I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.

I am afraid that my question is stupid.

Definitions in this book:

3.1 Definition $\mathcal{S}$-partition
Suppose $\mathcal{S}$ is a $\sigma$-algebra on a set $X.$ An $\mathcal{S}$-partition of $X$ is a finite collection $A_1,\dots,A_m$ of disjoint sets in $\mathcal{S}$ such that $A_1\cup\dots\cup A_m=X.$

3.2 Definition lower Lebesgue sum
Suppose $(X,\mathcal{S},\mu)$ is a measure space, $f:X\to [0,\infty]$ is an $\mathcal{S}$-measurable function, and $P$ is an $\mathcal{S}$-partition $A_1,\dots,A_m$ of $X.$ The lower Lebesgue sum $\mathcal{L}(f,P)$ is defined by $$\mathcal{L}(f,P)=\sum_{j=1}^m \mu(A_j)\inf_{A_j} f.$$

3.3 Definition integral of a nonegative function
Suppose $(X,\mathcal{S},\mu)$ is a measure space and $f:X\to [0,\infty]$ is an $\mathcal{S}$-measurable function. The integral of $f$ with respect to $\mu$, denoted $\int f d\mu$, is defined by $$\int f d\mu=\sup\{\mathcal{L}(f,P):P\text{ is an }\mathcal{S}\text{-partition of }X\}.$$

I wonder why the author requires $f$ is an $\mathcal{S}$-measurable function.
We can define the integral of $f:X\to [0,\infty]$ which is not an $\mathcal{S}$-measurable funtion.
I guess if $f$ is not an $\mathcal{S}$-measurable funtion, then the integral of $f$ doesn't have nice properties.

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    $\begingroup$ How do you define the inetgral of the characteristic function of a set $A$ if $A$ is not in the $\sigma-$ algebra? $\endgroup$ Commented May 29, 2023 at 4:47
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    $\begingroup$ @geetha290krm It seems the OP wants to define it with the same formulas. I think this definition implies $\int{\bf 1}_Ad\mu=\sup\{\mu(B)\mid B\in\mathcal S,\; B\subseteq A\}.$ $\endgroup$ Commented May 29, 2023 at 5:26
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    $\begingroup$ Outer measure is not additive which makes integral not additive either. $\endgroup$ Commented May 29, 2023 at 5:28
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    $\begingroup$ The key is that if $f$ is $\mathcal S$-measurable, then there is a increasing sequence of measurable simple functions which converges pointwise (uniformly if $f$ is bounded) to $f$. This gives nice properties to the definition of integral. This may not be the case if $f$ is not $S$-masurable. $\endgroup$
    – NewB
    Commented May 29, 2023 at 5:30
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    $\begingroup$ See my answer on another similar question. math.stackexchange.com/a/4014486/442 You are right. The property $\int(f+g) = \int f + \int g$ can fail for non-measurable functions. $\endgroup$
    – GEdgar
    Commented May 29, 2023 at 8:03


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