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.
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Math 6210 • Real Analysis
Spring 2016 • Question 1