# Why does the author define the integral of $f$ only for $f$ which is an $\mathcal{S}$-measurable function? Measure, Integration & Real Analysis Axler

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.

• How do you define the inetgral of the characteristic function of a set $A$ if $A$ is not in the $\sigma-$ algebra? Commented May 29, 2023 at 4:47
• @geetha290krm It seems the OP wants to define it with the same formulas. I think this definition implies $\int{\bf 1}_Ad\mu=\sup\{\mu(B)\mid B\in\mathcal S,\; B\subseteq A\}.$ Commented May 29, 2023 at 5:26
• Outer measure is not additive which makes integral not additive either. Commented May 29, 2023 at 5:28
• 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.
– NewB
Commented May 29, 2023 at 5:30
• 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. Commented May 29, 2023 at 8:03