[proofplan]
Throughout the proof, we write $\mathbb N := \{1,2,3,\dots\}$ for the set of positive integers. We first prove the identity for indicator functions, where it is exactly the defining equality $\mu(T^{-1}(A))=\mu(A)$. Linearity then gives the identity for nonnegative simple functions. For a general nonnegative measurable function, we approximate from below by an increasing sequence of nonnegative simple functions and pass to the limit by monotone convergence. Finally, a complex-valued $L^1$ function is decomposed into the positive and negative parts of its real and imaginary parts, and the nonnegative case is applied to each component.
[/proofplan]
[step:Prove the identity for indicators using measure preservation]
Let $A\in\mathcal E$, and define the indicator function $\mathbb 1_A:E\to\{0,1\}$ by
$\mathbb 1_A(x)=1$ if $x\in A$ and $\mathbb 1_A(x)=0$ if $x\notin A$. Since $T:E\to E$ is measurable, the preimage $T^{-1}(A)$ lies in $\mathcal E$, and for every $x\in E$ we have $\mathbb 1_A(T(x))=\mathbb 1_{T^{-1}(A)}(x)$. Therefore
\begin{align*}
\mathbb 1_A\circ T=\mathbb 1_{T^{-1}(A)}
\end{align*}
and the definition of the integral of an indicator function gives
\begin{align*}
\int_E \mathbb 1_A\circ T\,d\mu(x)=\mu(T^{-1}(A))=\mu(A)=\int_E \mathbb 1_A\,d\mu(x).
\end{align*}
[guided]
The first case to check is the one where the function records membership in a measurable set. Fix $A\in\mathcal E$, and define the indicator function $\mathbb 1_A:E\to\{0,1\}$ by setting $\mathbb 1_A(x)=1$ when $x\in A$ and $\mathbb 1_A(x)=0$ when $x\notin A$.
Because $T$ is measurable, the preimage $T^{-1}(A)$ belongs to $\mathcal E$. For each $x\in E$, the value $(\mathbb 1_A\circ T)(x)$ equals $1$ exactly when $T(x)\in A$, which is exactly when $x\in T^{-1}(A)$. Hence $\mathbb 1_A\circ T=\mathbb 1_{T^{-1}(A)}$.
The integral of an indicator is the measure of its set, so
\begin{align*}
\int_E \mathbb 1_A\circ T\,d\mu(x)=\mu(T^{-1}(A)).
\end{align*}
Since $T$ is measure-preserving, $\mu(T^{-1}(A))=\mu(A)$. Therefore
\begin{align*}
\int_E \mathbb 1_A\circ T\,d\mu(x)=\mu(A)=\int_E \mathbb 1_A\,d\mu(x).
\end{align*}
This proves the desired identity for indicators.
[/guided]
[/step]
[step:Extend the identity to nonnegative simple functions by linearity]
Let $s:E\to[0,\infty)$ be a nonnegative simple measurable function. Choose $m\in\mathbb N$, coefficients $a_1,\dots,a_m\in[0,\infty)$, and measurable sets $A_1,\dots,A_m\in\mathcal E$ such that
\begin{align*}
s=\sum_{j=1}^m a_j\mathbb 1_{A_j}.
\end{align*}
Then
\begin{align*}
s\circ T=\sum_{j=1}^m a_j(\mathbb 1_{A_j}\circ T).
\end{align*}
By finite linearity of the integral for nonnegative simple functions and the indicator case,
\begin{align*}
\int_E s\circ T\,d\mu(x)=\sum_{j=1}^m a_j\int_E \mathbb 1_{A_j}\circ T\,d\mu(x)=\sum_{j=1}^m a_j\int_E \mathbb 1_{A_j}\,d\mu(x)=\int_E s\,d\mu(x).
\end{align*}
[guided]
The next class of functions is built from indicators by finite sums. Let $s:E\to[0,\infty)$ be a nonnegative simple measurable function. By definition of a simple measurable function, there are $m\in\mathbb N$, coefficients $a_1,\dots,a_m\in[0,\infty)$, and measurable sets $A_1,\dots,A_m\in\mathcal E$ such that
\begin{align*}
s=\sum_{j=1}^m a_j\mathbb 1_{A_j}.
