Question: Fix \(\alpha\) with \(0<\alpha<1\) and repeat our "middle thirds" construction for the Cantor set except that now, at the \(n\) th stage, each of the \(2^{n-1}\) open intervals we discard from $[0,1]$ is to have length \((1-\alpha) 3^{-n}\). (We still want to remove each open interval from the "middle" of a closed interval in the current level - it is important that the closed intervals that remain turn out to be nested.) The limit of this process, a set that we ul name \(\Delta_{\alpha}\). is called a generalized Cantor set and is very much like the ordinary Cantor set. Note that \(\Delta_{\alpha}\) is uncountable, compact, nowhere dense, and so on, but has nonzero outer measure. Indeed, check that \(m^{*}\left(\Delta_{a}\right)=\alpha\). (See Chapter Two for an example.) [Hint: You only need upper estimates for \(m^{*}\left(\Delta_{\alpha}\right)\) and \(\left.m^{*}\left(\Delta_{a}^{c}\right) .\right]\)

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