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 $$\int_E s\ d\mu = \sum_{i=1}^n \alpha_i \mu(A_i\cap E)$$

enter image description here

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!


1 Answer 1

  • 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 $A_i\cap E=E$), $A_j\cap E=\varnothing$.

  • $\begingroup$ Thank you, that's very helpful! $\endgroup$ Commented Mar 4, 2021 at 8:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .