$f=\infty$ on a set of measure 0, then $\int_E f = 0$ Let $E$ be a set of measure zero and define $f = \infty$ on $E$. Show that $\int_E f = 0$.
This is out of Royden 4E, p 84.
I know how to prove this if $f=0$ on $E$.  But I'm curious, as stated, won't this result in a situation in which $\infty \cdot 0$.
 A: Royden's definition (at least in the Second Edition - I doubt that it has changed) of $\int_E f$, where $f$ is a nonnegative measurable function on measurable set $E$, is 
$$  \int_E f = \sup_{h \le f} \int_E h$$
where $h$ is a bounded measurable function such that $m\{x: h(x) \ne 0\}$ is finite.  So, if $h$ is such a function and $\mu(E) = 0$, what is $\int_E h$?
A: The Lebesgue integral doesn't "see" sets of measure zero. That is, if $X$ is any set and $A$ a set of measure zero, then 
$$
\int_X f = \int_{X \setminus A} f.
$$
For your question, we have
$$
\int_E f = \int_{E \setminus E} f = \int_\emptyset f = 0.
$$
A: The convention in measure theory is that $\infty \cdot0=0$.  That convention is taken specifically so that the $\int_E f d\mu = 0$ whenever $\mu(E)=0$ regardless of $f$.  
A: Another way would be to define $f_n=n$ on E and then apply the monotone convergence theorem. So $\int_{E}f=\lim_{n}\int_{E}f_n=\lim_{n}\int_{E}n=0$ , all the limits are as n goes to infinity.
A: With some simple theorems about integration over subsets we have that
$\qquad\displaystyle\int_E f \,d\mu = \int_X \chi_E\cdot f\,d\mu = \int_X 0 \,d\mu = 0 \cdot \mu(X) = 0$, 
where we've used that $f \cdot \chi_E = 0$ a.e. and that the integral of two a.e. equal functions agree.
Of course this all depends on the order that one develops measure theory in and thus what definitions one is working with. One can also define integrals over subsets by restriction of measure in a suitable way and then prove that 
$\qquad\displaystyle\int_E f\,d\mu = \int_X \chi_E\cdot f\,d\mu$.
