example of monadic algebra
The canonical example of a monadic algebra is what is known as a functional monadic algebra, which is explained in this entry.
Let $A$ be a Boolean algebra^{} and $X$ be a nonempty set. Then ${A}^{X}$, the set of all functions from $X$ into $A$, has a natural Boolean algebraic structure^{} defined as follows:
$$(f\wedge g)(x):=f(x)\wedge g(x),({f}^{\prime})(x):=f{(x)}^{\prime},1(x)=1$$ 
where $f,g:X\to A$ are functions, and $1:X\to A$ is just the constant function mapping everything to $1\in A$ (the abuse of notation here is harmless).
For each $f:X\to A$, let $f(X)\subseteq A$ be the range of $f$. Let $B$ be the subset of ${A}^{X}$ consisting of all functions $f$ such that $\bigvee f(X)$ and $\bigwedge f(X)$ exist, where $\bigvee $ and $\bigwedge $ are the infinite join and infinite meet operations^{} on $A$. In other words,
$$B:=\{f\in {A}^{X}\mid \bigvee f(X)\in A\text{and}\bigwedge f(X)\in A\}.$$ 
Proposition 1.
$B$ defined above is a Boolean subalgebra of ${A}^{X}$.
Proof.
We need to show that, (1): $1\in B$, (2): for any $f\in B$, ${f}^{\prime}\in B$, and (3): for any $f,g\in B$, $f\wedge g\in B$.

1.
$\bigvee 1(X)=\bigvee \{1\}=1$ and $\bigwedge 1(X)=\bigwedge \{1\}=1$ so $1\in B$

2.
Suppose $f\in B$. Then $\bigvee {f}^{\prime}(X)=\bigvee \{{f}^{\prime}(x)\mid x\in X\}=\bigvee \{f{(x)}^{\prime}\mid x\in X\}$. By de Morgan’s law on infinite joins, the last expression is ${(\bigwedge \{f(x)\mid x\in X\})}^{\prime}$, which exists. Dually, $\bigwedge {f}^{\prime}(X)$ exists by de Morgan’s law on infinite meets. Therefore, ${f}^{\prime}\in B$.

3.
Suppose $f,g\in B$. Then
$\bigwedge (f\wedge g)(X)$ $=$ $\bigwedge \{f(x)\wedge g(x)\mid x\in X\}$ $=$ $\bigwedge \{f(x)\mid x\in X\}}\wedge {\displaystyle \bigwedge \{g(x)\mid x\in X\}$ $=$ $\bigwedge f(X)}\wedge {\displaystyle \bigwedge g(X)},$ which exists because both $\bigwedge f(X)$ and $\bigwedge g(X)$ do. In addition,
$$\bigvee (f\wedge g)(X)=\bigvee \{f(x)\wedge g(x)\mid x\in X\}=\bigvee f(X)\wedge \bigvee g(X).$$ The last equality stems from the distributive law of infinite meets over finite joins. Since the last expression exists, $f\wedge g\in B$.
The three conditions are verified and the proof is complete^{}. ∎
Remark. Every constant function belongs to $B$.
For each $f\in B$, write ${f}^{\vee}:=\bigvee f(X)$ and ${f}^{\wedge}:=\bigwedge f(X)$. Define two functions ${f}^{\exists},{f}^{\forall}\in {A}^{X}$ by
$${f}^{\exists}(x):={f}^{\vee}\mathit{\hspace{1em}\hspace{1em}}\text{and}\mathit{\hspace{1em}\hspace{1em}}{f}^{\forall}(x):={f}^{\wedge}.$$ 
Since these are constant functions, they belong to $B$.
Now, we define operators $\exists ,\forall $ on $B$ by setting
$$\exists (f):={f}^{\exists}\mathit{\hspace{1em}\hspace{1em}}\text{and}\mathit{\hspace{1em}\hspace{1em}}\forall (f):={f}^{\forall}.$$ 
By the remark above, $\exists $ and $\forall $ are welldefined functions on $B$ (${f}^{\exists},{f}^{\forall}\in B$).
Proposition 2.
$\exists $ is an existential quantifier operator on $B$ and $\mathrm{\forall}$ is its dual.
Proof.
The following three conditions need to be verified:

•
$\exists (0)=0$: $\exists (0)(x)={0}^{\exists}(x)={0}^{\vee}=\bigvee 0(X)=\bigvee 0=0$.

•
$f\le \exists (f)$: $f(x)\le \bigvee f(X)={f}^{\vee}={f}^{\exists}(x)=\exists (f)(x)$.

•
$\exists (f\wedge \exists (g))=\exists (f)\wedge \exists (g)$:
$\exists (f\wedge \exists (g))(x)$ $=$ $\bigvee \left(f\wedge \exists (g)\right)(X)}={\displaystyle \bigvee \left(f(X)\wedge \exists (g)(X)\right)$ $=$ $\bigvee \left(f(X)\wedge \exists (g)(x)\right)}={\displaystyle \bigvee \left(f(X)\wedge \bigvee g(X)\right)$ $=$ $\bigvee f(X)}\wedge {\displaystyle \bigvee g(X)}=\exists (f)(x)\wedge \exists (g)(x)=\left(\exists (f)\wedge \exists (g)\right)(x).$
Finally, to see that $\forall $ is the dual of $\exists $, we do the following computations:
$\forall (f)(x)$  $=$  $\bigwedge \{f(x)\mid x\in X\}}={\displaystyle \bigwedge \{f{(x)}^{\prime \prime}\mid x\in X\}$  
$=$  ${\left({\displaystyle \bigvee \{f{(x)}^{\prime}\mid x\in X\}}\right)}^{\prime}={\left({\displaystyle \bigvee \{{f}^{\prime}(x)\mid x\in X\}}\right)}^{\prime}={\left(\exists ({f}^{\prime})\right)}^{\prime}(x),$ 
completing the proof. ∎
Based on Propositions^{} 1 and 2, $(B,\exists )$ is a monadic algebra, and is called the functional monadic algebra for the pair $(A,X)$.
Title  example of monadic algebra 

Canonical name  ExampleOfMonadicAlgebra 
Date of creation  20130322 17:51:55 
Last modified on  20130322 17:51:55 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  10 
Author  CWoo (3771) 
Entry type  Example 
Classification  msc 03G15 
Defines  functional monadic algebra 