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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)?

KCd
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user391838
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3 Answers3

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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 $\mu(B)\geq \mu(A)$.

Mundron Schmidt
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  • "Math processing error"? Is that b.c. I'm reating your comment on a phone? Some of my Latex isn't working, too, sorry about that. I don't understand what the ordered triple represents? And why are you Removing the same set (the empty set) multiple times from B? Why was that necessary? – user391838 Oct 03 '17 at 13:33
  • Latex is working. Might it be a problem with your phone? I don't know your background of the problem. If you are talking about measure, you should consider a measure space $(\Omega, \mathcal A, \mu)$ where $\Omega$ is your space, $\mathcal A\subset \mathcal P(\Omega)$ is a $\sigma$-algebra and $\mu$ your measure, which is defined as above by those two properties. And to use the countable additivity you need a collection. – Mundron Schmidt Oct 03 '17 at 21:40
  • I have a few questions about this proof. Why do we need to consider the set ${ A, A \backslash B, \emptyset, \emptyset, \ldots }$ instead of simply ${ A, A \backslash B }$? And I am also not sure why ${ A, A \backslash B, \emptyset, \emptyset, \ldots }$ is used instead of ${ A, A \backslash B, \emptyset }$. I would think these are the same, but from context I would say I am missing something. – Nikolas Koutroulakis Jan 12 '24 at 17:36
  • Follow up: After thinking about it, I am inferring that it was actually meant to be the case that ${ A, A\backslash B, \emptyset, \emptyset, \ldots } = { A, A \backslash B, \emptyset }$, and the purpose of referring to the collection in this way (and the reason for considering it at all) was so that the values could be plugged into the second condition in the definition, $\mu(\bigcup_{n=1}^{\infty}A_{n}) = \sum_{n=1}^\infty \mu(A_{n})$. Is that correct? – Nikolas Koutroulakis Jan 12 '24 at 17:44
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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).

AlvinL
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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 $m A \le mB$.

copper.hat
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