[proofplan] The proof is an immediate application of the defining equality for a [measure-preserving transformation](/page/Measure-Preserving%20Transformation). Since $N$ is measurable, the preimage $T^{-1}(N)$ is measurable, and measure preservation gives $\mu(T^{-1}(N))=\mu(N)$. The hypothesis that $N$ is null then gives the desired conclusion. [/proofplan] [step:Apply measure preservation to the null set $N$] Because $T:E\to E$ is measure-preserving, $T$ is measurable and satisfies \begin{align*} \mu(T^{-1}(A))=\mu(A) \end{align*} for every $A\in\mathcal E$. Taking $A=N$, which is allowed because $N\in\mathcal E$, gives $T^{-1}(N)\in\mathcal E$ and \begin{align*} \mu(T^{-1}(N))=\mu(N). \end{align*} [guided] The only input needed is the definition of a measure-preserving transformation. This definition has two relevant parts: first, $T:E\to E$ is measurable, so for every measurable set $A\in\mathcal E$ the preimage $T^{-1}(A)$ belongs to $\mathcal E$; second, the measure of each measurable set is preserved under preimage: \begin{align*} \mu(T^{-1}(A))=\mu(A). \end{align*} We now use this definition with the particular measurable set $A=N$. The hypothesis gives $N\in\mathcal E$, so the definition applies. Therefore $T^{-1}(N)\in\mathcal E$ and \begin{align*} \mu(T^{-1}(N))=\mu(N). \end{align*} [/guided] [/step] [step:Substitute the null measure hypothesis] By hypothesis, $\mu(N)=0$. Substituting this value into the equality obtained above yields \begin{align*} \mu(T^{-1}(N))=0. \end{align*} Thus the preimage of the null set $N$ is again a null set. [/step]