\end{align*}
Composing this finite sum with $T$ gives
\begin{align*}
s\circ T=\sum_{j=1}^m a_j(\mathbb 1_{A_j}\circ T),
\end{align*}
because composition distributes over finite pointwise linear combinations. For each $j\in\{1,\dots,m\}$, the indicator case gives
\begin{align*}
\int_E \mathbb 1_{A_j}\circ T\,d\mu(x)=\int_E \mathbb 1_{A_j}\,d\mu(x).
\end{align*}
Using finite linearity of the integral for nonnegative simple functions, we therefore obtain
\begin{align*}
\int_E s\circ T\,d\mu(x)=\sum_{j=1}^m a_j\int_E \mathbb 1_{A_j}\circ T\,d\mu(x)=\sum_{j=1}^m a_j\int_E \mathbb 1_{A_j}\,d\mu(x)=\int_E s\,d\mu(x).
\end{align*}
This proves the invariance identity for every nonnegative simple measurable function.
[/guided]
[/step]
[step:Pass from simple functions to nonnegative measurable functions]
Let $f:E\to[0,\infty]$ be measurable. By the standard simple-function approximation theorem for nonnegative [measurable functions](/page/Measurable%20Functions), there exists a sequence $(s_n)_{n\in\mathbb N}$ of nonnegative simple measurable functions $s_n:E\to[0,\infty)$ such that
\begin{align*}
0\le s_n(x)\le s_{n+1}(x)
\end{align*}
for every $n\in\mathbb N$ and every $x\in E$, and
\begin{align*}
\lim_{n\to\infty}s_n(x)=f(x)
\end{align*}
for every $x\in E$. Since $T:E\to E$ is measurable, each $s_n\circ T:E\to[0,\infty)$ is a nonnegative simple measurable function, and
\begin{align*}
0\le s_n(T(x))\le s_{n+1}(T(x))
\end{align*}
for every $n\in\mathbb N$ and every $x\in E$. Also,
\begin{align*}
\lim_{n\to\infty}s_n(T(x))=f(T(x))
\end{align*}
for every $x\in E$.
By the [Monotone Convergence Theorem](/theorems/509) applied to $(s_n)$ and to $(s_n\circ T)$, and by the simple-function identity from the previous step,
\begin{align*}
\int_E f\circ T\,d\mu(x)=\lim_{n\to\infty}\int_E s_n\circ T\,d\mu(x)=\lim_{n\to\infty}\int_E s_n\,d\mu(x)=\int_E f\,d\mu(x).
\end{align*}
This proves the asserted equality for every nonnegative measurable $f:E\to[0,\infty]$.
[guided]
We now move from simple functions to arbitrary nonnegative measurable functions. Let $f:E\to[0,\infty]$ be measurable. The standard simple-function approximation theorem for nonnegative measurable functions gives a sequence $(s_n)_{n\in\mathbb N}$ of nonnegative simple measurable functions
\begin{align*}
s_n:E\to[0,\infty)
\end{align*}
such that the sequence increases pointwise to $f$:
\begin{align*}
0\le s_n(x)\le s_{n+1}(x)
\end{align*}
for every $n\in\mathbb N$ and every $x\in E$, and
\begin{align*}
\lim_{n\to\infty}s_n(x)=f(x)
\end{align*}
for every $x\in E$.
Composing with $T$ preserves this pointwise order because $T(x)\in E$ for every $x\in E$. Thus
\begin{align*}
0\le s_n(T(x))\le s_{n+1}(T(x))
\end{align*}
for every $n\in\mathbb N$ and every $x\in E$. Also, taking the pointwise limit after evaluating at $T(x)$ gives
\begin{align*}
\lim_{n\to\infty}s_n(T(x))=f(T(x)).
\end{align*}
Equivalently, the sequence $(s_n\circ T)_{n\in\mathbb N}$ increases pointwise to $f\circ T$.
