# Confusing different definitions of Lebesgue Integral of Simple Functions

Formally, a simple function is a finite linear combination of indicator functions of measurable sets. More precisely, let $$(X, Σ)$$ be a measurable space. Let $$A_1, ..., A_n \in Σ$$ be a sequence of disjoint measurable sets, and let $$a_1, ..., a_n$$ be a sequence of real numbers. A simple function is a function $$f:X\to\mathbb{R}$$ of the form $$f=\sum^n_{i=1} a_i \space \chi_{A_i}.$$

If a measure $$\mu$$ is defined on the space $$(X, Σ)$$, the integral of $$f$$ with respect to $$\mu$$ is $$\sum^n_{i=1} a_i \space \mu(A_i)$$ if all summands are finite.

This the definition of Lebesgue integral of a simple function taken from Wikipedia.

My question is: Why we need a restriction such that choosing $$A_1, ..., A_n \in Σ$$ be a sequence of DISJOINT measurable sets?

For example $$f_1=2 \space \chi_{[0,2]}+4 \space \chi_{[1,3]}$$ and $$f_2=2 \space \chi_{[0,1)}+6 \space \chi_{[1,2]}+4 \space \chi_{(2,3]}$$ are the same simple function. But according to definition given above we can not integrate $$f_1$$.

In some other sources, the definition is given as follows.

$$f$$ is a simple measurable function if and only if it is written as a finite linear combination of characteristic functions of measurable sets.

Lebesgue integral of a simple function is defined by $$\int_{X}fd\mu:=\sum^n_{i=1} a_i \space \mu(A_i).$$ In this definition $$A_i$$ are not necessarily DISJOINT.

What is the difference between these two definitions?

• As the answer by @jürgensukuman already says, both definitions are equivalent. You may also note that in your concrete example both representations lead to same integral - one time as 2*2 + 4*2 = 12 and the other time as 2+6+4=12.
– Dirk
Commented May 3 at 12:53

## 1 Answer

A given simple function can have many representations, as you have noted. But, there is at least one representation using disjoint sets, sometimes called a canonical form of the simple function.

It doesn't really matter. Eventually you will prove/understand that the Lebesgue integral is linear and so whatever way you represent your nonnegative simple function will lead to the same integral. All you really have to define correctly is the integral of a characteristic function, then everything else will follow.

• Minor nitpick, it isn't really true that there is a unique representation using disjoint sets, because you can always divide one of the sets into more disjoint sets, but these parts will have the same coefficient. The "canonical" representation would be to take $f = \sum_{c\in \mathrm{range}(f)} c\cdot 1_{f^{-1}(\{c\})}$. This represents $f$ as a linear combination of indicator functions of disjoint sets in a minimal way. Commented May 3 at 13:19
• You're completely right, I've edited my answer. Commented May 3 at 13:25
• @AlexOrtiz Are there any advantages to using the definition given by the wiki? Why did they choose to give the definition in a more restrictive way instead of giving it in a more comprehensive and simple way? Commented May 4 at 6:00
• It's an equivalent definition and it's more intuitive if the sets are disjoint that we just add up the area of the rectangle on each set. Commented May 4 at 10:33