Exercise 2.6

Solution. Let $0<𝜖<1$. Let $E_0=[0,1]$, and let $E_1$ be $E_0$ with a central interval of length $𝜖∕2$ removed. Inductively, from each $E_{n-1}$ remove centrally situated intervals of length $𝜖∕(2\cdot 4^{n-1})$; note we are removing $2^{n-1}$ of these. Thus, the intersection $K={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_1^{\infty }{E}_n$ has measure $1-{\sum <!--\; FUNCTION\; APPLICATION\; -->}_1^{\infty }𝜖∕2^n=1-𝜖$. $K$ is closed and bounded hence compact by construction.

Now to show $K$ is totally disconnected, it suffices by [?, Thm. 2.47] to show that for every $x,y\in K$, there exists some $z\in R^1$ such that $x<z<y$ but $z\notin K$. But this is obvious, for $(x,y)$ contains some interval of length $𝜖∕(2\cdot 4^{n-1})$, for some $n$, which would have been removed at some stage in the construction.

Now we want to show if $v$ is lower semicontinuous and $v\le {\chi }_K$ for $K\subset R^1$ a totally disconnected compact set, then $v\le 0$. If $v$ is lower semicontinuous, then $v^{-1}((0,\infty ])$ must be open, and since $v\le {\chi }_K$, we see that $v^{-1}((0,\infty ])\subset K$. But $K$ is totally disconnected, so has empty interior, and so $v^{-1}((0,\infty ])=\varnothing$, that is, $v\le 0$. □