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I'm studying Lebesgue integral theory and I came across the concept of Simple function. And there are some definitions of simple function.

I wonder which definition of simple function I should recognize.

In this page https://en.wikipedia.org/wiki/Simple_function, it is said that $f$ is a simple function if there exist a sequence of Lebesgue-measurable sets $\{ E_k \}_{k=1}^{N}$ and a sequence of real numbers $\{ a_k \}_{k=1}^N$ s.t.

\begin{align} &\cdot f= \displaystyle\sum_{k=1}^N a_k \chi_{E_k} (x). \\ &\cdot E_i \cap E_j =\phi \ \text{for }i\neq j.\\ \end{align}

But in my class, I learned the definition of simple function as below.

$f$ is a simple function $\iff$ There exist a sequence of Lebesgue-measurable sets $\{ E_k \}_{k=1}^{N}$ and a sequence of real numbers $\{ a_k \}_{k=1}^N$ s.t.

\begin{align} &\cdot f= \displaystyle\sum_{k=1}^N a_k \chi_{E_k} (x). \\ &\cdot m(E_k)<\infty. \end{align}

Moreover, according to another website, $f: X \to \mathbb{R}$ is a simple function if there exist a sequence of Lebesgue-measurable sets $\{ E_k \}_{k=1}^{N}$ and a sequence of real numbers $\{ a_k \}_{k=1}^N$ s.t.

\begin{align} &\cdot f= \displaystyle\sum_{k=1}^N a_k \chi_{E_k} (x). \\ &\cdot X=\bigcup_{k=1}^N E_k. \\ &\cdot E_i \cap E_j =\phi. \\ \end{align}

Is each definition correct or equivalent to each other? I wonder which definition I should choose.

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  • $\begingroup$ Not every author agrees on definitions. However what you'll find is that for the situations you care about the distinctions in the definitions don't matter and they are equivalent. To me, any simple function should be integrable so I lean towards definition 2. You can always divvy up your sets $E_i$ so that they're distinct, so that's not a big deal. $\endgroup$ Jun 14, 2021 at 13:50
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    $\begingroup$ Just a note on the side: it's $\emptyset$ (\emptyset), not $\phi$. $\endgroup$ Jun 14, 2021 at 13:54
  • $\begingroup$ @Vercassivelaunos or $\varnothing$ (\varnothing). $\endgroup$ Jun 14, 2021 at 13:58

1 Answer 1

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The definitions are not equivalent in general. Definition 1 and 3 equivalently say that $f: X \to ℝ$ takes only finitely many values and is measurable. Definition 2 additionally demands that preimages of all values but $0$ have finite measure (which is trivial if $X$ has finite measure) – to ensure that it is integrable.

The extra condition also ensures that simple functions form an algebra (you can add, substract and multiply them, and also multiply them by a scalar). This together with the simple observation that every characteristic function of a set of finite measure is simple actually proves the equivalence of the definitions.

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