[proofplan]
We first verify that the disagreement event $\{Y \neq Z\}$ is measurable, using the assumed measurability of the diagonal in $E \times E$. Then we fix a measurable set $A \in \mathcal E$ and compare the two probabilities $\mu(A)$ and $\nu(A)$ through the indicator functions $\mathbb{1}_A \circ Y$ and $\mathbb{1}_A \circ Z$. These indicators can differ only on the disagreement event, so their integral difference is bounded by $\mathbb P(Y \neq Z)$. Taking the supremum over all measurable $A$ gives the total variation bound.
[/proofplan]
[step:Verify that the disagreement event is measurable]
Define the product [random variable](/page/Random%20Variable)
\begin{align*}
W : (\Omega,\mathcal F) \to (E \times E,\mathcal E \otimes \mathcal E), \qquad \omega \mapsto (Y(\omega),Z(\omega)).
\end{align*}
For every measurable rectangle $A \times B$ with $A,B \in \mathcal E$,
\begin{align*}
W^{-1}(A \times B) = Y^{-1}(A) \cap Z^{-1}(B) \in \mathcal F,
\end{align*}
so $W$ is measurable because $\mathcal E \otimes \mathcal E$ is generated by measurable rectangles.
Let
\begin{align*}
D := \{\omega \in \Omega : Y(\omega) \neq Z(\omega)\}.
\end{align*}
Since $\Delta_E \in \mathcal E \otimes \mathcal E$, its complement $(E \times E)\setminus \Delta_E$ also belongs to $\mathcal E \otimes \mathcal E$. Therefore
\begin{align*}
D = W^{-1}((E \times E)\setminus \Delta_E) \in \mathcal F.
\end{align*}
Thus $\mathbb P(D)$ is well-defined.
[/step]
[step:Bound the difference on an arbitrary measurable set]
Fix $A \in \mathcal E$. Define the indicator map
\begin{align*}
\mathbb{1}_A : (E,\mathcal E) \to (\mathbb R,\mathcal B(\mathbb R))
\end{align*}
by setting $\mathbb{1}_A(x)=1$ when $x \in A$ and $\mathbb{1}_A(x)=0$ when $x \notin A$. Since $A \in \mathcal E$, the map $\mathbb{1}_A$ is measurable. Then $\mathbb{1}_A \circ Y$ and $\mathbb{1}_A \circ Z$ are bounded real-valued [measurable functions](/page/Measurable%20Functions) on $(\Omega,\mathcal F)$, and by the definition of the laws $\mu$ and $\nu$,
\begin{align*}
\mu(A) = \mathbb P(Y^{-1}(A)) = \int_\Omega \mathbb{1}_A(Y(\omega))\,d\mathbb P(\omega)
\end{align*}
and
\begin{align*}
\nu(A) = \mathbb P(Z^{-1}(A)) = \int_\Omega \mathbb{1}_A(Z(\omega))\,d\mathbb P(\omega).
\end{align*}
Hence
\begin{align*}
|\mu(A)-\nu(A)| = \left|\int_\Omega \bigl(\mathbb{1}_A(Y(\omega))-\mathbb{1}_A(Z(\omega))\bigr)\,d\mathbb P(\omega)\right|.
\end{align*}
Applying the triangle inequality for the [Lebesgue integral](/page/Lebesgue%20Integral) with respect to $\mathbb P$ gives
\begin{align*}
|\mu(A)-\nu(A)| \leq \int_\Omega |\mathbb{1}_A(Y(\omega))-\mathbb{1}_A(Z(\omega))|\,d\mathbb P(\omega).
\end{align*}
If $\omega \in \Omega \setminus D$, then $Y(\omega)=Z(\omega)$, so
\begin{align*}
\mathbb{1}_A(Y(\omega))-\mathbb{1}_A(Z(\omega)) = 0.
\end{align*}
Since both indicators take values in $\{0,1\}$, for every $\omega \in \Omega$ one has
\begin{align*}
|\mathbb{1}_A(Y(\omega))-\mathbb{1}_A(Z(\omega))| \leq \mathbb{1}_D(\omega).
\end{align*}
Therefore
\begin{align*}
|\mu(A)-\nu(A)| \leq \int_\Omega \mathbb{1}_D(\omega)\,d\mathbb P(\omega) = \mathbb P(D).
\end{align*}
[guided]
Fix a measurable set $A \in \mathcal E$. The goal is to prove a bound for the single quantity $|\mu(A)-\nu(A)|$; the total variation norm will then be obtained by taking the supremum over all such $A$.
