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The proposition is from "Real and Complex Analysis" by Rudin.It states:

Let $s$ be a nonnegative measurable simple function on $X$ . For $E\in\mathfrak M$ (where $\mathfrak M$ is a $\sigma$-algebra in $X$), define $$\varphi(E)\equiv\int_{E} s\,d\mu\equiv\sum^{n}_{i=1}\alpha_i\mu(A_i\cap E)$$ Then $\varphi$ is a measure on $\mathfrak M$. ( $\mu$ is a positive measure.The meaning behind $\alpha$, and $A_i$ will addressed later...)

The proof of the proposition states.

Let $E$ be the union of disjoint sets $E_r$, i.e. $E_1, E_2,... $. If $s$ is as defined before

$$s=\sum^{n}_{i=1}\alpha_i\chi_{A_i}$$

where $\alpha_i$ are the distinct values of the simple function $s$, $A_i=\{x:s(x)=\alpha_i\}$, and $\chi_{A_i}$ is the characteristic function of $A_i$. $$\\$$The countable additivity of $\mu$ shows that $$\varphi(E)=\sum^{n}_{i=1}\alpha_i\mu(A_i\cap E)=\sum^{n}_{i=1}\alpha_i\sum^{\infty}_{r=1}\mu(A_i\cap E_r)$$ $$=\sum^{\infty}_{r=1}\sum^{n}_{i=1}\alpha_i\mu(A_i\cap E_r)=\sum_{r=1}^{\infty}\varphi(E_r)$$Also $\varphi(\emptyset)=0$, so that $\varphi$ is not identically $\infty$.

What I do not understand is the last statement: "Also $\varphi(\emptyset)=0$, so that $\varphi$ is not identically $\infty$."I do not quite grasp as to how the fact that $\varphi(\emptyset)=0$ leads to the fact that $\varphi$ is not identically infinity. Or is it a misunderstanding from my part? I was wondering if this statement was meant to emphasize the fact that $\varphi<\infty$ for at least $E\in\mathfrak M$. Or does it mean something else?

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    $\begingroup$ It means that $\phi(A)$ is not equal to $\infty$ for every $A \in \mathfrak{M}$. $\endgroup$ – Michael Albanese Sep 21 '14 at 1:57
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Until the last statement, the possibility that $\varphi$ is identically $\infty$ had not been ruled out; i.e. it could still be the case that $\varphi(A) = \infty$ for all $A \in \mathfrak{M}$. By demonstrating a set $A \in \mathfrak{M}$ with $\varphi(A) = 0$ (namely $A = \emptyset$), the author has shown that $\varphi$ is not identically infinity.

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