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# System prepended metadata

title: '[Need Some Disjoint Sets? (A Measure Theory Trick) but with proofs$\ldots$](https://www.math3ma.com/blog/need-some-disjoint-sets-a-measure-theory-trick)'

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# [Need Some Disjoint Sets? (A Measure Theory Trick) but with proofs$\ldots$](https://www.math3ma.com/blog/need-some-disjoint-sets-a-measure-theory-trick)

## Introduction: The Trick

[Math3ma](https://www.math3ma.com/) has a very good blog post regarding disjointification which offers intuition and why disjointification is an indispensable tool in Measure Theory.

Here, we intend to give of the same using induction. The disjointification trick as in [here](https://www.math3ma.com/blog/need-some-disjoint-sets-a-measure-theory-trick) is as follows:


> **Trick**: Given a countable collection of measurable sets $E_1,E_2,\ldots$ in some set $X$, consider their union $\bigcup_{i=1}^{\infty} E_i$. There exists an algorithm which allows us to create a new countable collection of measurable sets $F_1,F_2,\ldots$ such that the $F_i$ are pairwise disjoint and  have the same union as the original, i.e. $\bigcup_i F_i=\bigcup E_i$.

## But how do you construct the $F_i$?

*Here's how*: Define $F_1 = E_1$. Now for each $i>1$, define $F_i = E_i \setminus\left( \cup_{k=1}^{i-1} F_k\right)$. Now, the claim is that $F_i$ and $F_j$ are pairwise disjoint and $\cup_i F_i = \cup_i E_i$. Drawing a doodle as in the Math3ma article offers good to insight as to why $F_i$ as defined the way they are. But unfortunately the Math3ma article does not offer a "rigourous" proof of the trick works but just asks the reader to observe it.  

## How proving things for finite stuff allows us (sometime) to transcend to the countably infinite

We offer a proof here. We first show that 

> **Claim**: For any $n\ge 2$, $F_1 , F_2 , \ldots , F_n$ are pairwise disjoint and $\cup_{i=1}^n F_i = \cup_{i=1}^{n} E_i$.

But wait a minute! How will this help us to prove that $F_i$ and $F_j$ are pairwise disjoint for $i, j \in \mathbb N$ and $\cup_{i=1}^\infty F_i = \cup_{i=1}^{\infty} E_i$? Isn't this claim weaker?

Certaintly, it is but this "weak" claim is going to help us prove what we want. Let's assume our claim for the moment. Let $m,n \in \mathbb N$ with $m \ne n$. We can assume wlog that $m<n$. Our claim tells us that $F_1 , F_2, \ldots , F_m , \ldots , F_n$ are pairwise disjoint. Hence, we have that $F_m$ and $F_n$ are disjoint! Now, we prove that $\cup_{i\in \mathbb N} F_i = \cup_{i\in \mathbb N} E_i$. Let $x\in \cup_{i\in \mathbb N} F_i$. Then $x\in F_n$ for some $n \in \mathbb N$. By our claim, we have that $\cup_{i=1}^{n} F_i = \cup_{i=1}^{n} E_i \subseteq \cup_{i\in \mathbb N} E_i$. Thus, $\cup_{i=1}^\infty F_i \subseteq \cup_{i=1}^{\infty} E_i$. A similar argument shows that $\cup_{i=1}^\infty E_i \subseteq \cup_{i=1}^{\infty} F_i$. Hence, we are done!

## Proving the Claim via Induction

Now, we proceed to proof the claim:

**Claim**: For any $n\ge 2$, $F_1 , F_2 , \ldots , F_n$ are pairwise disjoint and $\cup_{i=1}^n F_i = \cup_{i=1}^{n} E_i$.

The base case is clear: $F_1 \cup F_2 = E_1 \cup E_2 \setminus E_1 = E_1 \cup E_2$ and $F_1 \cap F_2 = E_1 \cap (E_2 \setminus E_1) = \emptyset$.

For induction, assume that $F_1 , F_2 , \ldots , F_n$ are pairwise disjoint and $\cup_{i=1}^n F_i = \cup_{i=1}^{n} E_i$ for some $n \ge 2$.

Now, consider $F_1 , F_2, \ldots , F_n, F_{n+1}$. Since $F_1 , F_2 , \ldots , F_n$ are pairwise disjoint, it suffices to show that if $i<n+1$ then $F_i \cap F_{n+1} = \emptyset$ to prove pairwise disjointness of $F_1 , F_2, \ldots , F_n, F_{n+1}$. To do so, consider the following:

\begin{align}
F_i \cap F_{n+1} &\subseteq \left(\cup_{j=1}^{n} F_i\right) \cap F_{n+1} & \\
&= (\cup_{j=1}^{n} E_i ) \cap \left( E_{n+1} \setminus (\cup_{j=1}^{n} E_i )\right) & \text{(by induction hypothesis)} \\
&= \emptyset &(\text{since }A\cap (B\setminus A)=\emptyset)
\end{align}

It remains to prove that $\cup_{i=1}^{n+1} F_i = \cup_{i=1}^{n+1} E_i$. To do so, consider the following:

\begin{align}
F_{n+1} \cup \left(\cup_{i=1}^{n} F_i \right)&= \left(E_{n+1}\setminus \left( \cup_{i=1}^{n} E_i \right)\right) \cup \left( \cup_{i=1}^{n} E_i \right) & (\text{by induction hypothesis}) \\
&= \left( \cup_{i=1}^{n+1} E_i \right) & (\text{since }A\cup (B\setminus A)=A\cup B)
\end{align}

## Last Words

I will insist the reader to read to Math3ma article again! $\smile$