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$\textbf{Definition :}$ A set $X\in \mathbb{R}^n$ is Jordan measurable if exists $R$ a rectangle such that $A\subseteq R$ and the function $\chi_{A} : R \rightarrow \mathbb{R}$ is integrable.

Consider $Y\subseteq X$ where $X,Y$ are Jordan measurable. So exists a rectangle $R$ such that $X\subseteq R$ and $\chi_{X}$ is integrable. Then $\chi_{Y}$ is integrable?. Or equivalently : if $X$ is Jordan measurable then for all $R$ rectangle such that $X\subseteq R$ we have that $\chi_{X}$ is integrable?.

Intuitively, I would say yes, but I cannot prove it.

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The answer is yes and it can be proved using Lemma 13.1 in Analysis on Manifolds by Munkres.

If $f$ is a bounded function that vanishes outside of $R\cap R'$ where $R,R'$ are rectangles in $\mathbb{R}^n$, then $\int_R f = \int_{R'} f$, where one integral exists if and only if the other does.

In this case, take $f = \chi_X$ and let $R$ and $R'$ be any rectangles such that $X \subset R$ and $X \subset R'$. Since $X \subset R \cap R'$ it follows that $\int_R \chi_X = \int_{R'} \chi_X$.

RRL
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