[proofplan]
We prove the result directly from the definition of a transport plan. First we check that the pointwise convex combination $\gamma_t := (1-t)\gamma_0+t\gamma_1$ is a probability measure on the product measurable space. Then we compute its first and second marginals on arbitrary measurable sets and use the marginal identities for $\gamma_0$ and $\gamma_1$. These two marginal computations show that $\gamma_t$ has marginals $\mu$ and $\nu$, hence $\gamma_t \in \Pi(\mu,\nu)$.
[/proofplan]
[step:Define the convex combination as a measure on the product space]
Let $\mathcal C := \mathcal A \otimes \mathcal B$ denote the product $\sigma$-algebra on $X \times Y$. Define the set function \begin{align*}\gamma_t: \mathcal C \to [0,1], \qquad E \mapsto (1-t)\gamma_0(E) + t\gamma_1(E).\end{align*} Since $t \in [0,1]$, the coefficients $1-t$ and $t$ are non-negative. Therefore $\gamma_t(E) \geq 0$ for every $E \in \mathcal C$.
If $(E_n)_{n=1}^{\infty}$ is a pairwise disjoint sequence in $\mathcal C$, then countable additivity of $\gamma_0$ and $\gamma_1$ gives \begin{align*}\gamma_t\left(\bigcup_{n=1}^{\infty} E_n\right) = (1-t)\sum_{n=1}^{\infty}\gamma_0(E_n) + t\sum_{n=1}^{\infty}\gamma_1(E_n).\end{align*} By distributivity of non-negative scalar multiplication over convergent series of non-negative [real numbers](/page/Real%20Numbers), \begin{align*}\gamma_t\left(\bigcup_{n=1}^{\infty} E_n\right) = \sum_{n=1}^{\infty}\bigl((1-t)\gamma_0(E_n) + t\gamma_1(E_n)\bigr) = \sum_{n=1}^{\infty}\gamma_t(E_n).\end{align*} Thus $\gamma_t$ is a measure on $(X \times Y,\mathcal C)$. Since $\gamma_0$ and $\gamma_1$ are probability measures, \begin{align*}\gamma_t(X \times Y) = (1-t)\gamma_0(X \times Y) + t\gamma_1(X \times Y) = (1-t) + t = 1.\end{align*} Hence $\gamma_t \in \mathcal P(X \times Y,\mathcal A \otimes \mathcal B)$.
[guided]
We first have to verify that the expression $(1-t)\gamma_0+t\gamma_1$ is actually a probability measure, not just a formal linear combination. Let $\mathcal C := \mathcal A \otimes \mathcal B$ be the product $\sigma$-algebra on $X \times Y$, and define
\begin{align*}
\gamma_t: \mathcal C \to [0,1], \qquad E \mapsto (1-t)\gamma_0(E) + t\gamma_1(E).
\end{align*}
The condition $t \in [0,1]$ is exactly what makes this a convex combination: both coefficients $1-t$ and $t$ are non-negative. Since $\gamma_0(E) \geq 0$ and $\gamma_1(E) \geq 0$ for every $E \in \mathcal C$, it follows that $\gamma_t(E) \geq 0$.
Next we check countable additivity. Let $(E_n)_{n=1}^{\infty}$ be a pairwise disjoint sequence of sets in $\mathcal C$. Because $\gamma_0$ and $\gamma_1$ are measures on the same measurable space, they are countably additive on this sequence. Therefore
\begin{align*}
\gamma_0\left(\bigcup_{n=1}^{\infty} E_n\right) = \sum_{n=1}^{\infty}\gamma_0(E_n)
\end{align*}
and
\begin{align*}
\gamma_1\left(\bigcup_{n=1}^{\infty} E_n\right) = \sum_{n=1}^{\infty}\gamma_1(E_n).
