Stein and Shakarchi (2009) defines the Lebesgue integral of simple functions in the canonical form as below:

$$\int_{\mathbb{R}^d}\varphi(x)dx=\int_{\mathbb{R}^d}\sum_{k=1}^M c_k\chi_{F_k}(x)dx=\sum_{k=1}^M c_k\mu(F_k).$$

My Question:

  1. Interpreting the "definition" by Stein and Shakarchi (2009) is: when we "Lebesgue integrate" a simple function in the canonical form, we simply assign the measure on the characteristic function $F_k$ for each $k$ and sum them over. So this just becomes a finite sum of measures of each measurable set multiplied by some constant. Is this correct?

  2. Are we making this intermediate process?

$$\int_{\mathbb{R}^d}\varphi(x)dx=\int_{\mathbb{R}^d}\sum_{k=1}^M c_k\chi_{F_k}(x)dx=\sum_{k=1}^Mc_k\int_{\mathbb{R}^d} \chi_{F_k}(x)dx=\sum_{k=1}^M c_k\mu(F_k).$$

In other words,

$$\int_{\mathbb{R}^d} \chi_{F_k}(x)dx=\mu(F_k)?$$

Reference: Real Analysis: Measure Theory, Integration, and Hilbert Spaces. Elias M. Stein, Rami Shakarchi. Princeton University Press, 2009.

  • $\begingroup$ Well for example if $F_1=(0,1)$, $F_2=(1,2)$, $c_1=1$, and $c_2=2$ then $$ \int_{\mathbb R} \varphi(x)\ \mathsf dx = \sum_{k=1}^2 c_k\mu(F_k) = 1\cdot \mu((0,1)) + 2\cdot\mu((0,2)) = 5. $$ $\endgroup$
    – Math1000
    Sep 26, 2019 at 15:53
  • $\begingroup$ you have a typo in the RHS of the first expression, note that you changed $M$ by infinity. And yes, your interpretation is correct but just for simple measurable functions $\endgroup$
    – Masacroso
    Sep 26, 2019 at 16:52
  • $\begingroup$ @Masacroso I added a more specific question as that is what I originally was wondering about. $\endgroup$ Sep 27, 2019 at 0:20
  • $\begingroup$ @Masacroso Excellent. Thanks! $\endgroup$ Oct 3, 2019 at 18:00
  • $\begingroup$ @Masacroso Do you want to make your response to a formal answer? I will select your response. Thanks! $\endgroup$ Oct 3, 2019 at 18:10

1 Answer 1


As suggested I will write an answer instead of a comment. The equal sign in the next expression, as presented in the book, represent a definition

$$ \int_{\mathbb{R}^d}\sum_{k=1}^M c_k\chi_{F_k}(x)dx=\sum_{k=1}^M c_k\mu(F_k)\tag1 $$

Then there isn't an "intermediate process" at all because a definition doesn't need to be proved, it is just a way to give a name to some mathematical notation. In $\rm (1)$ what is written in the LHS are symbols that are stated as an equivalent representation to the expression on the RHS.

To avoid this kind of confusion, and be crystal clear that an equality sign represent a definition, mathematicians usually use a variation of the equal sign like $:=$, or $\overset{\rm def}{=}$, or $\overset{\triangle }{=}$, etc...

However, using $\rm (1)$, you can prove easily that the following equality holds using induction on $M$ because each $\chi_{F_k}$ is a simple function:

$$ \int_{\mathbb{R}^d}\sum_{k=1}^M c_k\chi_{F_k}(x)dx=\sum_{k=1}^Mc_k\int_{\mathbb{R}^d} \chi_{F_k}(x)dx\tag2 $$

NOTE: in mathematics an equal sign can represent many different things (but not necessarily mutually exclusive) depending on the context: it can represent an equation, an equality between formulas, a definition, and sometimes also set inclusion.


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