$f=\infty$ on a set of measure 0, then $\int_E f = 0$

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 $\int_E h$?

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 $$ \int_E f = \int_{E \setminus E} f = \int_\emptyset f = 0. $$

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 $f$.