In Introductory Combinatorics, by Brualdi, we have an explication on the Möbius Inversion. There's a line that I'm having a hard time believing, and would appreciate some explanation:
Let $n$ be a positive integer, and consider $(P(X_n), \subseteq)$ of all subsets of $X_n$, partially ordered by containment. We let $F: P(X_n) \rightarrow \mathbb{R}$ be a real-valued function.
We then define $G: P(X_n) \rightarrow \mathbb{R},$ where $$G(K) = \Sigma_{L\subseteq K}F(L)$$ where $K \subseteq X_N$ and the summation runs through all subsets of $K.$
Here is the claim that I'd like an explanation of: "Möbius inversion allows one to invert this equation and to recover $F$ from $G$, specifically, we have
$$F(K) = \Sigma_{L\subseteq K} (-1)^{|K|-|L|}G(L), (K \subseteq X_N)$$
Maybe this will help. Starting from the RHS we have
$$\sum_{L\subseteq K} (-1)^{|K|-|L|} G(L) = \sum_{L\subseteq K} (-1)^{|K|-|L|} \sum_{Q\subseteq L} F(Q)$$
Reversing the order of summation,
$$\begin{align*} & (-1)^{|K|} \sum_{Q\subseteq K} F(Q) \sum_{Q\subseteq L \subseteq K} (-1)^{|L|} \\ & = F(K) + (-1)^{|K|} \sum_{Q\subset K} F(Q) \sum_{Q\subseteq L \subseteq K} (-1)^{|L|} \\ & = F(K) + (-1)^{|K|} \sum_{Q\subset K} F(Q) \sum_{M \subseteq K\setminus Q} (-1)^{|Q|+|M|} .\end{align*}$$
Now $K\setminus Q$ is not the empty set which means that $|K\setminus Q|\ge 1$ so the inner sum becomes
$$(-1)^{|Q|} \sum_{M\subseteq K\setminus Q} (-1)^{|M|} = (-1)^{|Q|} \sum_{m=0}^{|K\setminus Q|} {|K\setminus Q|\choose m} (-1)^m = 0.$$
This leaves just $F(K)$ as claimed.