I find the Steiner Symmetrization is very useful in proving that the Hausdorff measure coincide with Lebesgue in the Euclidean space. However, I never saw anybody mention that the Steiner Symmetrization takes a measurable set to a measurable one. To know how this is done, see here.
Here is also a related problem.
For completeness, I'll sketch the proof in the book by Evans and Gariepy that Thomas mentioned. Let's consider symmetrization in the direction of the vector $e_n=(0,0,\dots, 1)$. Let $A^*$ be the Steiner symmetrization of $A$. So, the lines used in the symmetrization process are of the form $L_b =\{b+te_n :t\in\mathbb{R}\} $ where $b\in\mathbb{R}^{n-1}$.
Since the characteristic function $\chi_A$ is measurable, by Fubuni-Tonelli theorem the function $f(b) = \int_{L_b}\chi_A d\mathcal L^1$ is measurable with respect to the $(n-1)$-dimensional measure on $\mathbb{R}^{n-1}$. The set $A^*$ is bounded between two copies of the graph of $\frac12 f$. It remains to observe that for a measurable $g$, the set $$ \{x:x_n < g(x_1,\dots,x_{n-1}) \} $$ is measurable, being the pre-image of open half-space under the measurable map $x\mapsto x - g(x_1,\dots,x_{n-1})e_n$.
