Let $C \in \mathbb{R}^n$ be a Jordan mensurable set of measure zero. Show that $\int_A \chi_C = 0$, for every closed rectangle $A \supset C$. Let $C \in \mathbb{R}^n$ be a Jordan mensurable set of measure zero. Show that $\int_A \chi_C = 0$, for every closed rectangle $A \supset C$. A subset $A$ of $\mathbb{R}^n$ has measure 0 if for every $\epsilon > 0$ there is a cover $\\{U_1, U_2,... \\}$ of A by closed rectangles such that $\sum_{i=1}^{\infty}v(U_i) < \epsilon.$ A bounded set C is called Jordan-measurable if for every closed rectangle A such that $C \subset A$ we have that $\chi_C: A \to \mathbb{R}$ (given by $\chi_C(x) = 1$ if $x \in C$, and $\chi_C(x) = 0$ if $x \in A-C$) is integrable. In this case $v(C) = \int_A \chi_C$ * * * The only thing I know is that by hypothesis we have that $\int_A \chi_C = v(C)$. I am really bad in analysis. Any help would be great.

We can write A as $C\cup (A - C)$ and we can express the integral as $\int_A \chi_{C} = \int_C \chi_{C} \ + \int_{A-C}\chi_{C}$.

By the definition of $\chi_{C}$, we have that $\chi_{C}=0$ on $A-C$, so the second integral of the sum is zero. Then $\int_A \chi_{C} = \int_C \chi_{C} = v(C)$. Which is zero because C has measure zero (or alternatively we can show by the definition that $v(C)$ is less than $\epsilon$ for every $\epsilon > 0$