Proof of Sum-Union Measure Inequality

[mathmari]

Hello!

Let $(X, \mathcal{A}, \mu)$ a space of measure and $A_n$ a sequence of measurable sets such that each point of the space belongs to at most $M$ sets $A_n$.

Show that $$\sum_{n=1}^{+\infty} \mu (A_n) \leq M \mu \left ( \cup_{n=1}^{+\infty} A_n \right )$$

Could you give me some hints how we could do that?? (WOndering)

We have that $$\mu \left ( \cup_{n=1}^{+\infty} A_n \right ) \leq \sum_{n=1}^{+\infty} \mu (A_n)$$ but how we show the relation above?? (Wondering)

 

[GJA]

Hi mathmari,

This is going to be one of those arguments where if we were to discuss it in person it would probably make sense, but may be a bit tough to follow written out. If it's the latter, let me know and I can try to elaborate further.

FYI: Staring at the unit interval using \(\displaystyle A_{1}=[0,3/4]\); \(\displaystyle A_{2}=[1/4,1]\); \(\displaystyle A_{3}=[1/4,3/4]\) was helpful to me to set up the general proof. One must be careful, though, because a few things are true for this example which are not necessarily the case for the problem statement:

1) There are only finitely many sets;

2) The union of the sets is the whole space.

Anyways, here we go. I will use \(\displaystyle A-B\) to denote the difference of two sets (Note: This will preserve measurability; i.e. if \(\displaystyle A\) and \(\displaystyle B\) are measurable, then so is \(\displaystyle A-B\)). The key is to understand what comes next.

• \(\displaystyle A_{i_{1}}-\bigcup_{i\neq i_{1}}A_{i}\) where \(\displaystyle i_{1}\in \{1,2,3,\ldots\}\) consists of points that are only in \(\displaystyle A_{i_{1}}\)
• \(\displaystyle A_{i_{1}}\cap A_{i_{2}}-\bigcup_{i\neq i_{1},i_{2}}A_{i}\) where \(\displaystyle i_{1},i_{2}\in \{1,2,3,\ldots\}\) consists a points that are only in \(\displaystyle A_{i_{1}}\cap A_{i_{2}}\)
• etcetera...
• \(\displaystyle A_{i_{1}}\cap\ldots\cap A_{i_{M}}-\bigcup_{i\neq i_{1},\ldots,i_{M}}A_{i}\) where \(\displaystyle i_{1},\ldots,i_{M}\in\{1,2,3,\ldots\}\) consists of points that only belong to \(\displaystyle A_{i_{1}}\cap\ldots\cap A_{i_{M}}.\)

Essentially what the above does is partition/break up each of the A's into little pieces. Running through this process on the unit interval example above helps illustrate what's going on.

Now, first note that all of the sets above are mutually disjoint (even if they are listed in different bullet points), and that their union is actually equal to \(\displaystyle \cup_{i=1}^{\infty}A_{i}\). Furthermore, the process above stops at all possible intersections of size M because of the given assumption (i.e. that each point in the union of the A's can belong to at most M of the A's).

Now let \(\displaystyle B_{i}\) denote the set obtained by "unioning" the sets in a particular bullet point. For example,

\(\displaystyle B_{1}=\bigcup_{i_{1}=1}^{\infty}\Big(A_{i_{1}}-\bigcup_{i\neq i_{1}}A_{i}\Big)\)

This next bit may require more explanation, but the equality below follows from the fact that the sets that form the B's are disjoint and can be rearranged to reform the original A's, and the inequality comes from the fact that each B is contained in the union of the original A's:

\(\displaystyle \sum_{n=1}^{\infty}\mu(A_{n})=\mu(B_{1})+\cdots +\mu(B_{M})\leq \mu(\cup A_{n})+\cdots +\mu(\cup A_{n})=M\mu(\cup A_{n})\)

Let me know if anything seems unclear/not quite right.Edit: I think there is a small issue with getting the counting of intersections correct. Also, there is a slight issue when we union the B's together because we will lose needed multiple countings of intersections, but I think the general idea here is what we want. Too tired to clean it up right now.