0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\epsilon > 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.