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

Proof

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Wikipedia defines

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

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To prove the monotonicity property, the author has made use of the finite additivity property and the fact that measures are non-negative.

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*}

and we are done.

Hopefully this helps!

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They just mean a union of two disjoint (i.e. with empty intersection) sets. Ignore the definition in Wikipedia for this case.

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