[proofplan]
We prove that the superlevel sets of $f^*$ have exactly the same measure as the corresponding superlevel sets of $f$. The key point is the inverse relation between the decreasing function $d_f$ and its generalized inverse $f^*$: for $s\in(0,\mu(E))$, one has $f^*(s)>t$ exactly when $s<d_f(t)$. The only endpoint issue is handled by the right-continuity of $d_f$, which follows from continuity of finite measures from below.
[/proofplan]
[step:Establish monotonicity, right-continuity, and tail decay of the distribution function]
Let $M:=\mu(E)$. For $0\le a\le b$, the inclusion
\begin{align*}
\{x\in E:f(x)>b\}\subset \{x\in E:f(x)>a\}
\end{align*}
gives $d_f(b)\le d_f(a)$, so $d_f$ is nonincreasing.
Fix $t\ge0$. Let $(a_n)_{n\ge1}$ be the sequence in $[0,\infty)$ defined by
\begin{align*}
a_n:=t+\frac{1}{n}.
\end{align*}
For each $n\in\mathbb N$, define
\begin{align*}
A_n:=\{x\in E:f(x)>a_n\}\in\mathcal E.
\end{align*}
Since $a_n\downarrow t$, the sets $(A_n)_{n\ge1}$ are increasing and satisfy
\begin{align*}
\bigcup_{n=1}^{\infty}A_n=\{x\in E:f(x)>t\}.
\end{align*}
By continuity from below of the finite measure $\mu$,
\begin{align*}
\lim_{n\to\infty}d_f(a_n)=\lim_{n\to\infty}\mu(A_n)=\mu(\{x\in E:f(x)>t\})=d_f(t).
\end{align*}
We now upgrade this sequential limit to full right-continuity. Let $u>t$. Choose $n\in\mathbb N$ such that $u<t+1/n$. Since $d_f$ is nonincreasing and $t<u<t+1/n$, we have
\begin{align*}
d_f(t)\ge d_f(u)\ge d_f(t+1/n).
\end{align*}
Given $\delta>0$, choose $n\in\mathbb N$ such that $d_f(t)-d_f(t+1/n)<\delta$. Then every $u\in(t,t+1/n)$ satisfies
\begin{align*}
0\le d_f(t)-d_f(u)\le d_f(t)-d_f(t+1/n)<\delta.
\end{align*}
Hence $d_f(u)\to d_f(t)$ as $u\downarrow t$, so $d_f$ is right-continuous at $t$.
We also record the tail decay needed to define $f^*$ by an infimum over a nonempty set. For each $m\in\mathbb N$, define
\begin{align*}
B_m:=\{x\in E:f(x)>m\}\in\mathcal E.
\end{align*}
The sets $(B_m)_{m\ge1}$ are decreasing, and because $f$ takes values in $[0,\infty)$ rather than $[0,\infty]$,
\begin{align*}
\bigcap_{m=1}^{\infty}B_m=\varnothing.
\end{align*}
Since $\mu(B_1)\le\mu(E)<\infty$, continuity from above of finite measures gives
\begin{align*}
\lim_{m\to\infty}d_f(m)=\lim_{m\to\infty}\mu(B_m)=0.
\end{align*}
Therefore, for every $s\in(0,M)$, there exists $m\in\mathbb N$ such that $d_f(m)\le s$.
[guided]
We first isolate the only analytic fact about $d_f$ that will be needed later. Define $M:=\mu(E)$. If $0\le a\le b$, then every point satisfying $f(x)>b$ also satisfies $f(x)>a$. Hence
\begin{align*}
\{x\in E:f(x)>b\}\subset \{x\in E:f(x)>a\}.
\end{align*}
Applying the measure $\mu$ to this inclusion gives
\begin{align*}
d_f(b)=\mu(\{x\in E:f(x)>b\})\le \mu(\{x\in E:f(x)>a\})=d_f(a).
\end{align*}
Thus $d_f$ is nonincreasing.
Now fix $t\ge0$. To prove right-continuity, we approach $t$ from above. Define a sequence
\begin{align*}
a_n:=t+\frac{1}{n}
\end{align*}
for $n\in\mathbb N$, and define measurable sets
\begin{align*}
A_n:=\{x\in E:f(x)>a_n\}.
\end{align*}
Since $a_n\downarrow t$, the condition $f(x)>a_n$ becomes weaker as $n$ increases, so $(A_n)_{n\ge1}$ is an increasing sequence of measurable sets. Its union is exactly the strict superlevel set at $t$:
\begin{align*}
\bigcup_{n=1}^{\infty}A_n=\{x\in E:f(x)>t\}.
\end{align*}
Indeed, if $f(x)>t$, then $f(x)>t+1/n$ for all sufficiently large $n$; the reverse inclusion follows from $a_n>t$ for every $n$.
Continuity from below for measures applies to the increasing sequence $(A_n)_{n\ge1}$ and gives
\begin{align*}
\mu\left(\bigcup_{n=1}^{\infty}A_n\right)=\lim_{n\to\infty}\mu(A_n).
