Integration and Tonelli's Theorem

Question

Let $f$ be a nonnegative continuous function on $R$. We define $\psi(t) =m(\{x: f(x)>t\})$ for $t\geq 0$. I want to show that $\int_0^\infty \psi(t) \, dt$ exists and equal to $\int_{-\infty}^\infty f(x) \, dx$.

Please note that our integration is the Lebesgue integral. Intuitively I can see that the conclusion holds, and basically it's a consequence of the definition of the Lebesgue integration. I might be wrong.

Also, note that the given function is measurable because it's continuous. I've seen somewhere that this equality is called the distribution formula. Is it a standard theorem? Anyway, can we prove this result using the Tonelli theorem? I guess Tonelli's theorem will work. Can you please give me a little hint? Thanks so much.

Answer 1

The key step is to write $\psi(t) = \int_{-\infty}^\infty \chi_{f(x) > t} \;dx$ where $\chi_{f(x) > t}$ is $1$ if $f(x) > t$ and $0$ otherwise. Then when you swap the order of integration, you get

$$ \int_{-\infty}^\infty \int_0^\infty \chi_{f(x) > t} \;dt\, dx = \int_{-\infty}^{\infty} f(x) \;dx. $$

Just think about each step carefully.

Answer 2

$\int_0^{\infty} \psi (t)dt=\int_0^{\infty} \int I_{\{x: f(x) >t\}}(x) dx dt=\int_0^{\infty} \int I_{\{x: f(x) >t\}}(x) dt dx= \int f(x)dx$ because $\int I_{\{x: f(x) >t\}}(x) dt $ is the length of the interval from $0$ to $f(x)$.