A question about non-negative measurable function Here is the proposition that I see in Rudin's Real and Complex analysis
Suppose $f$ is a measurable function into $[0,\infty]$ and $E$ is a measurable set with respect to the measure $\mu$
Let $c$ be a constant, $0 \leq c < \infty$
Then, $\int_Ecfd\mu = c\int_Efd\mu$ 
I am wondering why we do not allow c to be $\infty$.
If $c$ is allowed to be $\infty$, then $\int_Ecfd\mu = c\int_Efd\mu$  is not true anymore?
 A: It is also true for $c=\infty$, as can be seen by using the monotone convergence theorem and by applying the claim to $c_n =n$.
I don't have the book here right now, so I can't tell you exactly why Rudin does not allow $\infty$ in that Proposition. Maybe it is (more or less) immediately after the definition of the integral, so that he does not have the monotone convergence theorem at hand yet. Note that Rudin does not allow simple functions to take the value $\infty$ (IIRC), so that the proof for $c=\infty$ using just the definition of the integral is not directly obvious.
A: The answer comes I think from the definition involving simple functions, see Rudin Real and Complex Analysis definition 1.16. Rudin specifically makes the remark "Note that we explicitly exclude $\infty$ from the values of a simple function." I think this is his justification, seeing as integrals of non-negative functions are the supremum of integrals over simple functions. The issue is in the usual proof, where we divide by $c$.
Edit: Consider $c=\infty$, $f$ measurable, and consider $\int_E cfd\mu$. Then we want $\int_E cf=\sup\{\int_E sd\mu, 0\leq s\leq cf, s \text{ simple}\}$. Well, since $cf=\infty$ whenever $f\neq 0$, then every $s$ simple satisfies this property. In particular, if $\mu(E)>0$, and $f\neq 0$ on a set of positive measure then we can find an $s$ such that $\int_E s=M$ for any $M>0$. This implies $\int_E cf=\sup\{M>0\}=\infty$. But if $f=0$ a.e., then $cf=0$ a.e. as well, so $\int_E cfd\mu=0$ Similarly, consider $c\int_E fd\mu$. What if $\mu(E)>0$ and $\int_E fd\mu=0$? Then $f=0$ a.e. Well, this is fine, so $c\int_E fd\mu=\infty\cdot 0=0$. If $f\neq 0$ on a set of positive measure, then $\int_E f>0$, so $c\int_E f=\infty$. Thus they do agree. So I don;t actually think there's an issue with extending to $c=\infty$.
