Is $1^{\infty}$ equal to something in the extended real line? I'm begginig to study measure theory and the extended real line $\mathbb{R}\cup\{\infty\}\cup\{-\infty\}$. I noted that there are useful definitions on the value of some the "illegal" things that were called indeterminated forms of calculus when solving limits, such that $0.\infty = 0$.
In my list there isn't a mention to de indeterminated form $1^{\infty}$, so I wonder if it is defined or obtained in the real extended line.
Thanks.
(Also, I'm not sure what tag I'm supposed to use. I put measure theory since I didn't find "extended-real-numbers" or something similar)
 A: In measure theory the algebra of indeterminate forms is chosen to make theorems and definitions more straightforward to state. 
For example we define the integral of a characteristic function ${\bf 1}_A$ as $\displaystyle \int {\bf 1}_A d \mu = \mu(A)$. For $A$ with infinite measure this gives $\displaystyle \int {\bf 1}_A d \mu = \infty$. We also define the integral of the zero function $\bf 0$ as $\displaystyle \int{\bf 0} \, d \mu = 0$ since what else would it be? 
Everything is fine so far since none of the definitions require us to multiply infinity by anything. That is until we want to express the fact that integration is linear. In other words. . . .
$\displaystyle \int c{\bf 1}_A d \mu = c \int {\bf 1}_A d \mu$.
Now consider what happens for $c=0$. The LHS becomes $\displaystyle \int{\bf 0} \, d \mu = 0$. The RHS becomes $0 \cdot \infty$. So to keep our definitions consistent in this case we must have $0 \cdot \infty = 0$. 
If we wanted $0 \cdot \infty = \infty$ then in order to express how integrals are linear we'd have to include some special cases and the whole thing would quickly become a mess when trying to express higher-order theorems.
A similar thing happens with the lemma $A \subset B \implies \sup A \le \sup B$. In order not to include the special case when $A$ or $B$ is empty it is sometimes convenient to define $\sup \{\cdot\} = - \infty$. That makes the lemma true regardless of whether the sets are empty or nonempty.
The reason you don't see something similar for $1^\infty$ is that measure theory doesn't have much use for functions like $A \mapsto 1^{\mu(A)}$ which might give $1^\infty$ when $A$ has infinite measure. But if we did then we would define $1^\infty$ however best suits the rest of our definitions.
