Looking for a non-$\sigma$-finite measure for a counter example

Question

This is just the context:

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 $f$, and $E$ is measurable.

This is the question :

$$\int_X f(x) d\mu= \int_0^{+\infty} \mu(\{x: f(x)>t\})\ dt$$

What is a possible counter example for that equality? We want it to fail for some non-$\sigma$-finite measure.

Answer 1

Edited and rewritten: I'm pretty confident that $\sigma$-finiteness of $\mu$ is not needed to prove the second equality!

Claim: Let $\mu$ be a measure on a measurable space $X$. ($\mu$ may or may not be $\sigma$-finite.) Let $f: X \to [0, \infty)$ be a positive $\mu$-measurable function. Then:

• For all $t \in [0, \infty)$, the set $\{ x \in X : f(x) > t \}$ is $\mu$-measurable.
• The function $t \mapsto \mu ( \{ x \in X : f(x) > t \} )$ is Lebesgue-measurable.
• $\int_X f(x) d\mu(x) = \int_0^\infty \mu ( \{ x \in X : f(x) > t \} ) dt$.

Proof: The first bullet point holds by the definition of what it means for $f$ to be measurable.

The second bullet point holds because the function $t \mapsto \mu\{ x \in X : f(x) > t \}$ is monotonically decreasing, and all monotonic functions are Lebesgue measurable.

For the third bullet point, observe that there exists a sequence of positive simple ($\mu$-measurable) functions $s_n$ such that $ s_n(x)\nearrow f(x)$ as $ n\to \infty$ for all $x\in X$. It's clear that the equality that we want to prove is true when $f$ is replaced with any of the $s_n$'s. All that then remains is to apply the Monotone Convergence Theorem on both sides of the equality.