Please help me prove that if $A \subseteq B$, then $m(A) \leq m(B)$ (That $m$ is monotonic).

How would you prove this? Can we say $m^*(A \cup B) \leq m^*(A) + m^*(B)$ where $m^*(A \cup B) + m^*(A) = m^*(B)$? Do we need some points in $\mathbb{Q}$ or $\mathbb{R}$? Please help.


Let $A,B$ be in a $\sigma$-algebra $\Sigma$ of $X$. Then $B \cap A = X - ((X - A) \cup (X - B))$ by De Morgan's law and so $B \cap A$ is in $\Sigma$ too (and we see $\Sigma$ is closed under intersections). Then $B \cap (X - A\cap B) = B - A$ and so set differences are also within $\Sigma$.

Countable additivity says that if $A_1 \cap A_2 = \emptyset$, then $m(A_1) + m(A_2) = m(A_1 \cup A_2)$. $A$ and $B-A$ are disjoint. $m(A) + m(B-A) = m(A \cup (B - A)) = m(B)$ where the last step uses the fact $A \subseteq B$. Because $m(B-A) \geq 0$, we see that $m(A) \leq m(B)$.

  • $\begingroup$ Oh I just realised you were showing this for a $\sigma$-algebra. I'll edit with more details $\endgroup$ – muzzlator Mar 20 '13 at 9:10
  • $\begingroup$ Thanks. That was the idea that I had. Thanks for showing it with details with sigma algebra. I was working that one out. $\endgroup$ – 9959 Mar 20 '13 at 9:34

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