# A question on the construction of the Lebesgue integral in Rudin

In Rudin's Real and Complex Analysis (page 19, definition 1.23) we take a measurable function $$f:X\rightarrow [0,\infty]$$ in which $$(X,\mathcal M,\mu)$$ is a measure space with $$\sigma-$$algebra $$\mathcal M$$ and measure $$\mu$$. We define the integral of this function over $$E\in \mathcal M$$ to be: $$\int_E f\text{ }d\mu=\sup\int_E s\text{ }d\mu \tag{1},$$ in which $$s$$ is a simple function, $$s:X\rightarrow [0,\infty)$$, defined as: $$s(x)=\sum_{i=1}^n \alpha_i\chi_{A_i}, \tag{2}$$ in which $$\chi_{A_i}(x)$$ is the characteristic function on $$X$$. Rudin states that the supremum in $$(1)$$ is being taken over all simple measurable functions such that $$0 \leq s\leq f$$, I do not know what this means. Is this a pointwise comparison of the functions between $$0$$ and $$f$$?

Standard mandatory disclaimer: I am a physicist not a mathematician.

• Yes. In other words, it is a supremum over all functions $s$ such that $s$ and $f-s$ are non-negative functions. Commented Nov 8, 2020 at 17:39
• @popoolmica So when we integrate any function $f$ we are sort of approximating it as a (generally large) simple function? Commented Nov 8, 2020 at 17:40
• I'm not sure why you say generally large. However, yes, you consider all simple functions which are smaller than $f$. That is the right definition since, it turns out, to get closer and closer to the supremum, you have to pick $s$ to approximate $f$ better and better. This is at least how I think of this mechanism. Commented Nov 8, 2020 at 17:46
• @popoolmica I see, thanks for your help! Commented Nov 8, 2020 at 17:47

$$1). S$$ is the collection of simple functions such that $$s(x)\le f(x)$$ for all $$x\in X$$.
$$2).\ I:= \{\int_X s:s\in S\}$$ is a set of numbers, with each $$\int_X s$$ defined as in your question.
$$3).\$$ Since $$I$$ is a set of numbers, we can take $$\sup I=\sup\{\int_Xs:s\in S\}.$$ The result is a number in the extended reals, and this is the integral of $$f$$ by definition.
• So we take the function $f$ on $X$ and "approximate" it as simple functions acting on the measurable sets in $X$, then integrate those "approximations" and take the largest result to be the integral of $f$? Commented Nov 8, 2020 at 17:46
• @Charlie Yes, exactly. It might help to actually do the calculation for an easy function like $y=x^2$ and with the simple functions defined for example, in baby Rudin. Commented Nov 8, 2020 at 17:53