[proofplan] The proof is an expansion of the two definitions involved. The pushforward measure $T_\#\mu$ is defined by evaluating $\mu$ on preimages under $T$, while measure preservation is exactly the assertion that those preimage measures agree with the original measures of all measurable sets. We prove each implication by evaluating the relevant measures on an arbitrary set $A\in\mathcal E$. [/proofplan] [step:Expand the pushforward definition under the assumption that $T$ preserves $\mu$] Assume first that $T$ is measure-preserving. By definition, this means that for every $A\in\mathcal E$, \begin{align*} \mu(T^{-1}(A))=\mu(A). \end{align*} Since $T:(E,\mathcal E)\to(E,\mathcal E)$ is measurable, $T^{-1}(A)\in\mathcal E$ for every $A\in\mathcal E$, so the pushforward measure $T_\#\mu$ on $(E,\mathcal E)$ is well-defined by \begin{align*} (T_\#\mu)(A)=\mu(T^{-1}(A)). \end{align*} Therefore, for every $A\in\mathcal E$, \begin{align*} (T_\#\mu)(A)=\mu(T^{-1}(A))=\mu(A). \end{align*} Thus $T_\#\mu$ and $\mu$ agree on every measurable set $A\in\mathcal E$, so $T_\#\mu=\mu$. [guided] Assume that $T$ is measure-preserving. The definition of measure preservation says precisely that every measurable set has the same measure as its full preimage under $T$: for every $A\in\mathcal E$, \begin{align*} \mu(T^{-1}(A))=\mu(A). \end{align*} We now compare this condition with the definition of the pushforward measure. Because $T:(E,\mathcal E)\to(E,\mathcal E)$ is measurable, whenever $A\in\mathcal E$, its preimage $T^{-1}(A)$ also belongs to $\mathcal E$. Hence the expression $\mu(T^{-1}(A))$ is defined for every measurable set $A$. The pushforward measure $T_\#\mu$ is the measure on $(E,\mathcal E)$ defined by \begin{align*} (T_\#\mu)(A)=\mu(T^{-1}(A)) \end{align*} for every $A\in\mathcal E$. Combining the pushforward definition with measure preservation gives, for every $A\in\mathcal E$, \begin{align*} (T_\#\mu)(A)=\mu(T^{-1}(A))=\mu(A). \end{align*} Two measures on the same measurable space are equal exactly when they assign the same value to every measurable set. Therefore $T_\#\mu=\mu$ as measures on $(E,\mathcal E)$. [/guided] [/step] [step:Recover measure preservation from equality of the pushforward measure] Conversely, assume that \begin{align*} T_\#\mu=\mu. \end{align*} Let $A\in\mathcal E$ be arbitrary. By the definition of the pushforward measure, \begin{align*} \mu(T^{-1}(A))=(T_\#\mu)(A). \end{align*} Since $T_\#\mu=\mu$, evaluating both measures on $A$ gives \begin{align*} (T_\#\mu)(A)=\mu(A). \end{align*} Hence \begin{align*} \mu(T^{-1}(A))=\mu(A). \end{align*} Because this holds for every $A\in\mathcal E$, the map $T$ is measure-preserving. [/step]