You need to distinguish between facts and convention. A convention is useful if it helps to organize a number of facts. In the context of measure theory, there are several facts that have the appearance of "$0\times \infty =0$" that are proven from the definitions, and thus the convention allows for an easier way to state facts.
A couple of examples:
- If $\mu(X)=0$ and $f(x)=+\infty$ for all $x\in X$, then the definition of $\int_X f\,d\mu$ leads to a value of $0$. On the other hand, this could be thought of as "$\mu(X)\times \infty=0\times \infty$", integrating a constant function.
- If $f(x)=0$ for all $x\in X$ and $\mu(X)=\infty$, then the definition of $\int_X f\,d\mu$ leads to a value of $0$. On the other hand, this could be thought of as "$\mu(X)\times 0=\infty\times 0$", integrating a constant function.
As a consequence of 1. and 2. above, adopting the convention $0\times \infty=0$ (along with the more intuitive $a\times \infty=\infty$ if $a>0$) allows the following result to be stated:
$$\int_{X} C\,d\mu=\mu(X)\times C,$$
for all constants $C\in[0,\infty]$ regardless of the size of $\mu(X)$. The convention is used only in stating and using the result in condensed form, not in proving it.
Another example that you highlight: If $E$ has measure zero, so does $E\times \mathbb R$. This is not proved using the convention. I would think of it as $$\mu(E\times \mathbb R)=\mu\left(\bigcup_n E\times [-n,n]\right)\leq\sum_n\mu(E\times[-n,n])=\sum_n 0=0,$$ where the equation $\mu(E\times[-n,n])=0$ requires justification, but nothing with infinite measure is involved.