Proof, monotonicity of measurable sets (lebesgue measure) I am trying to prove the following:
(Monotonicity) If $A \subset B$ , then
$m(A) \le m(B)$.
Now, I've drawn some pictures and defined several things, including several different identities for a measurable/lebesgue set. However, I am not sure what exactly would prove this. It seems pretty obvious, especially once a picture is drawn.
..........A..........B
(------(------)--------------)
So $A \subset B \implies \forall x \in A \in B$ (which isnt the correct way to say that).
Now, $m(A) = l(A)$. A Lebesgue measure is the difference of the endpoints of a set (when a set is bounded). I don't think we can be sure that B is a bounded set but since $A \subset B $, I think A is bounded. I assume I can take to the endpoints to be the infimum and supremum of A (although by Def of inf., sup., they are not necessarily endpoints of a set.
So maybe I could say something like,
$sup(A), inf(A) \in A$ and
$m(A) = l(sup(A)-inf(A))$.
Now, $\forall x \in A \in B $, so $sup(A), inf(A) \in B$
Also, $m(B) = l(sup(B) - inf(B)),
So,
$inf(B) \le inf(A) \le sup(A) \le sup(B)$.
So, $inf(B) \le m(A) \le sup(B)$
Therefore, $m(A) \le m(B)$
I am all over the place here, can somebody please help me streamline or tell me what I shouldnt be assuming (and what I can)?
 A: I suppose you have a measure space $(\Omega,\mathcal A, \mu)$ and you like to prove the monotonicity of the measure $\mu$?
Reading your arguments, I suppose you consider $\Omega=\mathbb R$ and you restrict yourself to intervals $A,B\subset \mathbb R$. But the monotonicity of a measure comes directly from the definition of a measure.

Let be $\Omega$ a set and $\mathcal A$ a $\sigma$-algebra over $\Omega$. A function $\mu:\mathcal A\to [0,\infty]$ is called a measure if
  $$\mu(\emptyset)=0$$
  and for a countable collection $\{A_n\}_{n=1}^\infty$ of pairwise disjoint sets in $\mathcal A$ holds
  $$
\mu\left(\bigcup_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty\mu(A_n).
$$
  The second property is called countable additivity.

You just need these two properties and you can prove the monotonicity for each element in $\mathcal A$. They don't need to be intervals!
Hint: Consider the collection $\{A, B\setminus A,\emptyset,\emptyset,\ldots\}$.(If $A,B\in\mathcal A$ we get $B\setminus A\in\mathcal A$ since $\mathcal A$ is a $\sigma$-algebra)
Solution:

From $A\subset B$ we conclude $B=A\cup (B\setminus A)$ and using the properties of a measure yields \begin{align}\mu(B)&=\mu\left(A\cup(B\setminus A)\cup\emptyset\cup\emptyset\cup\ldots\right)=\mu(A)+\mu(B\setminus A)+\mu(\emptyset)+\mu(\emptyset)+\ldots\\&=\mu(A)+\mu(B\setminus A).\end{align}Since $\mu(B\setminus A)\geq0$ we conclude $\mu(B)\geq \mu(A)$.

A: Let $(X,\Sigma,m)$ be a measure space. A measure is countably additive, therefore finitely additive i.e for every two disjoint sets $A,B\in\Sigma\quad$ $m(A\cup B) = m(A)+m(B)$.
Assume $A\subseteq B$, then $A\cup(B\setminus A) = B$. By additivity
$$m(B)  = m(A)+m(B\setminus A)$$
Since a measure of a set is non-negative, the result follows.
A thing to consider. If $A,B\in\Sigma$, is $B\setminus A\in\Sigma$? (Otherwise the above would be nonsensical).
A: One definition of $m$ is in terms of an outer measure $m^*$ and then
one defines the measurable sets and for these we have $m=m^*$.
Usually we have something like $m^* A = \inf \{ \sum_k l(I_k) \, | \, A \subset \cup_k I_k\}$, where the $I_k$ are some core sets (such
as intervals) and the union may be at most countably infinite.
If $A \subset B$ we see that $\{ \{ I_k \}_k \, | \, B \subset \cup_k I_k\} \subset \{ \{ I_k \}_k \, | \, A \subset \cup_k I_k\}$ and
so $m^* A \le m^* B$.
Hence for measurable sets we have $m A \le mB$.
