# Evaluating $\int_{E_{ij}} (s+t)\ d\mu$ in Proposition $1.25$ (Rudin's RCA)

I am unsure how $$\int_{E_{ij}} (s+t)\ d\mu$$ has been evaluated in Proposition 1.25, and I've attached a picture of the same for context:
(Please scroll down to below the image to see my question and thoughts!)

$$s,t$$ are simple measurable functions, i.e. they take the form $$s = \sum_{i=1}^n \alpha_i \chi_{A_i}, t = \sum_{i=1}^m \beta_i \chi_{B_i}$$ where $$A_i = \{x: s(x) = \alpha_i\}$$ and $$B_i = \{x:t(x) = \beta_i\}$$. $$\chi_S$$ denotes the indicator function of set $$S$$, for every $$S$$. On the previous page, Rudin defined $$\int_E s\ d\mu = \sum_{i=1}^n \alpha_i \mu(A_i\cap E)$$

Firstly, let's think about $$s+t$$. Since $$s,t$$ are simple measurable, so is $$s+t$$. In fact, since $$s$$ takes $$n$$ distinct values, and $$t$$ takes $$m$$ distinct values, $$s+t$$ can take at most $$mn$$ distinct values, right? Let's say these are $$\gamma_i$$ for $$i\in\{1,2,\ldots,p\}$$ where $$p\le mn$$. Now they define $$E_{ij} = A_i \cap B_j$$.

So, by definition, one must have $$\int_{E_{ij}} (s+t)\ d\mu = \sum_{k=1}^p \gamma_k \mu(C_k\cap E_{ij})$$ where $$C_k = \{x: (s+t)(x) = \gamma_i\}$$. So we must show $$\sum_{k=1}^p \gamma_k \mu(C_k\cap E_{ij}) \stackrel{??}{=} (\alpha_i + \beta_j) \mu(E_{ij})$$

• Is it right to say that $$C_k \cap E_{ij} = \varnothing$$ for all $$k$$ for which $$\gamma_k \neq \alpha_i + \beta_j$$? If yes, then I think that solves it and we obtain $$(\alpha_i + \beta_j) \mu(E_{ij})$$ on the RHS as required.

• Is similar reasoning employed when claiming in the next line that $$\int_{E_{ij}} s\ d\mu = \alpha_i\mu(E_{ij})$$ and $$\int_{E_{ij}} t\ d\mu = \beta_j\mu(E_{ij})$$?

Thank you!

• Yes, $$C_k \cap E_{ij} = \varnothing$$ for all $$k$$ for which $$\gamma_k \neq \alpha_i + \beta_j$$. This is because $$x\in C_k \cap E_{ij}$$ implies $$s(x)=\alpha_i$$, $$t(x)=\beta_j$$ and $$(s+t)(x)=\gamma_k$$ but $$\gamma_k \neq \alpha_i + \beta_j$$, which is not possible.
• Yes, a similar reasoning can be applied to these situations. In fact, one may observe that the simple functions are constant on these sets. Take a look at the general definition: $$\int_E s\ d\mu = \sum_{i=1}^n \alpha_i \mu(A_i\cap E)$$ except for $$A_i$$ where $$\alpha_i$$ is equal to that constant (in which case $$A_i\cap E=E$$), $$A_j\cap E=\varnothing$$.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 44, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}