Measure is Monotone

Theorem
Let $$\left({X, \Sigma, \mu}\right)\ $$ be a measure space.

Then:
 * $$\forall A, B \in \Sigma: A \subseteq B \implies \mu \left({A}\right) \le \mu \left({B}\right)$$

Proof
Let $$A \subseteq B$$, and let $$\mu$$ be a measure.

From Set Difference Union Intersection we have that:
 * $$B = \left({B \setminus A}\right) \cup \left({A \cap B}\right)$$

From Subset Equivalences we have that:
 * $$A \subseteq B \implies A \cap B = A$$

So:
 * $$A \subseteq B \implies B = \left({B \setminus A}\right) \cup A$$

From Set Difference with Intersection we have that:
 * $$\left({B \setminus A}\right) \cap A = \varnothing$$

So $$B$$ is the union of the two disjoint sets $$B \setminus A$$ and $$A \cap B$$.

A measure is by definition an additive function.

So, by the definition of additive function:
 * $$\mu \left({B}\right) = \mu \left({B \setminus A}\right) + \mu \left({A}\right)$$

But as, by definition of measure, $$\mu \left({B \setminus A}\right) \ge 0$$, it follows that:
 * $$\mu \left({B}\right) \ge \mu \left({A}\right)$$

hence the result.