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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

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  • $\begingroup$ What is the extended integral? $\endgroup$ – Calvin Khor Feb 13 at 17:57
  • $\begingroup$ I believe it means the same as improper integral. This is the definition: Let $U\subseteq \mathbb{R}^n$ be an open set, and $f:U\to \mathbb{R}$ be non negative continuous function. Let $\tilde K=\{K\subseteq U: \text{K is compact}, \text{Jordan measureable}\}$. Define the extended integral of $f$ over $U$ to be $\int_A f:= \underset{K\in \tilde K}{\sup} \int_K f$ if the supremum exists. $\endgroup$ – AColoredReptile Feb 13 at 18:09
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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,

$$\int_Kf = \int_U f$$

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  • $\begingroup$ Alright thanks. I assume $C_k$ is a compact exhaustion of $U$? Also this comment "The Riemann integral of f over K must exist even if the boundary ∂K does not have zero content." I'm not sure what issue you're saying won't matter. Since $K$ is compact and $f$ is continuous don't we have integrability? Why does the boundary of $K$ matter? $\endgroup$ – AColoredReptile Feb 13 at 20:33
  • $\begingroup$ @AColoredReptile: You're welcome. (1) You are correct $\cup C_k = U$. (2) Suppose that $f$ is continuous on $U$ but does not vanish outside $K$ and $K$ is not Jordan measurable (rectifiable). The integral $\int_K f$ is defined as $\int_Q f \chi_K$ where $Q$ is a rectangle containing $K$. It could happen that even though $f$ is continuous, the function $f \chi_K$ which vanishes outside $K$ is not and the integral does not exist. So I think the assumption that $f$ is vanishing outside $K$ implies integrability without further assumptions about $K$ -- which are not given here $\endgroup$ – RRL Feb 13 at 20:50
  • $\begingroup$ That is to say unless we know that for $p \in \partial K$ we have $f(x) \to 0$ as $x \to p$ from inside $K$, we would not be sure that $f\chi_K$ is continuous on $\partial K$. If not and $\partial K$ is not of content zero then the Riemann integral $\int_K f $ would not exist. If a hypothesis here is that $K$ is Jordan measurable, then all of this does not matter. $\endgroup$ – RRL Feb 13 at 20:57

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