question about the statement on Royden, real analysis 3rd, page 79.

I am very confused about the following statement on Royden, real analysis (3rd), page 79.

It follows this proposition that, if $\varphi = \sum^{n}_{i=1} a_i \chi_{E_i}$, then $\int \varphi = \sum a_i mE_i$, and so the restriction of Lemma 1 that the sets $E_i$ be disjoint is unnecessary.

However, according to my understanding about the proof of the mentioned lemma and the proposition, the requirement of disjoint sets is necessary. Any help on why "the restriction of Lemma 1 that the sets $E_i$ be disjoint is unnecessary"?

Here are the lemma and the proposition mentioned above.

Lemma: Let $\varphi = \sum^{n}_{i=1} a_i \chi_{E_i}$, with $E_j \cap E_j = \emptyset$ for $i \neq j$. Suppose each set $E_i$ is a measurable set of finite measure. Then $\int \varphi = \sum^{n}_{i=1} a_i \chi_{E_i}$.

Proposition: Let $\varphi$ and $\psi$ be simple functions which vanish outside a set of finite measure. Then $\int (a \varphi + b \psi) = a \int \varphi + b \int \psi$, and if $\varphi \geq \int \psi$ a.e., then $\int \varphi \geq \psi$.

And I attached the proofs of lemma and the proposition below.

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• You might be simply getting confused by the language. The proof depended very much on the sets being disjointed. Yes. But now that you have completed the proof you can deduce a new valid statement, namely that the lemma is true even if the sets are not disjointed. There is another way that mathematicians can argue this kind of thing. State the Lemma in general (without disjointed assumption). Then start the proof with a special case, assuming disjointed. Then continue the proof to show that the general case works. Long proof but maybe easier for students to follow. – B. S. Thomson Jan 18 '16 at 17:05
• @B. S. Thomson and LutzL. Thanks for your comments. However, I still did not get the idea. I tried to remove the condition in the lemma that any two sets are disjoint, and tried to prove it, but I failed to reach the conclusion. And I cannot convince myself it is possible, as if the two measurable sets are not disjoint, the sum of the measures of the two sets should be greater than the measure of the union of the two sets. Is there anything I missed or misunderstand? Thanks in advance. – chenxin Jan 18 '16 at 17:29
• Easier to see with two. Take $a\chi_A + b\chi_B$ and integrate. If we know the result only when the sets are disjoint we have to reduce this problem to that situation. Write $E_1=A \setminus B$, $E_2=B\setminus A$ and $E_3=A\cap B$. Figure out how to write $a\chi_A + b\chi_B$ as a linear combination of $\chi_{E_1}$, $\chi_{E_2}$, and $\chi_{E_3}$ . Once you know the formula for two you can write a general version (as LutzL did) or use induction. – B. S. Thomson Jan 18 '16 at 17:53
• @B.S.Thomson. Thanks for your comments. They are very useful for me to get a deep understanding for the lemma and proposition. So, now my understanding is that if we have a function like $\varphi = \sum^{n}_{i=1} a_i \chi_i$, it is OK to decompose the function into several simple functions, then follow the proof of the proposition, the conclusion of the lemma will be reached. LutzL's formula gives the way about how to find those simple functions. Is my understanding correct? Thanks. – chenxin Jan 18 '16 at 18:47
• Yes. You just needed a nudge. – B. S. Thomson Jan 18 '16 at 18:56

You can decompose the collection of $E_i$ into disjoint atomic sets $A_j$. Then $$φ=\sum_{j=1}^N\left(\sum_{i:A_j\subset E_i}a_i\right)χ_{A_j}$$ and the value formula of the integral can be assembled in the backwar direction giving the claim.