# Definition of Lebesgue integral for simple functions

I am reading lecture notes on Measure theory and Integration and I ran into the following definition and lemma which confuses me a lot.

Definition 1: Let $$(X,\mathcal{M},\mu)$$ be a $$\sigma$$-finite measure space. Let $$f:X\to \mathbb{R}$$ be a measurable function which takes on only finitely many values and any non-zero value is achieved on the set of finite measure. Then function $$f(x)$$ is called simple.

Definition 1': The function $$f(x)$$ is simple, if $$f(x)=\sum \limits_{k=1}^{n}c_k\chi_{E_k}(x),$$ where $$E_k\in \mathcal{M}$$, $$E_k\cap E_j=\varnothing$$ for $$k\neq j$$ and $$\mu(E_k)<\infty$$ for $$c_k\neq 0$$.

Remark: We note that any simple function can be written in the form $$f(x)=\sum \limits_{i=1}^{m}a_i\chi_{F_i}(x)$$ $$a_1 and $$\sqcup_{i=1}^{m}F_i=X$$. This is called canonical representation of simple function.

Definition 2: Let $$f:X\to \mathbb{R}$$ is a simple function and $$f(x)=\sum \limits_{i=1}^{m}a_i\chi_{F_i}(x)$$ is the canonical representation of $$f(x)$$. In this case we define Lebesgue integral of simple function $$f(x)$$ as follows: $$(L)\int \limits_{X}f(x)d\mu=\int \limits_{X}f(x)d\mu:=\sum \limits_{i=1}^{m}a_i\mu(F_i) \qquad \qquad(*)$$ (here we formally assume that $$0\times \infty=0$$).

Lemma 1(unchanged). The value of Lebesgue integral of simple function $$f(x)$$ does not depend on representation of $$f(x)$$ in the form $$(*)$$.

• I'm not sure I follow your question. If say, $c_1=c_2$, then you can define $E':=E_1\cup E_2$, and use that instead, to make $f$ canonical ($c_1<c_2,...$). In general any finite linear combination of indicator functions on measurable sets can be standardized by first making the sets disjoint (e.g. $E'_1=E_1, E'_2=E_2\backslash E_1, ...$) and then deriving the constants. Commented Apr 30, 2020 at 19:40
• @AlexR., thanks a lot for your reply! Probably you are right since my question is not so clear. I will edit it.
– RFZ
Commented Apr 30, 2020 at 19:46
• At least I think that's what the author means in the Remark. Commented Apr 30, 2020 at 19:48
• @AlexR., please take a look at my edit
– RFZ
Commented Apr 30, 2020 at 19:49
• @AlexR., I think we can define the simple function directly as in the remark. The definition where we say that some coefficients in $\sum a_k\chi_{F_k}$ may be equal seems to me meaningless.
– RFZ
Commented Apr 30, 2020 at 19:54

This is not really an answer, just some observations.

I think the presentation is a little confusing.

There are three characters in play.

Let me give my own name to the three things: (i) simple functions (Definition 1, finite range, finite support, the $$E_k$$ are disjoint), (ii) partition simple functions (the $$E_k$$ form a partition of $$X$$ and the $$c_k$$ need not be distinct) and (iii) 'ordered' canonical simple functions (as in the Remark, the $$c_k$$ are ordered and the $$E_k$$ are 'maximal').

(I have no idea why the author introduces the notion of ordered $$c_k$$.)

The author defines the integral in terms of (ii), the partition simple functions. Note that this representation is not unique. The $$c_k$$ may be repeated and hence the $$E_k$$ are not necessarily maximal.

For example, $$1_{[0,2]} + 0\cdot 1_{[0,2]^c} = 1_{[0,1)} + 1_{[1,2]} + 0\cdot 1_{[0,2]^c}$$ are two (partition simple function) representations of the same simple function. Obviously we have $$1 \cdot m([0,2]) + 0 \cdot \infty = 1 \cdot m([0,1)) + 1 \cdot m([2,1]) + 0 \cdot \infty$$, but the result still needs to be established formally so that Definition 2 makes sense.

So, a priori, the definition is ambiguous. Hence the author needs Lemma 1 to establish that any of the partition simple function representations gives the same value for the integral.

(As an aside, note that Lemma 1 is the first step in showing that the integral is linear.)

Personally I would prefer to define the integral in terms of a canonical representation (I am willing to forgo the ordering :-)) and then show that any other representation yields the same integral.

• Excellent! I read your answer but did not pay attention to the details but I guess this is what I want. Can I clarify some moments of your answer?
– RFZ
Commented Apr 30, 2020 at 23:09
• After day of thinking i came to conclusion that it is better to define Lebesgue integral in terms of canonical representation.
– RFZ
Commented Apr 30, 2020 at 23:10
• I would agree. ${}$ Commented Apr 30, 2020 at 23:11
• I modified the content of my topic. Please take a look. I guess it looks much better. I have few question: 1) I did not change the Lemma. How to formulate this Lemma in better way?
– RFZ
Commented Apr 30, 2020 at 23:55
• Looks good. I see now what the ordering $a_1<...< a_n$ does, it forces the $E_k$ to be maximal. I did not realise this earlier. Commented May 1, 2020 at 0:01