measure-preserving transformations are spectrally isomorphic If $(X_{1}, \mathcal{B}_{1}, m_{1})$ and $(X_{2}, \mathcal{B}_{2}, m_{2})$ are probability spaces together with measure-preserving transformations
$T_{1}:X_{1}\to X_{1}$,$T_{2}:X_{2}\to X_{2}$.
How prove this?
$T_{1} \simeq T_{2}$$ \implies T_{1} \simeq_{e} T_{2}$
Where:
1)We say $T_{1}$ is isomorphic to $T_{2}$ ($T_{1} \simeq T_{2}$)
If there exist $M_{1}\in\mathcal{B}_{1}$,$M_{2}\in\mathcal{B}_{2}$ with $m_{1}(M_{1})=1= m_{2}(M_{2})$ such that 
(i)$T_{1}M_{1}\subseteq M_{1}$ ,$T_{2}M_{2}\subseteq M_{2}$ and
(ii)There is an invertible measure-preserving transformation $\phi:M_{1}\to M_{2}$ with $\phi T_{1}(x)=T_{2}\phi (x)$ for all $x\in M_{1}$
2)We say $T_{1}$ is spectrally isomorphic to $T_{2}$ ($T_{1} \simeq_{e} T_{2}$)
 if there is a linear operator  $W:L^{2}(m_{2})\to L^{2}(m_{1})$ such that 
(i) $W$ is invertible
(ii) $\langle Wf,Wg\rangle=\langle f,g\rangle$ for all $f,g \in L^{2}(m_{2})$
(iii)  $U_{T_{1}} W=W  U_{T_{2}}$
Remarks : $T:X\to X$ be a measure-preserving transformation on the probability space $(X, \mathcal{B}, m)$ we have defined the operator $U_{T}:L^{2}(m)\to L^{2}(m)$, $U_{T}f=foT$ the operator $U_{T}$ is linear and $\langle U_{T}f,U_{T}g\rangle=\langle f,g\rangle$ for all $f,g \in L^{2}(m)$
 A: Take $f \in L^2(m_2)$You just need to consider $$W f (x) = \begin{cases}f \circ \phi(x) & \text{ if } x \in M_1\\
0 & \text{ if } x \notin M_1
\end{cases}$$ 
Note that $Wf(x)= f \circ \phi 1_{M_1}(x)$
claim: $W: L^2(m_2) \to L^2(m_1)$ is the isometry you are looking for
0) It is linear: $W(f+\alpha g)(x)= f\circ \phi (x)1_{ M_1}(x) + \alpha g \circ phi(x)1_{ M_1}(x) = W f + \alpha W g$
1) It is invertible: $W^{-1}g (y)= g \circ \phi^{-1}(y)1_{M_2}(y)$. 
Indeed, $W^-1 W f = f(x) 1_{M_1}(x) = f(x)$ (since $m_1(M_1)= 1$ the functions differ only in a null set)
2) It is an isometry: take $f = 1_A$ $g = 1_B$ $ $$\langle Wf,Wg \rangle_1 = \int 1_A \circ \phi 1_B \circ \phi \, dm_1  = m_1(\phi^{-1}(A) \cap \phi^{-1}(B)) = m_1(\phi^{-1}(A \cap B)) = m_2(A \cap B) = \int 1_A  1_B  \, dm_2 = \langle f,g \rangle_2 $$
This is valid for any simple function $f_n, g_n$ and you can extend it to arbitrary function in $L^2$ using the continuity of the inner product and the density of the simple functions in $L^2(m_2)$
3) $$U_{T_1}W(f) = U_{T_1}(f \circ \phi)  =(f \circ \phi \circ T_1) = (f \circ T_2\circ \phi  )  = W(f \circ T_2) = W U_{T_2} (f) $$
Therefore $U_{T_1}W = W U_{T_2} $
