I'm following Royden's textbook and he mentions this in one line but doesn't elaborate. I'll post my attempt at a proof and would appreciate any corrections/feedback.

So far only the weaker lemma has been proved. That is, if $\{E_i\}$ are all disjoint and $\varphi = \sum a_iX_{E_i}$, then $$\int\varphi = \sum_{i=1}^{N}(a_i\;\mu(E_i))$$

Now suppose that $E_i$ aren't necessarily disjoint. I create two collections of sets $\{A_i\}$ and $\{B_i\}$ whose members are all disjoint from each other. Let $$A_i = E_i - \bigcup_{i \neq n}E_n$$ $$B_i = \bigcup_{n}E_i \cap E_n - \bigcap_{p,q \neq i}E_p \cap E_q$$

Furthermore $\mu(E_i) = \mu(A_i \cup B_i) = \mu(A_i) + \mu(B_i)$ Since $A_i$ and $B_i$ are disjoint. Also $a_iX_{E_i} = a_iX_{A_i} + a_iX_{B_i}$. Therefore,

$$\int \varphi = \int \sum_{i=1}^{N}(a_i\;X_{E_i}) = \int\sum_{i=1}^{N}(a_iX_{A_i}) + \int\sum_{i=1}^{N}(a_iX_{B_i}) $$

Now since the collections $\{A_i\}$ and $\{B_i\}$ are already disjoint, I can use the old lemma to show

$$\int \varphi = \sum_{i=1}^{N}(a_i\;\mu(A_i)) + \sum_{i=1}^{N}(a_i\;\mu(B_i)) = \sum_{i=1}^{N}(a_i\; \mu(A_i \cup B_i))$$

Finally since $\mu(A_i \cup B_i) = \mu(E_i)$, I have

$$\int \varphi = \sum_{i=1}^{N}(a_i\; \mu(E_i))$$

In particular I'm not sure about how I constructed the disjoint sets $A_i$ and $B_i$. I don't think that part is correct.

  • $\begingroup$ How is $\int\phi$ actually defined? $\endgroup$ – drhab Feb 23 '16 at 12:09
  • $\begingroup$ In my textbook if $\varphi = \sum a_i X_{E_i}$ is a simple function in canonical form, then $\int \varphi$ is defined to be $\sum a_i \mu(E_i)$. This is then strengthened to include collections of disjoint sets not in canonical form. It is then strengthened again to include collections of non-disjoint sets, which I am trying to prove. $\endgroup$ – Jake Browning Feb 23 '16 at 13:31

Hint: prove the following not quite difficult lemma:

If $A_{1},\dots,A_{n}$ are disjoint measurable sets that cover the whole space then for every tuple $\langle r_{1},\dots,r_{n}\rangle\in\mathbb{R}^{n}$ function $\psi=\sum_{i=1}^{n}r_{i}1_{A_{i}}$ is a simple function with $\int\psi d\mu=\sum_{i=1}^{n}r_{i}\mu A_{i}$.

Note that this representation is not very far from the canonical one which is used to define the integral. It is not completely the same though, so a little work must be done.

If $\chi=\sum_{j=1}^{m}s_{j}1_{B_{j}}$ is a sortlike function and the $B_j$ form a sortlike collection of measurable sets then the sets $A_{i}\cap B_{j}$ are disjoint and cover the space, and $\psi+\chi$ takes value $r_{i}+s_{j}$ on that set.

Applying the lemma results in: $$\int\psi+\chi d\mu=\sum_{i=1}^{n}\sum_{j=1}^{m}\left(r_{i}+s_{j}\right)\mu\left(A_{i}\cap B_{j}\right)=\sum_{i=1}^{n}r_{i}\mu A_{i}+\sum_{j=1}^{m}s_{j}\mu B_{j}=\int\psi d\mu+\int\chi d\mu$$

Every simple function has such a characterization so actually it has been shown now that the integral is additive on simple functions.

Then for $\phi=\sum_{k=1}^{N}a_{k}1_{E_{k}}$ we have: $$\int\phi d\mu=\sum_{k=1}^{N}\int a_{k}1_{E_{k}}d\mu=\sum_{k=1}^{N}a_{k}\mu\left(E_{k}\right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.