[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*}[/step]

[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: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*}[/step]

[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: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 $f:E\to[0,\infty]$.[/step]

[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: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]