\end{align*}
Substituting the definitions of $A_n$ and $d_f$, we obtain
\begin{align*}
d_f(t)=\mu(\{x\in E:f(x)>t\})=\lim_{n\to\infty}d_f\left(t+\frac{1}{n}\right).
\end{align*}
We now justify the last sentence with the squeeze argument. Let $u>t$. Choose $n\in\mathbb N$ such that $u<t+1/n$. Because $d_f$ is nonincreasing and $t<u<t+1/n$, the values of $d_f$ satisfy
\begin{align*}
d_f(t)\ge d_f(u)\ge d_f(t+1/n).
\end{align*}
Now let $\delta>0$. From the convergence already proved, choose $n\in\mathbb N$ such that
\begin{align*}
d_f(t)-d_f(t+1/n)<\delta.
\end{align*}
Then every $u\in(t,t+1/n)$ satisfies
\begin{align*}
0\le d_f(t)-d_f(u)\le d_f(t)-d_f(t+1/n)<\delta.
\end{align*}
This proves $d_f(u)\to d_f(t)$ as $u\downarrow t$, which is exactly right-continuity at $t$.
We also need one well-definedness fact for the generalized inverse defining $f^*$. Namely, for each $s\in(0,M)$, the set
\begin{align*}
\{a\in[0,\infty):d_f(a)\le s\}
\end{align*}
must be nonempty. To prove this, define
\begin{align*}
B_m:=\{x\in E:f(x)>m\}
\end{align*}
for $m\in\mathbb N$. These sets are measurable and decreasing in $m$. Since $f$ is finite-valued with codomain $[0,\infty)$, no point $x\in E$ can satisfy $f(x)>m$ for every $m\in\mathbb N$, so
\begin{align*}
\bigcap_{m=1}^{\infty}B_m=\varnothing.
\end{align*}
The hypothesis $\mu(E)<\infty$ gives $\mu(B_1)<\infty$, so continuity from above applies to the decreasing sequence $(B_m)_{m\ge1}$. Hence
\begin{align*}
\lim_{m\to\infty}d_f(m)=\lim_{m\to\infty}\mu(B_m)=\mu(\varnothing)=0.
\end{align*}
Therefore, if $s\in(0,M)$, we may choose $m\in\mathbb N$ with $d_f(m)\le s$. This proves that the set over which $f^*(s)$ takes its infimum is nonempty.
[/guided]
[/step]
[step:Identify the superlevel set of $f^*$ with an interval]
Fix $t\ge0$ and $s\in(0,M)$. Define
\begin{align*}
S_s:=\{a\in[0,\infty):d_f(a)\le s\}.
\end{align*}
By the tail decay proved in the preceding step, $S_s$ is nonempty. Thus the quantity $f^*(s)=\inf S_s$ is the infimum of a nonempty subset of $[0,\infty)$. We prove
\begin{align*}
f^*(s)>t \quad \Longleftrightarrow \quad s<d_f(t).
\end{align*}
First suppose $f^*(s)>t$. If $d_f(t)\le s$, then $t\in\{a\in[0,\infty):d_f(a)\le s\}$, and therefore
\begin{align*}
f^*(s)=\inf\{a\in[0,\infty):d_f(a)\le s\}\le t,
\end{align*}
contradicting $f^*(s)>t$. Hence $s<d_f(t)$.
Conversely, suppose $s<d_f(t)$. By right-continuity of $d_f$ at $t$, there exists $\varepsilon>0$ such that
\begin{align*}
d_f(t+\varepsilon)>s.
\end{align*}
If $0\le a\le t+\varepsilon$, then monotonicity gives
\begin{align*}
d_f(a)\ge d_f(t+\varepsilon)>s.
\end{align*}
Thus no $a\in[0,t+\varepsilon]$ belongs to the set $\{r\in[0,\infty):d_f(r)\le s\}$. It follows that
\begin{align*}
f^*(s)=\inf\{r\in[0,\infty):d_f(r)\le s\}\ge t+\varepsilon>t.
\end{align*}
Therefore $f^*(s)>t$.
Consequently,
\begin{align*}
\{s\in(0,M):f^*(s)>t\}=\{s\in(0,M):s<d_f(t)\}.
\end{align*}
Since $0\le d_f(t)\le M$, the right-hand side is the interval $(0,d_f(t))$, with the convention that this interval is empty when $d_f(t)=0$.
[/step]
[step:Compute the Lebesgue measure of the identified interval]
From the preceding step,
\begin{align*}
\mathcal L^1(\{s\in(0,M):f^*(s)>t\})=\mathcal L^1((0,d_f(t))).
\end{align*}
By the defining length property of one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure),
\begin{align*}
\mathcal L^1((0,d_f(t)))=d_f(t).
\end{align*}
Finally, by the definition of $d_f$,
\begin{align*}
d_f(t)=\mu(\{x\in E:f(x)>t\}).
\end{align*}
Combining these equalities gives
\begin{align*}
\mathcal L^1(\{s\in(0,\mu(E)):f^*(s)>t\})=\mu(\{x\in E:f(x)>t\}).
\end{align*}
This proves the asserted equimeasurability for every $t\ge0$.
[/step]
