Let a percolation measure be a measure on $\{0,1\}^n$. We have a natural partial order on $\{0,1\}^n$ given by comparing all coordinates. An event $A$ is called increasing if for all $ \omega \in A $ we have for every $\eta \geq \omega$ that $\eta \in A$. Now a percolation measure $\mu$ satisfies the FKG inequality if for all increasing events $A, B$ we have that $\mu(A \cap B) \geq \mu(A) \mu(B)$. Now, define the union of two percolation measures as the measure with the law coming from sampling each of the two measures independently and taking the union of the 1s. 


Q: Does the union of two percolation measures satisfying the (FKG) inequality still satisfy (FKG)? 



Does the union of two percolation measures satisfying the (FKG) inequality still satisfy (FKG)?
Welcome to MathOverflow! The answer to your question is yes. Indeed, let $X$ and $Y$ be independent random elements of the set $\{0,1\}^n$ whose distributions are percolation measures, which latter thus satisfy the [Harris--FKG inequality, Proposition 2.3](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=2ahUKEwjUnoSox9zjAhXYHM0KHfC0C1YQFjABegQIBBAB&url=http%3A%2F%2Fwww.ihes.fr%2F~duminil%2Fpubli%2F2017percolation.pdf&usg=AOvVaw2VxEQnQ7slBZi7kxwwstBd): $Ef(X)g(X)\ge Ef(X)\, Eg(X)$ and $Ef(Y)g(Y)\ge Ef(Y)\, Eg(Y)$ for all bounded nondecreasing functions $f$ and $g$. We have to show that for any such functions $f$ and $g$ we have 
$$Ef(X\vee Y)g(X\vee Y)\ge Ef(X\vee Y)\, Eg(X\vee Y),\tag{1}$$ 
where $X\vee Y$ is coordinate-wise maximum of $X$ and $Y$. 
Note that (i) your union of the measures is the distribution of $X\vee Y$ and (ii) the inequality $P(X\vee Y\in A\cap B)\ge P(X\vee Y\in A)P(X\vee Y\in B)$ for increasing sets $A$ and $B$ is a special case of (1), with $f$ and $g$ being the indicator functions of $A$ and $B$. 


To prove (1), let $F(y):=Ef(X\vee y)$ and $G(y):=Eg(X\vee y)$ for $y\in\{0,1\}^n$. Then
\begin{align}
Ef(X\vee Y)g(X\vee Y)&=\sum\_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)g(X\vee y) \\
&\ge\sum\_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)\,Eg(X\vee y) \\
&=\sum\_{y\in\{0,1\}^n} P(Y=y)F(y)\,G(y) \\ 
&=EF(Y)G(Y) \\ 
&\ge EF(Y)\,EG(Y) \\ 
&= Ef(X\vee Y)\, Eg(X\vee Y),
\end{align}
so that (1) is proved. The first inequality in the above multiline display follows by the Harris--FKG inequality for $X$ and nondecreasing functions $x\mapsto f(x\vee y)$ and $x\mapsto g(x\vee y)$, and the second inequality there is an instance of the Harris--FKG inequality for $Y$ and nondecreasing functions $F$ and $G$. 


