4

Studying the Lebesgue Integral I am moving back and forth from different books (...I know, bad habit!), and I could not really figure out why sometime, when dealing with the definition of simple functions, some authors require the family of subsets to be disjoint, where others do not.

I actually found an explanation by Professor Tao, however, I did not really get it. Here there is the quotation:

"In this definition, we did not require the $E_1,\dots, E_k$ to be disjoint. However, it is easy enough to arrange this, basically by exploiting Venn diagrams (or, to use fancier language, finite boolean algebras). Indeed, any $k$ subsets $E_1 , \dots , E_k$ of $\mathbb{R}$ partition $\mathbb{R}$ into $2^k$ disjoint sets, each of which is an intersection of $E_i$ or the complement $\mathbb{R}\setminus E_i$ for $i = 1,\dots , k$ (and in particular, is measurable). The (complex or unsigned) simple function is constant on each of these sets, and so can easily be decomposed as a linear combination of the indicator function of these sets." (Terence Tao)

Could somebody explain it?

Thank you for your time.

Kolmin
  • 4,249
  • 1
    Seems like $2k$ should be $2^k$. For example, given any subsets $E_1,E_2,E_3$ of $\mathbb R$, you can partition $\mathbb R$ into $$E_1 \cap E_2 \cap E_3$$ $$E_1^c \cap E_2 \cap E_3$$ $$E_1 \cap E_2^c \cap E_3$$ $$E_1^c \cap E_2^c \cap E_3$$ $$E_1 \cap E_2 \cap E_3^c$$ $$E_1^c \cap E_2 \cap E_3^c$$ $$E_1 \cap E_2^c \cap E_3^c$$ $$E_1^c \cap E_2^c \cap E_3^c$$ Every $x \in \mathbb R$ is in exactly one of these eight sets. –  Dec 14 '15 at 17:44
  • Thanks a lot for the reply. You are absolutely right. Now, I corrected the typo! Could you please expand a bit on how we can actually use this procedure to get rid of disjointness? – Kolmin Dec 14 '15 at 17:51
  • 1
    I don't think moving back and forth between different books is a bad habit. – littleO Dec 14 '15 at 18:32
  • 1
    @littleO: I am not sure. As a self-taught, it can be bad, because you keep on flying from a book to another, without sticking to a specific one, to really know it inside-out, But, if you are aware of this potential problem, then it can be useful, because it forces you to see things from different perspectives. Maybe, it does depend on the point of mathematical evolution at which somebody stands. – Kolmin Dec 14 '15 at 18:36

1 Answers1

3

Tao's statement is a bit compressed. He is stating a proposition:

Given measurable sets $E_i$ and a function $f$ constant on each one and zero outside their union, we can find disjoint measurable sets $F_i$ such that

a) $f$ is still constant on each $F_i$,
b) each $F_i$ is the intersection of a collection of $E_i$'s and complements $\mathbb{R} \setminus E_i$,
c) the union of the $F_i$'s is the union of the $E_i$'s.

Tao gives no proof, presumably because he thinks the proof would be obvious to his intended audience.

For my weak brain it helps to look at two and three sets first (and to draw a few Venn diagrams). If we have $E_1$ and $E_2$, then $E_1\cap E_2$, $E_1 \setminus E_2$, and $E_2 \setminus E_1$ fit the bill. Note too that $E_1 \setminus E_2 =$ $E_1 \cap (\mathbb{R} \setminus E_2)$. If we are given $E_1$, $E_2$, and $E_3$, we may assume wlog that $E_1$ and $E_2$ are disjoint (otherwise first make them so). Then we can make the $F_i$'s be $E_1 \cap E_3$, $E_2 \cap E_3$, $E_1 \setminus E_3$, and $E_2 \setminus E_3$, and $E_3 \setminus (E_1\cup E_2)$. The argument for three sets seems to set us up for an inductive one whereby if we know how to make $n$ sets disjoint, we can also make $n+1$ sets disjoint. Because we build each set in the $F_i$'s as a subset of at least one $E_i$, $f$ is constant on each $F_i$.

Community
  • 1
ForgotALot
  • 3,971
  • 2
  • 15
  • 17