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0 1 $begingroup$ Another fun qualifying exam practice question! Suppose $E$ is a given set and $O_n$ is the open set defined as $O_n = left{x: operatorname{dist}(x,E) < frac1nright}$ a) Show that if $E$ is compact then $m(E) = lim_{n to infty} m(O_n)$ What I have so far: Define $B_k = bigcap^k_{i=0}O_n = O_k$ Then $B_{k+1} subseteq B_k$ Also $lim_{n rightarrow infty}B_k = bigcap^infty_{i=0}O_n = E$ (Since $E$ is closed and bounded? I'm not really sure why this is true!) So $B_k$ converges downward to $E$ . Since measures are monotone, $m(B_k)$ converges downward to $m(E)$ So $lim_{k rightarrow infty}m(B_k) =lim_{k rightarrow infty}m(bigcap^k_{n=1}O_n) =lim_{k rightarrow infty}m(O_k)=m(E)$ Basically I'm wondering how compactness plays into this argument. b) Show that the conclu