I was reading the introductary chapter on the book Geometric integration theory (Birkhäuser, Cornerstones series 2008, Krantz & Parks) and stumbled upon the following theorem/proposition about Steiner Symmetrization:
Proposition 1.6.14. If T is a compact subset of $R^N$ and if S is obtained from T by Steiner symmetrization, then S is compact.
Proof: Let V be an $(N - 1)$ dimensional vector subspace of $R^N$, and suppose that S is the result of Steiner symmetrization of T with respect to V. It is clear that the boundedness of T implies the boundedness of S. To see that S is closed, consider any sequence of points $p_1, p_2,...$ in S that converges to some point p. Each $p_i$ lies in a line $l_i$ perpendicular to V, and we know that $dist(p_i,V) \le \frac{1}{2}L^1(l_i \cap S) = \frac{1}{2}L^1(l_i \cap T)$ We also know that the line perpendicular to V and containing p must be the limit of the sequence of lines $l_1, l_2,...$ . Further, we know that $dist(p,V) = \lim \limits_{i \to \infty} dist(p_i, V)$. The inequality $\lim \limits_{i} sup\ L^1(l_i \cap T) \le L^1(l \bigcap T)$ would allow us to conclude that $dist(p,V) = \lim \limits_{i \to \infty} dist(p_i, V) \le \lim \limits_{i} sup\ L^1(l_i \bigcap T) \le L^1(l \cap T)$ and thus that $p \in S$.
All good so far
To obtain the inequality $\lim \limits_{i} sup\ L^1(l_i \cap T) \le L^1(l \bigcap T)$, we let $q_i$ be the vector parallel to V that translates $l_i$ to $l$, and we apply lemma 1.6.13, with N replaced by 1 and with $l$ identified with $\mathbb{R}$, to the sets $A_i = \tau_{q_i}(l_i \cap T)$, which are the translates of the sets $l_i \cap T$. We can take $A=l \cap T$, because $T$ is closed.
The lemma it is refering to is the following:
Lemma 1.6.13. Fix $0 < M < \infty$ If $A$ and $A_1,A_2,...$ are closed subsets of $\mathbb{R^N} \cap \mathbb{\bar B}(0,M)$ such that $\bigcap_{i_0 = 1}^{\infty} \overline{[\bigcup_{i = i_0}^{\infty} A_i]} \subseteq A$, then $\lim \limits_{i} sup\ L^N(A_i) \le L^N(A).$
The problem I have regarding proposition 1.6.14 is that I don't know how we can show that $\bigcap_{i_0 = 1}^{\infty} \overline{[\bigcup_{i = i_0}^{\infty} A_i]} \subseteq A$ for the sets in question. I tried looking for other sources, but couldn't find anything.
Edit: If someone knows of an alternative proof of the theorem that is fine aswell