[proofplan] We prove both directions from the definition of the pushforward measure. In the forward direction, equality of measures gives equality of integrals first for indicator functions, then for simple functions, and finally for bounded [measurable functions](/page/Measurable%20Functions) by uniform approximation by simple functions. In the reverse direction, we test the assumed integral identity on indicators $\mathbb{1}_B$ of measurable sets $B \in \mathcal{B}$; this recovers exactly the defining formula $(T_{\#}\mu)(B)=\mu(T^{-1}(B))$ and hence proves equality of measures. [/proofplan]

[step:Derive the integral identity from equality of pushforward measures]Assume $T_{\#}\mu=\nu$. By definition of pushforward measure, for every set $B \in \mathcal{B}$, \begin{align*} \nu(B) = (T_{\#}\mu)(B) = \mu(T^{-1}(B)). \end{align*} Let $s: Y \to \mathbb{R}$ be a bounded $\mathcal{B}$-measurable [simple function](/page/Simple%20Function). Choose a finite representation \begin{align*} s = \sum_{k=1}^{m} a_k \mathbb{1}_{B_k}, \end{align*} where $m \in \mathbb{N}$, $a_k \in \mathbb{R}$, and $B_k \in \mathcal{B}$ for each $k \in \{1,\dots,m\}$. Since $T$ is $\mathcal{A}$-$\mathcal{B}$ measurable, $T^{-1}(B_k) \in \mathcal{A}$ for each $k$, and \begin{align*} s \circ T = \sum_{k=1}^{m} a_k \mathbb{1}_{T^{-1}(B_k)}. \end{align*} Using the definition of the integral of a simple function and the equality $\nu(B_k)=\mu(T^{-1}(B_k))$, we obtain \begin{align*} \int_Y s(y)\,d\nu(y) = \sum_{k=1}^{m} a_k \nu(B_k). \end{align*} \begin{align*} \sum_{k=1}^{m} a_k \nu(B_k) = \sum_{k=1}^{m} a_k \mu(T^{-1}(B_k)). \end{align*} \begin{align*} \sum_{k=1}^{m} a_k \mu(T^{-1}(B_k)) = \int_X s(T(x))\,d\mu(x). \end{align*} Now let $f: Y \to \mathbb{R}$ be bounded and $\mathcal{B}$-measurable. Since $T$ is measurable, $f \circ T: X \to \mathbb{R}$ is $\mathcal{A}$-measurable. For each $n \in \mathbb{N}$, choose a bounded $\mathcal{B}$-measurable simple function $s_n: Y \to \mathbb{R}$ such that \begin{align*} \sup_{y \in Y}|s_n(y)-f(y)| \leq \frac{1}{n}. \end{align*} The simple-function identity gives \begin{align*} \int_Y s_n(y)\,d\nu(y)=\int_X s_n(T(x))\,d\mu(x). \end{align*} Because $\nu(Y)=1$ and $\mu(X)=1$, the uniform error estimate gives \begin{align*} \left|\int_Y f(y)\,d\nu(y)-\int_Y s_n(y)\,d\nu(y)\right| \leq \int_Y |f(y)-s_n(y)|\,d\nu(y) \leq \frac{1}{n}. \end{align*} \begin{align*} \left|\int_X f(T(x))\,d\mu(x)-\int_X s_n(T(x))\,d\mu(x)\right| \leq \int_X |f(T(x))-s_n(T(x))|\,d\mu(x) \leq \frac{1}{n}. \end{align*} Letting $n \to \infty$ yields \begin{align*} \int_Y f(y)\,d\nu(y)=\int_X f(T(x))\,d\mu(x). \end{align*}[/step]

