Value of the function $\phi=c_1 \chi_{I_1} + c_2 \chi_{I_2}$

Question

Lets say I have the function $\phi=c_1 \chi_{I_1} + c_2 \chi_{I_2}$

When $I_1$ and $I_2$ are disjont, why does $\phi$ take the constant value of $c_1$ at the point of $I_1$ and the constant value of $I_2$ at the point of $I_2$?

When $I_1$ and $I_2$ have common points, why does $\phi$ take the value of $c_1+c_2$ at all points of $I_1 \cap I_2$?

Answer 1

Break it down:

In the first case, $I_1 \cap I_2 = \emptyset$. if $x$ is a real number, these are the possibilities:

• $x \in I_1, x \in I_2$: Since $I_1 \cap I_2 = \emptyset$, this is impossible.
• $x \in I_1, x \notin I_2$: since $x \in I_1, \chi_1(x) = 1$. Since $x\notin I_2$, $\chi_2(x) = 0$. So $$\phi(x) = c_1\chi_1(x) + c_2\chi_2(x) = c_1\cdot 1 + c_2\cdot 0 = c_1 + 0 = c_1$$
• $x \notin I_1, x \in I_2$: In this case $\chi_1(x) = 0, \chi_2(x) = 1$. So can you calculate $\phi(x)$ like I did above?
• $x \notin I_1, x \notin I_2$: In this case $\chi_1(x) = 0, \chi_2(x) = 0$. So can you calculate $\phi(x)$ like I did above?

Now in the second case, $I_1 \cap I_2 \ne \emptyset$, so there exists values of $x$ for which $\chi_1(x) = 1$ and $\chi_2(x) = 1$. So can you calculate $\phi(x)$ in this case too?