since $f$ is continuous $f$ is integrable on $U$ and the set $S$ above is jordan measureable as there are 0 discontinuities. Since $f(x)=0, \forall x\in U\setminus K$, then $\mu(S)=0=\int_{U\setminus K} f$ and by additivity of domain $$\int_K f+\int_{U\setminus K}f=\int_U f$$
$$\iff$$ $$\int_K f+0=\int_U f$$$$\iff$$
$$\int_K f=\int_U f$$
I don't think this is correct though because I didn't use compactness anywhere, I also am not sure if my use of this theorem is correct. My guess is maybe that compactness was supposed to be used as integrability is typically defined on rectangles? And maybe I've somewhat missed justifying that the integral $\int_{U\setminus K} f$ exists?
Actually looking at this now, I'm not sure what justification I have for the extended integral $\int_U f$ existing which is another problem. As the theorem I have which says:
Let $A\subseteq\mathbb{R}^{n+1}$ be an open set and $f:A\to \mathbb{R}$ a continuous function. If $(K_i)_{i=1}^\infty$ is any compact exhaustion of A, then:
$\text{(extened)}\int_A f$ exists if and only if the sequence $\int_{K_i} \vert f \vert$ is bounded
Best Answer
The extended integral is a generalization of the improper Riemann integral to sets in $\mathbb{R}^n$ ($n > 1$). There is a very specific definition. For a nonnegative function $f$ the extended integral over $U$ is defined as
$$\int_U f = \sup_D \int_D f$$
where $D$ ranges over every compact rectifiable subset of $U$. For a general function it is defined as the difference of the extended integrals of the positive and negative parts, $f^+$ and $f^-$,
$$\int_Uf = \int_U f^+ - \int_Uf^-$$
In this case, we have the very strong conditions that $f$ is continuous on $U$ as well as $f(x) = 0$ for $x \in U \setminus K$. Consequently, $f$ must vanish on $\partial K$ since for any point $p \in \partial K$ we have $\lim_{x \to p, x \notin K}f(x) = 0.$ The Riemann integral of $f$ over $K$ must exist even if the boundary $\partial K$ does not have zero content.
Also continuity of $f$ implies that $|f|$, $f^+ = (|f|+f)/2$, and $f^- = (|f| - f)/2$ are all continuous and vanish on $\partial K$.
Taking any increasing and exhausting sequence of compact rectifiable sets $C_k$ where $K \subset C_1 \subset C_2 \subset \ldots$ we have
$$\int_{C_i} f^+ = \int_Kf^+ + \int_{C_i \setminus K} f^+ = \int_K f^+$$
which implies
$$\int_K f^+ = \lim_{i \to \infty} \int_{C_i} f^+ = \int_U f^+$$
The same result holds for $f^-$ and, therefore,
$$\int_Kf = \int_U f$$