Fix $t>0$ et $y\in \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$. This is equivalent to say that
$$c_d|y|^d<|\{|f|>t\}|,\qquad \qquad c_d=|B(0,1)|. $$
Assume that
$$\int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds=0.$$
Then $y\not \in \{|f| > s\}^{*}$ for all $s\geq t$.
That is, for all $s\geq t$ we have
$$c_d|y|^d\geq |\{|f|>s\}| $$
The Fatou lemma implies
$$c_d|y|^d\geq |\{|f|\geq t\}|\geq |\{|f|>t\}|.$$
This is in contradiction with the fact that $c_d|y|^d<|\{|f|>t\}|$. Therefore we deduce $$\int_{t}^{\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds>0.$$
On the other hand, one can check that for every, $0<s< t$ one has $$\left\{x \in \mathbb{R}^n: |f(x)| > t \right\}\subset \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}$$ this entails that,
\begin{equation}\label{eq-inclu t-s}\tag{I}
\left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}\subset \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}~~~\textrm{for all $s\in (0,t)$}.
\end{equation}
this implies that,$$ \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) =1 ~~~s\in (0,t)$$
From definition of $f^{*}$, if $y\in \{|f|>t\}^*$ then we have
$$\begin{align*}
f^{*}(y) &:= \int_{0}^{+ \infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\
&= \int_{0}^{t} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds+ \int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\
& = \int_{0}^{t} ds+\int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds >t.
\end{align*}$$
Whence, $$\left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}\subset \left\{ x \in \mathbb{R}^n:f^{*}(x)> t \right\}.$$
Conversely, suppose, $y\notin \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}$. If $t\leq s$ then analogously to relation \eqref{eq-inclu t-s} we get
$$y\notin \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}.$$
That is for all $s\geq t$ we get $$\mathbf{1}_{\{ | f| > s\}^*}(y)=0$$
Hence necessarily for all $s>0$ such that
$ y\in \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}$ one has $0<s\leq t$. This means that,
$$\sup\left\{s>0 : y\in \left\{x \in \mathbb{R}^n: |f(x)| > s \right\}^{*}\right\}\leq t. $$
We then deduce that,
$$\begin{align*}
f^{*}(y) &:= \int_{0}^{+ \infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds\\
&= \int_{0}^{t} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds+ \underbrace{\int_{t}^{+\infty} \mathbf{1}_{\left\{ | f| > s \right\}^*}(y) ds}_{=0}\leq t
\end{align*}$$
that is $f^*(y)\leq t$ or that $y\notin \left\{x \in \mathbb{R}^n: f^*(x) > t \right\}$. We've just prove that,
\begin{equation}\label{eq}\tag{II}
\Bbb R^n\setminus \left\{x \in \mathbb{R}^n: |f(x)| > t \right\}^{*}\subset \Bbb R^n\setminus\left\{x \in \mathbb{R}^n: f^*(x) > t \right\}.
\end{equation}
Which ends the prove by taking the complementary.
