It can be shown, for a $\sigma$-finite measure $\mu$ on a measurable spaces $X$, the Lebesgue measure $\lambda$ on $[0,+\infty)$, and a measurable function $f:X\to [0,+\infty)$, that: $$\int_X f(x) d\mu= \mu\times\lambda\ (E)$$ Where $E=\{(x,t): t<f(x)\}$ is the 'area under the graph' of $f$.
If we drop the $\sigma$-finiteness of $\mu$, what is a possible counter example?
It seems like we can drop that condition and still get away with $+\infty=+\infty$ .
Or I am missing some 'weird' non-$\sigma$-finite measure.
$\textbf{EDIT}$: $\mu\times \lambda$ is not well defined when $\mu$ is not $\sigma$-finite, then what is a counter example for the following useful equality: $$\int_X f(x) d\mu= \int_0^{+\infty} u(\{x: f(x)>t\})\,dt$$