Multiple integral on a zero measured set

Let A \subseteq \mathbb{R}^n be a Jordan measurable set of zero measure. Prove that \bigintsss \limits_{A} 1 \, {\rm d} \bar{x}= 0.

Solution

Since \mu(A)=0 there exists a sequence U_n of closed rectangles of \mathbb{R}^n such that A \subseteq \bigcup \limits_{n} U_n and \forall \epsilon>0 it is \sum \limits_{n} \mathcal{V} \left( U_n \right)< \epsilon where \mathcal{V}(U_n) is the volume of the rectangle U_n. Then foreach \epsilon>0

    \begin{align*} 0 &\leq \int \limits_{A} 1 \, {\rm d} \bar{x} \\ &\leq \int \limits_{\bigcup \limits_{n} U_n} 1 \, {\rm d} \bar{x} \\ &\leq \sum_{n} \int \limits_{U_n} 1\, {\rm d}\bar{x} \\ &= \sum_{n} \mathcal{V} \left ( U_n \right )\\ &< \epsilon \end{align*}

 

Read more

Leave a Reply