\end{align*}
Substituting these identities into the definition of $\gamma_t$ gives
\begin{align*}
\gamma_t\left(\bigcup_{n=1}^{\infty} E_n\right) = (1-t)\sum_{n=1}^{\infty}\gamma_0(E_n) + t\sum_{n=1}^{\infty}\gamma_1(E_n).
\end{align*}
Since all terms are non-negative, scalar multiplication and addition may be distributed through the series, yielding
\begin{align*}
\gamma_t\left(\bigcup_{n=1}^{\infty} E_n\right) = \sum_{n=1}^{\infty}\bigl((1-t)\gamma_0(E_n) + t\gamma_1(E_n)\bigr) = \sum_{n=1}^{\infty}\gamma_t(E_n).
\end{align*}
Thus $\gamma_t$ is a measure. Finally, because each $\gamma_i$ is a probability measure on $X \times Y$, we have $\gamma_i(X \times Y)=1$ for $i \in \{0,1\}$. Hence
\begin{align*}
\gamma_t(X \times Y) = (1-t)\gamma_0(X \times Y) + t\gamma_1(X \times Y) = (1-t) + t = 1.
\end{align*}
So $\gamma_t$ is a probability measure on $(X \times Y,\mathcal A \otimes \mathcal B)$.
[/guided]
[/step]
[step:Compute the first marginal of the convex combination]
Let $\pi_X: X \times Y \to X$ be the first coordinate projection, defined by $\pi_X(x,y)=x$. To prove that the first marginal of $\gamma_t$ is $\mu$, let $A \in \mathcal A$. Since $\gamma_0,\gamma_1 \in \Pi(\mu,\nu)$, their first marginals are both $\mu$, so \begin{align*}\gamma_0(A \times Y) = \mu(A)\end{align*} and \begin{align*}\gamma_1(A \times Y) = \mu(A).\end{align*} Using the definition of $\gamma_t$ on the measurable set $A \times Y \in \mathcal A \otimes \mathcal B$, we obtain \begin{align*}\gamma_t(A \times Y) = (1-t)\gamma_0(A \times Y) + t\gamma_1(A \times Y).\end{align*} Substituting the two marginal identities gives \begin{align*}\gamma_t(A \times Y) = (1-t)\mu(A) + t\mu(A) = \mu(A).\end{align*} Since this holds for every $A \in \mathcal A$, the first marginal of $\gamma_t$ is $\mu$.
[/step]
[step:Compute the second marginal of the convex combination]
Let $\pi_Y: X \times Y \to Y$ be the second coordinate projection, defined by $\pi_Y(x,y)=y$. To prove that the second marginal of $\gamma_t$ is $\nu$, let $B \in \mathcal B$. Since $\gamma_0,\gamma_1 \in \Pi(\mu,\nu)$, their second marginals are both $\nu$, so \begin{align*}\gamma_0(X \times B) = \nu(B)\end{align*} and \begin{align*}\gamma_1(X \times B) = \nu(B).\end{align*} Using the definition of $\gamma_t$ on the measurable set $X \times B \in \mathcal A \otimes \mathcal B$, we get \begin{align*}\gamma_t(X \times B) = (1-t)\gamma_0(X \times B) + t\gamma_1(X \times B).\end{align*} Substituting the second marginal identities gives \begin{align*}\gamma_t(X \times B) = (1-t)\nu(B) + t\nu(B) = \nu(B).\end{align*} Since this holds for every $B \in \mathcal B$, the second marginal of $\gamma_t$ is $\nu$.
[/step]
[step:Conclude that the convex combination is a transport plan]
We have shown that $\gamma_t$ is a probability measure on $(X \times Y,\mathcal A \otimes \mathcal B)$, that its first marginal is $\mu$, and that its second marginal is $\nu$. By the definition of $\Pi(\mu,\nu)$, this means \begin{align*}\gamma_t \in \Pi(\mu,\nu).\end{align*} Since $\gamma_t=(1-t)\gamma_0+t\gamma_1$, the desired conclusion follows.
[/step]
