In Richard Bass' Real Analysis for Graduate Students (v. 3.1), there is the following problem (my phrasing):
Let $(X,\mu)$ be $\sigma$-finite, $f\in L^1(\mu)$ be nonnegative, and $\epsilon\in\mathbb{R}^+$. Show that there exists $A\subseteq X$ of finite measure with $$\epsilon+\int_A{f\,d\mu}>\int_X{f\,d\mu}$$
The solution is a simple application of $\sigma$-finitude.
But: the result seems to hold for the non-$\sigma$-finite measures I know. (Those are the counting measure on an uncountable set and the measure which is infinite except on null sets.) In each case, the integrability condition allows a reduction to the $\sigma$-finite case. Is $\sigma$-finitude necessary?
Let $Y = \{x: f(x) > 0 \}$. Note that $Y = \bigcup_{n=1}^\infty Y_n$ where $Y_n = \{x: f(x) > 1/n\}$ has finite measure. Now $\int_X f(x)\; d\mu = \int_Y f(x)\; d\mu$, and the restriction of $\mu$ to $Y$ is $\sigma$-finite. Thus the general case is reduced to the $\sigma$-finite case.