I have read " Convexity, H.G.Eggleston, 1958'' and in page 91, theorem 43, it proves that a closed, bounded and convex set $\mathcal{X}$ is still closed bounded (and convex) under Steiner symmetrization with respect to one plane $\beta$ (or with respect to its normal direction $a$ and I denote the transformation as $S_a$ here). Here I want to know why $S_a(\mathcal{X} )$ is closed (thus compact in $\mathbb{R}^n$)?
More explicitly, in the book, it uses $\limsup_{i\to \infty}\lambda(p_i)\leq \lambda(p)$ as $i\to \infty$, where $\lambda(p)$ is the Lebgesgue measure of the segment whose mid-point at $p$ lies in $S_a(\mathcal{X})$, but I don't know why it is true. I know the Lebesgue measure is upper semi-continous with respect to the Hausdorff distance but does it works in this situation?
Recently I've noticed that this question can be done like this: By one of Blashke's theorem, for a sequence $(p_n)$ with $\lambda(p_n)\to \limsup_i \lambda(p_i)\ (n\to \infty)$, we have a limit $K$ of a subsequence of lines with respect to $p_i$, and we can simply argue that $K$ is contained in the line with respect to $p$ and since Lebesgue measure is upper semi-continous with respect to the Hausdorff distance, it's done. And a more easy way to prove it is to use the definition of Lebesgue measure and some knowledge of compactness.