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Question

Math 6210 • Real Analysis

Spring 2016 • Question 1

For any subset \(A\) of \(\mathbb R\) we define its Lebesgue outer measure \[ \lambda ^*(A)=\inf \left \{ \sum _{i=0}^{\infty }|b_i-a_i| \;\middle |\; A\subset \bigcup _{i=0}^{\infty }(a_i,b_i) \right \}. \] Recall that given an outer measure \(\phi ^*:\mathcal P(X)\to [0,\infty ]\), we say a set \(A\) is \(\phi ^*\)-measurable if for all \(Y\in \mathcal P(X)\) we have \[ \phi ^*(Y)=\phi ^*(Y\cap A)+\phi ^*(Y\setminus A). \] Show that if \(\lambda ^*(\partial E)=0\), then \(E\) is \(\lambda ^*\)-measurable.

Raw LaTeX

For any subset \(A\) of \(\mathbb R\) we define its Lebesgue outer
measure
\[
\lambda^*(A)=\inf\left\{
\sum_{i=0}^{\infty}|b_i-a_i|
\;\middle|\;
A\subset\bigcup_{i=0}^{\infty}(a_i,b_i)
\right\}.
\]
Recall that given an outer measure
\(\phi^*:\mathcal P(X)\to[0,\infty]\), we say a set \(A\) is
\(\phi^*\)-measurable if for all \(Y\in\mathcal P(X)\) we have
\[
\phi^*(Y)=\phi^*(Y\cap A)+\phi^*(Y\setminus A).
\]
Show that if \(\lambda^*(\partial E)=0\), then \(E\) is
\(\lambda^*\)-measurable.