Define the indicator map
\begin{align*}
\mathbb{1}_A : (E,\mathcal E) \to (\mathbb R,\mathcal B(\mathbb R))
\end{align*}
by setting $\mathbb{1}_A(x)=1$ when $x \in A$ and $\mathbb{1}_A(x)=0$ when $x \notin A$. Because $A \in \mathcal E$, this indicator map is measurable. Since $Y$ and $Z$ are measurable maps from $(\Omega,\mathcal F)$ to $(E,\mathcal E)$, the compositions $\mathbb{1}_A \circ Y$ and $\mathbb{1}_A \circ Z$ are bounded real-valued measurable functions on $(\Omega,\mathcal F)$.
The laws $\mu$ and $\nu$ are the pushforward measures $\mu=\mathbb P \circ Y^{-1}$ and $\nu=\mathbb P \circ Z^{-1}$. Therefore
\begin{align*}
\mu(A) = \mathbb P(Y^{-1}(A)) = \int_\Omega \mathbb{1}_A(Y(\omega))\,d\mathbb P(\omega)
\end{align*}
and
\begin{align*}
\nu(A) = \mathbb P(Z^{-1}(A)) = \int_\Omega \mathbb{1}_A(Z(\omega))\,d\mathbb P(\omega).
\end{align*}
Subtracting these identities gives
\begin{align*}
|\mu(A)-\nu(A)| = \left|\int_\Omega \bigl(\mathbb{1}_A(Y(\omega))-\mathbb{1}_A(Z(\omega))\bigr)\,d\mathbb P(\omega)\right|.
\end{align*}
The triangle inequality for integration with respect to the probability measure $\mathbb P$ yields
\begin{align*}
|\mu(A)-\nu(A)| \leq \int_\Omega |\mathbb{1}_A(Y(\omega))-\mathbb{1}_A(Z(\omega))|\,d\mathbb P(\omega).
\end{align*}
Now we identify where the integrand can be nonzero. Let
\begin{align*}
D := \{\omega \in \Omega : Y(\omega) \neq Z(\omega)\}.
\end{align*}
The preceding step proved that $D \in \mathcal F$, so the indicator $\mathbb{1}_D : (\Omega,\mathcal F) \to (\mathbb R,\mathcal B(\mathbb R))$ is measurable and $\int_\Omega \mathbb{1}_D(\omega)\,d\mathbb P(\omega)=\mathbb P(D)$. If $\omega \in \Omega \setminus D$, then $Y(\omega)=Z(\omega)$, and so the two indicator values are equal:
\begin{align*}
\mathbb{1}_A(Y(\omega)) = \mathbb{1}_A(Z(\omega)).
\end{align*}
Thus the absolute difference of the indicators vanishes on $\Omega \setminus D$. On $D$, the absolute difference is at most $1$, because both indicators take only the values $0$ and $1$. Combining these two cases gives the pointwise estimate
\begin{align*}
|\mathbb{1}_A(Y(\omega))-\mathbb{1}_A(Z(\omega))| \leq \mathbb{1}_D(\omega)
\end{align*}
for every $\omega \in \Omega$.
Integrating this pointwise inequality with respect to $\mathbb P$ gives
\begin{align*}
|\mu(A)-\nu(A)| \leq \int_\Omega \mathbb{1}_D(\omega)\,d\mathbb P(\omega) = \mathbb P(D).
\end{align*}
This proves that the probability discrepancy on the particular set $A$ can only be caused by outcomes on which the coupled variables $Y$ and $Z$ disagree.
[/guided]
[/step]
[step:Take the supremum over measurable sets]
The previous step proves that for every $A \in \mathcal E$,
\begin{align*}
|\mu(A)-\nu(A)| \leq \mathbb P(D).
\end{align*}
Taking the supremum over all $A \in \mathcal E$ and using the stated convention for total variation distance gives
\begin{align*}
\|\mu-\nu\|_{\mathrm{TV}} = \sup_{A \in \mathcal E} |\mu(A)-\nu(A)| \leq \mathbb P(D).
\end{align*}
Since $D=\{\omega \in \Omega : Y(\omega) \neq Z(\omega)\}$, this is precisely
\begin{align*}
\|\mu-\nu\|_{\mathrm{TV}} \leq \mathbb P(Y \neq Z).
\end{align*}
[/step]
