Let $f \colon \mathbb{Z}^2 \to [0, 1]$ be a function such that for any integers $x$ and $y$,
$$f(x, y) = \frac{f(x - 1, y) + f(x, y - 1)}{2}.$$

We will prove that the only functions satisfying this condition are constant functions.

First, we use induction on $n$ to show that
$$f(x, y) = \frac{f(x - n, y) + \binom{n}{1} f(x - n + 1, y - 1) + \cdots + \binom{n}{1} f(x - 1, y - n + 1) + f(x, y - n)}{2^n}.$$

Next, consider the function $f(x - 1, y + 1)$:
$$f(x - 1, y + 1) = \frac{f(x - 1 - n, y + 1) + \binom{n}{1} f(x - n, y) + \cdots + \binom{n}{1} f(x - 2, y - n + 2) + f(x - 1, y + 1 - n)}{2^n}.$$

Subtracting these two expressions, we get:
$$2^n \left( f(x, y) - f(x - 1, y + 1) \right) = f(x - 1 - n, y + 1) + \left( 1 - \binom{n}{1} \right) f(x - n, y) + \left( \binom{n}{1} - \binom{n}{2} \right) f(x - n + 1, y - 1) + \cdots + \left( \binom{n}{1} - 1 \right) f(x - 1, y + 1 - n) + f(x, y - n).$$

Since $f(x, y) \in [0, 1]$, the right-hand side is bounded by $\binom{n}{\left\lfloor \frac{n}{2} \right\rfloor}$. Thus,
$$2^n \left( f(x, y) - f(x - 1, y + 1) \right) \leq \binom{n}{\left\lfloor \frac{n}{2} \right\rfloor}.$$

For large $n$, this inequality becomes impossible because $2^n$ grows exponentially while $\binom{n}{\left\lfloor \frac{n}{2} \right\rfloor}$ grows polynomially. Therefore, we must have $f(x, y) = f(x - 1, y + 1)$.

From the initial equation, it follows that $f(x, y) = f(x - n, y) = f(x, y - n)$ for all $n$. Hence, $f(x, y)$ must be constant for all $(x, y) \in \mathbb{Z}^2$.

Thus, the only functions $f \colon \mathbb{Z}^2 \to [0, 1]$ that satisfy the given condition are constant functions. Therefore, the answer is:
$$\boxed{f(x, y) = C \text{ for some constant } C \in [0, 1]}.$$