Proof of linearity of abstract integration using simple functions

Question

Proof of linearity of abstract integrals for simple functions by Prof. David Gamarnik and Prof. John Tsitsiklis - Lecture notes of Fundamentals of Probability 6-436j OCW MIT

In the proof provided above, it seems very confusing to me that the sum of two simple functions $g = \sum_{i=1}^n a_i \mathbf{1}_{A_i}$

and $h=\sum_{i=1}^m b_i \mathbf{1}_{B_i}$,

where $\mathbf{1}_{S}$ is the indicator function,

is given by $f = \sum_{i=1}^n \sum_{j=1}^m (a_i+b_j) \mathbf{1}_{A_i\cap B_j}$.

For example, let $g = 3\times\mathbf{1}_{\omega\geq 0}$ and $h = 2\times\mathbf{1}_{\omega< 0}$. If $\omega = 2$, then $g(2) = 3, h(2) = 0$, while $f(2) = 0 $ since the intersection of ${\omega\geq 0}$ and ${\omega< 0}$ is empty.

Moreover, in the third equality in the proof where $\sum_{j=1}^m\mu(A_i\cap B_j)$ is said to be $ = \mu(A_i)$ by finite additivity, how do we know that the union of all intersections of $B_j$'s with $A_i$ will cover all of $A_i$ ?

Thanks in advance!

Link to lecture notes

Answer 1

As your example show, writing $g+h$ as $\sum_{i=1}^n \sum_{j=1}^m (a_i+b_j) \mathbf{1}_{A_i\cap B_j}$ may not be the simplest way to write it, because there are a lot of empty sets. Nevertheless, in this way, we are sure that we wrote the sum as a linear combination of indicator function of disjoint sets. In order to verify its validity, if we take an element $x$ in the set of definition, there are several cases:

• $x$ belongs to some $A_i$ and some $B_j$: in this case, the double sum is $a_i+b_j=f(x)+g(x)$.
• $x$ does not belong to any $A_i$ but to some $B_j$: in this case, the double sum is $b_j=f(x)+g(x)$.
• $x$ does not belong to any $B_j$ but to some $A_i$: in this case, the double sum is $a_i=f(x)+g(x)$.
• • $x$ does not belong to any $A_i$ and $x$ does not belong to any $B_j$: in this case, the double sum is $0=f(x)+g(x)$.

In writing $h$ as a simple function, we can assume without loss of generality that $\bigcup_{j=1}^m B_j$ equals to whole space, say $X$; otherwise we add $b_{m+1}=0$ and $B_{m+1}= X\setminus\bigcup_{j=1}^m B_j$.

Answer 2

When writing this way you should use partitions into measurable sets ${\cal A} = (A_1,...,A_n)$ and ${\cal B} = (B_1,...,B_m)$. (so that $\cup_i A_i=X$ and $\cup_j B_j=X$). Then the refinement $$ {\cal A} \wedge {\cal B} = \{ A\cap B : A\in {\cal A}, B\in {\cal B}\}$$ is again a partition and the function $g+h$ is a simple function with respect to this refined partition. Using this writing systematically, the set of simple functions become an algebra which you then may use in monotone/dominated/etc... convergence.

Another point of view is that each measurable simple function $g$ takes finitely many values (possibly including zero) and ${\cal A}$ is the collection of (measurable) pre-images of all these values and ${\cal A}$ forms a partition.