The support of the Schwarz rearrangement of a compactly supported function

Question

Let $E \subset \mathbb{R}^N$ a measurable set of finite measure. We define $E^\ast$, the symmetric rearrangement of $E$ to be $$ E^\ast := B_R(0), R = \left(\frac{|E|}{|B|}\right)^{\frac{1}{N}}. $$ Notice that $|E| = |E^\ast|$. Consider $f : \mathbb{R}^N \to \mathbb{R}$ a Borel measurable function that vanish at infinity, that is, for every $t > 0$, the sets $\{|f| > t\}$ are of finite measure. Then its spherically symmetric and decreasing rearrangement, $f^\ast$, is defined on $\mathbb{R}^N$ by $$ f^\ast(x) = \int_0^\infty \chi_{\{|f| > t\}^\ast} (x) dt. $$ Now assume $f$ has support in the annulos $A_{\epsilon_1, \epsilon_2} = \{x \in \mathbb{R} : \epsilon _1< |x| < \epsilon_2\}$. I am trying to discover what should be the support of $f^\ast$. I know how to prove that, at least, $\text{supp}(f^\ast) \subset B_{\epsilon_2}$. To see this, notice first that, for all $t > 0$ $$ \{|f| > t\} := \{x \in \mathbb{R}^N : |f(x)| > t\} = \{x \in \mathbb{R}^N : \epsilon_1 <|x| <\epsilon_2 \text{ and } |f(x)| > t\}, $$ because it is impossible to happen $|f(x)| > t$ for $|x| \leq \epsilon_1$ or $|x| \geq \epsilon_2$. Thus $$ \{|f| > t\} \subset B_{\epsilon_2}. $$ So, $|\{|f| > t\}| \leq |B_{\epsilon_2}| = |B| \epsilon_2^{N} $. Therefore, if $|x| \geq \epsilon_2$ then $$ |B| |x|^N \geq |B| \epsilon_2^N \geq |\{|f| > t\}|, \quad \forall t > 0. $$ From which we obtain $$ |x| \geq \left( \frac{|\{|f| > t\}|}{|B|}\right)^{\frac{1}{N}} $$ In other words, $x \not\in \{|f| > t\}^\ast = B_R, \quad R = \left( \frac{|\{|f| > t\}|}{|B|}\right)^{\frac{1}{N}}.$ Finally $$ \chi_{\{|f| > t\}^\ast}(x) = 0, \quad \forall t > 0 $$ and $$ f^\ast(x) = \int_0^{+\infty} \chi_{\{|f| > t\}^\ast}(x) dt = 0 ,\quad |x| \geq \epsilon_2. $$ Now I wonder if I can conclude that $f^\ast(x) = 0$ for $|x| \leq \epsilon_1$, but I don't know how to prove it.