Understanding the proof of Corollary 6 on page 76 in Royden "Fourth Edition".

Question

The corollary and part of its proof are given below:

But I do not understand why the last statement is true, could anyone explains this for me please?

Answer 1

As $A$ and $B$ are disjoint, one has $\chi_{A\cup B}=\chi_A+\chi_B$. This follows easily from the definition of the indicator function $\chi$. For any set $S$, $$ \chi_S(x)=\begin{cases} 0&\text{if }x\not\in S\\ 1&\text{if }x\in S \end{cases}. $$

Now, $\chi_{A\cup B}=\chi_A+\chi_B$ is the same as $\chi_{A\cup B}(x)=\chi_A(x)+\chi_B(x)$ for every $x$. There are only three cases to check and I will leave them to you:

• $x\in A$;

• $x\in B$;

• $x\not\in A\cup B$.

Answer 2

By definition, if $\chi_{A}(x) =1 $ if $ x\in A $ and $\chi_{A}(x) =0 $ otherwise

As A and B are disjoint, we must have $\chi_{A \cup B} =\chi_{A}+\chi_{ B} $, implying $f \times \chi_{A \cup B} =f \times (\chi_{A}+\chi_{ B}) =f \times \chi_{A}+ f \times \chi_{ B}$