[proofplan] The proof is obtained by unfolding the definitions. The pushforward measure $T_{\#}\mu$ is the measure on $(Y,\mathcal B)$ defined by evaluating $\mu$ on preimages under $T$. Thus equality $T_{\#}\mu=\nu$ is exactly the assertion that the two measures assign the same value to every measurable set $B\in\mathcal B$. [/proofplan] [step:Unfold the pushforward measure on arbitrary measurable sets] Because $T:(X,\mathcal A)\to (Y,\mathcal B)$ is measurable, for every $B\in\mathcal B$ the preimage $T^{-1}(B)$ belongs to $\mathcal A$. Therefore the set function $T_{\#}\mu:\mathcal B\to [0,1]$ is defined by \begin{align*} (T_{\#}\mu)(B)=\mu(T^{-1}(B)) \end{align*} for every $B\in\mathcal B$. [guided] The only point that must be checked before writing the formula for $T_{\#}\mu$ is measurability. Since $T:(X,\mathcal A)\to (Y,\mathcal B)$ is measurable, every measurable set $B\in\mathcal B$ has measurable preimage $T^{-1}(B)\in\mathcal A$. Hence $\mu(T^{-1}(B))$ is defined. The pushforward measure $T_{\#}\mu$ is the measure on $(Y,\mathcal B)$ obtained by assigning to each measurable set $B\in\mathcal B$ the $\mu$-mass of its preimage under $T$. Therefore, for every $B\in\mathcal B$, \begin{align*} (T_{\#}\mu)(B)=\mu(T^{-1}(B)). \end{align*} This formula is the whole content of the mass balance condition: mass in a target set $B$ is computed by measuring all source points that $T$ sends into $B$. [/guided] [/step] [step:Derive the mass balance identity from the transport condition] Assume that $T$ is a transport map from $\mu$ to $\nu$. By definition, this means \begin{align*} T_{\#}\mu=\nu \end{align*} as measures on $(Y,\mathcal B)$. Therefore, for every $B\in\mathcal B$, \begin{align*} (T_{\#}\mu)(B)=\nu(B). \end{align*} Using the definition of $T_{\#}\mu$ from the previous step gives \begin{align*} \mu(T^{-1}(B))=\nu(B). \end{align*} Thus the displayed mass balance identity holds for every $B\in\mathcal B$. [/step] [step:Recover equality of measures from the mass balance identity] Conversely, assume that for every $B\in\mathcal B$, \begin{align*} \mu(T^{-1}(B))=\nu(B). \end{align*} By the definition of the pushforward measure, for every $B\in\mathcal B$, \begin{align*} (T_{\#}\mu)(B)=\mu(T^{-1}(B)). \end{align*} Combining the two equalities gives \begin{align*} (T_{\#}\mu)(B)=\nu(B) \end{align*} for every $B\in\mathcal B$. Since two measures on the same measurable space are equal exactly when they agree on every measurable set, we obtain \begin{align*} T_{\#}\mu=\nu. \end{align*} Hence $T$ is a transport map from $\mu$ to $\nu$. [/step]