# Disjoint union for monotonicity

Let $$\mu$$ be a measure on $$(\Omega,\mathcal{F})$$. Show that if $$A\subset B$$, then $$\mu(A) \le \mu(B)$$.

Proof

Wikipedia defines

I don't see how that "disjoint union" works here. Is this the wrong definition of "disjoint union", or am I missing something?

More precisely, given a collection of pairwise disjoint sets $$A_{k}\in\mathcal{F}$$, where $$1\leq k\leq n$$, one has that \begin{align*} \mu\left(\bigcup_{k=1}^{n}A_{k}\right) = \sum_{k=1}^{n}\mu(A_{k}) \end{align*}
At your case, one has that $$A_{1} = A$$ and $$A_{2} = B\cap A^{c}$$. Consequently, given that $$A\subseteq B$$, one has that \begin{align*} \mu(B) = \mu(A\cup(B\cap A^{c})) = \mu(A) + \mu(B\cap A^{c}) \geq \mu(A) \end{align*}