$(\alpha \star \beta)([x, z]) = \sum_{x \leq y \leq z} \alpha(x, y) \beta(y, z)$

A linear algebra way to look at this is to consider $|P| x |P|$ matrices over $K$
where the rows and columns are indexed by $P$.
The a function $\alpha: P \times P \rightarrow K$
can be written as the elements of the $P \times P$ matrix.
Then this convolution-like operator $\star$ is simply matrix multiplication.
We have three natural functions:
(1) The characteristic function, which is the identity for $\star$:
$\delta([x, z]) \equiv 1 \texttt{ if } x = z \texttt{; } 0 \texttt{ otherwise }$

(2) the zeta function, which plays the role of the constant $1$:
$\zeta([x, z]) \equiv 1 \texttt{ if } x \leq z \texttt{; } 0 \texttt{ otherwise }$

(3) The inverse of the zeta function, the mobius function, a tool for mobius inversion:
$\begin{aligned}
&\mu([x, z]) = 1 \\
&\mu([x, z]) = - \sum_{x \leq y < z} \mu([x, y]) \\
\end{aligned}$

The mobius inversion theorem for posets states that $\zeta$ and $\mu$ as
defined above are convolutional inverses. that is, $\zeta \star \mu = \delta$.
This allows us to prove:
$\begin{aligned}
&g([x, z]) = \sum_{x \leq y \leq z} f([x, y]) \\
&g([x, z]) = \sum_{x \leq y \leq z} f([x, y]) \cdot 1 \\
&g([x, z]) = \sum_{x \leq y \leq z} f([x, y]) \cdot \zeta(y, z) \\
&g = f \star \zeta \\
&g \star \mu = f \star \zeta \star \mu \\
&g \star \mu = f \star \delta \\
&g \star \mu = f
\end{aligned}$

We have managed to find $f$ in terms of $g$, when previously we had $g$
in terms of $f$.
$F(i) = \sum_{0 \leq i \leq n} f(i) \iff f(k) = F(k) - F(k-1)$

by setting up mobius inversion on the usual partial order for the natural
numbers. For simplicity, I'll show the example on $[0, 1, 2, 3, 4]$. The example
immediately generalizes.
- We have the partial (well, total) order $P$: $0 < 1 < 2 < 3 < 4$.
- We are given a function $f(\cdot)$ we wish to integrate. We define an auxiliary function $fi([x, y]) = f(y)$ which evaluates $f$ on the right endpoint.
- We can now define $F([x, z])$ as the sum of $f$ from $x$ to $z$:

$\begin{aligned}
&F([x, z]) \equiv \sum_{x \leq y \leq z} f(y) \\
&= \sum_{x \leq y \leq z} fi([x, y]) \\
&= \sum_{x \leq y \leq z} fi([x, y]) \cdot \zeta(y, z) \\
&= fi \star \zeta
\end{aligned}$

- This tells us that $f(n) = fi([0, n]) = (F \star \mu)([0, n])$:

$\begin{aligned}
&f(n) = fi([0, n]) \equiv (F \star mu)[0, n] \\
&= \sum_{0 \leq x \leq n} F([0, x]) \mu([x, n])
\end{aligned}$

- We note that we need to know the values of $\mu([x, n])$ for a
*fixed*n, for*varying*x. Let us attempt to calculate $\mu([0, 4]), \mu([1, 4]), \mu([2, 4]), \mu([3, 4]), \mu([4, 4])$and see if this can be generalized:

$\begin{aligned}
& \mu([4, 4]) = 1 \text{ By definition} \\
& \mu([3, 4]) = - \left (\sum_{3 \leq x < 4} \right) \text{ By definition } \\
\end{aligned}$