# Proving that $\{E_i\}$ need not be disjoint for the Lebesgue integral $\int\varphi = \sum a_i \mu(E_i)$

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.

• How is $\int\phi$ actually defined? – drhab Feb 23 '16 at 12:09
• 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. – 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: $$\int\phi d\mu=\sum_{k=1}^{N}\int a_{k}1_{E_{k}}d\mu=\sum_{k=1}^{N}a_{k}\mu\left(E_{k}\right)$$