[guided]Assume $T_{\#}\mu=\nu$. The definition of the pushforward measure says that $T_{\#}\mu$ is the probability measure on $(Y,\mathcal{B})$ defined by \begin{align*} (T_{\#}\mu)(B)=\mu(T^{-1}(B)) \end{align*} for every $B \in \mathcal{B}$. Therefore the hypothesis $T_{\#}\mu=\nu$ gives, for every $B \in \mathcal{B}$, \begin{align*} \nu(B)=\mu(T^{-1}(B)). \end{align*} The integral identity is first checked on simple functions, because a simple function is a finite linear combination of indicator functions. Let $s: Y \to \mathbb{R}$ be a bounded $\mathcal{B}$-measurable simple function, and write \begin{align*} s=\sum_{k=1}^{m} a_k\mathbb{1}_{B_k}, \end{align*} where $m \in \mathbb{N}$, $a_k \in \mathbb{R}$, and $B_k \in \mathcal{B}$. Since $T$ is measurable, each preimage $T^{-1}(B_k)$ belongs to $\mathcal{A}$, so the composition $s\circ T: X \to \mathbb{R}$ is the simple function \begin{align*} s\circ T=\sum_{k=1}^{m} a_k\mathbb{1}_{T^{-1}(B_k)}. \end{align*} Using the definition of integration for simple functions, \begin{align*} \int_Y s(y)\,d\nu(y)=\sum_{k=1}^{m}a_k\nu(B_k). \end{align*} Since $\nu(B_k)=\mu(T^{-1}(B_k))$ for each $k$, \begin{align*} \sum_{k=1}^{m}a_k\nu(B_k)=\sum_{k=1}^{m}a_k\mu(T^{-1}(B_k)). \end{align*} The right-hand side is exactly the simple-function integral of $s\circ T$ over $X$: \begin{align*} \sum_{k=1}^{m}a_k\mu(T^{-1}(B_k))=\int_X s(T(x))\,d\mu(x). \end{align*} Now let $f: Y \to \mathbb{R}$ be bounded and $\mathcal{B}$-measurable. The composition $f\circ T$ is $\mathcal{A}$-measurable because $T$ is $\mathcal{A}$-$\mathcal{B}$ measurable and $f$ is $\mathcal{B}$-measurable. Since $f$ is bounded and measurable, it can be uniformly approximated by bounded measurable simple functions: for each $n \in \mathbb{N}$, choose $s_n: Y \to \mathbb{R}$ simple and $\mathcal{B}$-measurable such that \begin{align*} \sup_{y \in Y}|s_n(y)-f(y)|\leq \frac{1}{n}. \end{align*} Applying the simple-function identity to $s_n$ gives \begin{align*} \int_Y s_n(y)\,d\nu(y)=\int_X s_n(T(x))\,d\mu(x). \end{align*} Because both $\mu$ and $\nu$ are probability measures, the uniform error controls the integral error: \begin{align*} \left|\int_Y f(y)\,d\nu(y)-\int_Y s_n(y)\,d\nu(y)\right|\leq \frac{1}{n}. \end{align*} Likewise, since $|f(T(x))-s_n(T(x))|\leq 1/n$ for every $x \in X$, \begin{align*} \left|\int_X f(T(x))\,d\mu(x)-\int_X s_n(T(x))\,d\mu(x)\right|\leq \frac{1}{n}. \end{align*} Thus both sides of the simple-function identity converge to the corresponding integrals for $f$. Passing to the limit as $n\to\infty$ gives \begin{align*} \int_Y f(y)\,d\nu(y)=\int_X f(T(x))\,d\mu(x). \end{align*}[/guided]

[step:Recover equality of measures by testing on indicators] Conversely, assume that for every bounded $\mathcal{B}$-measurable function $f: Y \to \mathbb{R}$, \begin{align*} \int_Y f(y)\,d\nu(y)=\int_X f(T(x))\,d\mu(x). \end{align*} Let $B \in \mathcal{B}$ be arbitrary, and define the indicator function $\mathbb{1}_B: Y \to \mathbb{R}$ by setting $\mathbb{1}_B(y) = 1$ if $y \in B$ and $\mathbb{1}_B(y) = 0$ otherwise. The function $\mathbb{1}_B$ is bounded and $\mathcal{B}$-measurable, so it is an admissible [test function](/page/Test%20Function). Since $T$ is measurable, $T^{-1}(B)\in\mathcal{A}$ and \begin{align*} \mathbb{1}_B(T(x))=\mathbb{1}_{T^{-1}(B)}(x) \end{align*} for every $x \in X$. Applying the assumed identity to $f=\mathbb{1}_B$ gives \begin{align*} \nu(B)=\int_Y \mathbb{1}_B(y)\,d\nu(y)=\int_X \mathbb{1}_B(T(x))\,d\mu(x). \end{align*} \begin{align*} \int_X \mathbb{1}_B(T(x))\,d\mu(x)=\int_X \mathbb{1}_{T^{-1}(B)}(x)\,d\mu(x)=\mu(T^{-1}(B)). \end{align*} Thus $\nu(B)=\mu(T^{-1}(B))=(T_{\#}\mu)(B)$ for every $B \in \mathcal{B}$. Therefore $\nu=T_{\#}\mu$, completing the proof. [/step]