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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 $$f(x)=\sum \limits_{i=1}^{m}a_i\chi_{F_i}(x)$$ $a_1<a_2<\dots<a_m$ 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 $(*)$.

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  • $\begingroup$ 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. $\endgroup$
    – Alex R.
    Commented Apr 30, 2020 at 19:40
  • $\begingroup$ @AlexR., thanks a lot for your reply! Probably you are right since my question is not so clear. I will edit it. $\endgroup$
    – RFZ
    Commented Apr 30, 2020 at 19:46
  • $\begingroup$ At least I think that's what the author means in the Remark. $\endgroup$
    – Alex R.
    Commented Apr 30, 2020 at 19:48
  • $\begingroup$ @AlexR., please take a look at my edit $\endgroup$
    – RFZ
    Commented Apr 30, 2020 at 19:49
  • $\begingroup$ @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. $\endgroup$
    – RFZ
    Commented Apr 30, 2020 at 19:54

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – RFZ
    Commented Apr 30, 2020 at 23:09
  • $\begingroup$ After day of thinking i came to conclusion that it is better to define Lebesgue integral in terms of canonical representation. $\endgroup$
    – RFZ
    Commented Apr 30, 2020 at 23:10
  • $\begingroup$ I would agree. ${}$ $\endgroup$
    – copper.hat
    Commented Apr 30, 2020 at 23:11
  • $\begingroup$ 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? $\endgroup$
    – RFZ
    Commented Apr 30, 2020 at 23:55
  • $\begingroup$ 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. $\endgroup$
    – copper.hat
    Commented May 1, 2020 at 0:01

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