[proofplan]
We use tightness of the two fixed marginal probability measures to choose compact sets $K_X \subset X$ and $K_Y \subset Y$ that capture almost all of $\mu$ and $\nu$. The product $K_X \times K_Y$ is compact in $X \times Y$. For an arbitrary coupling $\gamma \in \Pi(\mu,\nu)$, the complement of $K_X \times K_Y$ is contained in the union of two coordinate cylinders, whose $\gamma$-masses are exactly controlled by the two marginals. This gives one compact set capturing at least $1-\varepsilon$ mass for every coupling simultaneously.
[/proofplan]
[step:Choose compact sets capturing the two marginal measures]
Fix $\varepsilon > 0$. Since $X$ and $Y$ are Polish spaces and $\mu,\nu$ are Borel probability measures, the standard tightness theorem for Borel probability measures on Polish spaces applies (citing a result not yet in the wiki: every Borel probability measure on a Polish space is tight). Hence there exist compact sets $K_X \subset X$ and $K_Y \subset Y$ such that
\begin{align*}
\mu(X \setminus K_X) < \frac{\varepsilon}{2}
\end{align*}
and
\begin{align*}
\nu(Y \setminus K_Y) < \frac{\varepsilon}{2}.
\end{align*}
Since $K_X$ and $K_Y$ are compact, their finite product
\begin{align*}
K := K_X \times K_Y \subset X \times Y
\end{align*}
is compact in the [product topology](/page/Product%20Topology) on $X \times Y$.
[/step]
[step:Bound the complement of the product compact set by coordinate cylinders]
Let $\gamma \in \Pi(\mu,\nu)$ be arbitrary. Define the Borel coordinate cylinders
\begin{align*}
A_X := (X \setminus K_X) \times Y
\end{align*}
and
\begin{align*}
A_Y := X \times (Y \setminus K_Y).
\end{align*}
If $(x,y) \in (X \times Y) \setminus (K_X \times K_Y)$, then either $x \notin K_X$ or $y \notin K_Y$. Therefore
\begin{align*}
(X \times Y) \setminus K \subset A_X \cup A_Y.
\end{align*}
By [monotonicity and subadditivity](/theorems/1081) of the probability measure $\gamma$,
\begin{align*}
\gamma((X \times Y) \setminus K) \leq \gamma(A_X \cup A_Y) \leq \gamma(A_X) + \gamma(A_Y).
\end{align*}
[guided]
Let $\gamma \in \Pi(\mu,\nu)$ be fixed. The compact set we want to test is
\begin{align*}
K := K_X \times K_Y.
\end{align*}
A point $(x,y) \in X \times Y$ fails to lie in $K_X \times K_Y$ precisely when at least one coordinate fails to lie in its compact set. Thus we introduce the two coordinate cylinders
\begin{align*}
A_X := (X \setminus K_X) \times Y
\end{align*}
and
\begin{align*}
A_Y := X \times (Y \setminus K_Y).
\end{align*}
These sets are Borel because $K_X$ and $K_Y$ are compact, hence closed, hence Borel, and products of Borel sets are Borel in the product Borel $\sigma$-algebra. The pointwise implication is
\begin{align*}
(x,y) \in (X \times Y) \setminus (K_X \times K_Y) \implies (x,y) \in A_X \cup A_Y.
\end{align*}
Equivalently,
\begin{align*}
(X \times Y) \setminus K \subset A_X \cup A_Y.
\end{align*}
Because $\gamma$ is a probability measure on $X \times Y$, it is monotone with respect to set inclusion and subadditive on finite unions. Applying these two measure properties gives
\begin{align*}
\gamma((X \times Y) \setminus K) \leq \gamma(A_X \cup A_Y) \leq \gamma(A_X) + \gamma(A_Y).
\end{align*}
This reduces the problem from controlling mass outside a product compact set to controlling the masses of two coordinate events, which are exactly the events measured by the marginals.
[/guided]
[/step]
[step:Use the marginal constraints to estimate the two cylinder masses]
Let
\begin{align*}
\pi_X: X \times Y \to X
\end{align*}
and
\begin{align*}
\pi_Y: X \times Y \to Y
\end{align*}
denote the coordinate projections. Since $\gamma \in \Pi(\mu,\nu)$, its pushforward marginals satisfy $(\pi_X)_\#\gamma = \mu$ and $(\pi_Y)_\#\gamma = \nu$. Also,
\begin{align*}
A_X = \pi_X^{-1}(X \setminus K_X)
\end{align*}
and
\begin{align*}
A_Y = \pi_Y^{-1}(Y \setminus K_Y).
\end{align*}
By the definition of pushforward measure,
\begin{align*}
\gamma(A_X) = \gamma(\pi_X^{-1}(X \setminus K_X)) = \mu(X \setminus K_X)
\end{align*}
and
\begin{align*}
\gamma(A_Y) = \gamma(\pi_Y^{-1}(Y \setminus K_Y)) = \nu(Y \setminus K_Y).
\end{align*}
Therefore
\begin{align*}
\gamma((X \times Y) \setminus K) \leq \mu(X \setminus K_X) + \nu(Y \setminus K_Y) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
\end{align*}
[/step]
[step:Conclude tightness uniformly over all couplings]
The compact set $K = K_X \times K_Y$ depends only on $\varepsilon$, $\mu$, and $\nu$, not on the particular coupling $\gamma$. Since the preceding estimate holds for every $\gamma \in \Pi(\mu,\nu)$, for every $\varepsilon > 0$ there exists a compact set $K \subset X \times Y$ such that
\begin{align*}
\sup_{\gamma \in \Pi(\mu,\nu)} \gamma((X \times Y) \setminus K) \leq \varepsilon.
\end{align*}
Replacing the initial choice of $\varepsilon$ by any smaller positive number gives the strict form in the definition of tightness. Hence $\Pi(\mu,\nu)$ is tight as a family of probability measures on $X \times Y$.
[/step]