Now we apply the Monotone Convergence Theorem twice. It applies to $(s_n)$ because the functions are nonnegative measurable and increase pointwise to $f$. It applies to $(s_n\circ T)$ because the functions are nonnegative measurable and increase pointwise to $f\circ T$. Therefore
\begin{align*}
\int_E f\,d\mu(x)=\lim_{n\to\infty}\int_E s_n\,d\mu(x)
\end{align*}
and
\begin{align*}
\int_E f\circ T\,d\mu(x)=\lim_{n\to\infty}\int_E s_n\circ T\,d\mu(x).
\end{align*}
For each $n\in\mathbb N$, the simple-function identity proved above gives
\begin{align*}
\int_E s_n\circ T\,d\mu(x)=\int_E s_n\,d\mu(x).
\end{align*}
Substituting this identity into the limiting formula yields
\begin{align*}
\int_E f\circ T\,d\mu(x)=\lim_{n\to\infty}\int_E s_n\circ T\,d\mu(x)=\lim_{n\to\infty}\int_E s_n\,d\mu(x)=\int_E f\,d\mu(x).
\end{align*}
This proves the nonnegative measurable case.
[/guided]
[/step]
[step:Decompose a complex integrable function into nonnegative parts]
Let $f:E\to\mathbb C$ be measurable and assume $f\in L^1(E,\mathcal E,\mu)$. Let $\operatorname{Re}:\mathbb C\to\mathbb R$ and $\operatorname{Im}:\mathbb C\to\mathbb R$ denote the real-part and imaginary-part maps. Define $u,v:E\to\mathbb R$ by $u=\operatorname{Re}\circ f$ and $v=\operatorname{Im}\circ f$; these maps are measurable because $\operatorname{Re}$ and $\operatorname{Im}$ are continuous and $f$ is measurable. Define the positive and negative parts of $u$ and $v$ by $u^+(x)=\max\{u(x),0\}$, $u^-(x)=\max\{-u(x),0\}$, $v^+(x)=\max\{v(x),0\}$, and $v^-(x)=\max\{-v(x),0\}$ for $x\in E$. Then
\begin{align*}
f=u^+-u^-+i v^+-i v^-.
\end{align*}
Since $|u|\le |f|$ and $|v|\le |f|$, the functions $u^+,u^-,v^+,v^-$ are measurable and satisfy $0\le u^+,u^-,v^+,v^-\le |f|$. Because $f\in L^1(E,\mathcal E,\mu)$, each of $u^+,u^-,v^+,v^-$ has finite integral with respect to $\mu$. Applying the nonnegative case to these four functions gives
\begin{align*}
\int_E u^\pm\circ T\,d\mu(x)=\int_E u^\pm\,d\mu(x)
\end{align*}
and
\begin{align*}
\int_E v^\pm\circ T\,d\mu(x)=\int_E v^\pm\,d\mu(x).
\end{align*}
For every $x\in E$,
\begin{align*}
(f\circ T)(x)=(u^+\circ T)(x)-(u^-\circ T)(x)+i(v^+\circ T)(x)-i(v^-\circ T)(x).
\end{align*}
Hence
\begin{align*}
|(f\circ T)(x)|\le (u^+\circ T)(x)+(u^-\circ T)(x)+(v^+\circ T)(x)+(v^-\circ T)(x).
\end{align*}
The function $f\circ T$ is measurable as a composition of measurable maps, and the displayed bound shows that it is dominated by an integrable nonnegative function. Therefore $f\circ T\in L^1(E,\mathcal E,\mu)$. By linearity of the complex [Lebesgue integral](/page/Lebesgue%20Integral),
\begin{align*}
\int_E f\circ T\,d\mu(x)=\int_E u^+\circ T\,d\mu(x)-\int_E u^-\circ T\,d\mu(x)+i\int_E v^+\circ T\,d\mu(x)-i\int_E v^-\circ T\,d\mu(x).
\end{align*}
Using the four equalities above,
\begin{align*}
\int_E f\circ T\,d\mu(x)=\int_E u^+\,d\mu(x)-\int_E u^-\,d\mu(x)+i\int_E v^+\,d\mu(x)-i\int_E v^-\,d\mu(x)=\int_E f\,d\mu(x).
\end{align*}
This proves the $L^1$ complex-valued case and completes the proof.
[/step]
