A proof of Burnside's lemma

Theorem. Let $G$ be a finite group acting on a set $X$. Write $\operatorname{Fix} g = \{ x \in X : g \cdot x = x\}$ for $g \in G$ and $\operatorname{Orb} x = \{ g \cdot x : g \in G\}$ for $x \in X$. (The latter is called an orbit of $x$.) If $X/G$ denotes the set of orbits on $X$, then \begin{align*} |X/G| = \frac{1}{|G|} \sum_{g \in G} \left| \operatorname{Fix} g \right|. \end{align*}

The proof is essentially counting the set $G\times X$ in two ways. Let $\gamma$ be a uniform distribution over $G$ under the law $\mathsf{P}$. The probabilistic analogue of the Lagrange's theorem is as follows:

Lemma. $\gamma.x$ has the uniform distribution over the orbit $\operatorname{Orb} x$ for each $x\in X$.

Proof of Lemma. $\gamma \cdot x$ takes values only in $\operatorname{Orb} x$, and for each $y \in \operatorname{Orb} x$ there exists $h \in G$ for which $x = h \cdot y$. Since each map $g \mapsto hg$ is a bijection on $G$, we know that $h\gamma$ is also uniform over $G$, thus has the same law as $\gamma$. Therefore $$ \mathsf{P}[\gamma \cdot x = y] = \mathsf{P}[h\gamma \cdot x = h \cdot y] = \mathsf{P}[\gamma \cdot x = x], $$ which proves that $\gamma \cdot x$ is uniform over $\operatorname{Orb} x$. ////

Returning to the original proof, notice that $X/G$ is a partition of $X$. So \begin{align*} \frac{1}{|G|} \sum_{g \in G} \left|\operatorname{Fix} g\right| &= \mathsf{E}[\left|\operatorname{Fix} \gamma\right|] = \mathsf{E}\left[ \sum_{x\in X} \mathbf{1}_{\{\gamma \cdot x = x\}} \right] = \sum_{x\in X} \mathsf{P}[\gamma \cdot x = x] \\ &= \sum_{O \in X/G} \sum_{x \in O} \mathsf{P}[\gamma \cdot x = x] = \sum_{O \in X/G} \underbrace{\sum_{x \in O} \frac{1}{|O|}}_{=1} = |X/G|. \end{align*}