I have troubles understanding one step in the solution for this task:
Let $(\Omega,\mathcal{A},\mu)$ be a $\sigma$-finite measure and $f:\Omega \rightarrow [0,\infty]$ be a measurable function. Let $E_t:=\{\omega \in \Omega: f(\omega)>t \}$ for each $t>0$. Prove that $$\int_{\Omega} f d\mu =\int_{]0,\infty[} \mu(E_t) d\lambda_1(t)$$
Solution: Note that $f$ is a nonnegative measurable function and for all $t>0$, $$\mu(E_t)=\int_{\Omega} \chi_{\{f(\omega)>t\}} \mu(d\omega)$$
Therefore by Tonelli: $$\int_{]0,\infty[} \mu(E_t) d\lambda_1(t)=\int_{\Omega} \left( \int_{]0,\infty[}\chi_{\{f(\omega)>t\}} d\lambda_1(t)\right) \mu(d\omega)$$
Now here is the step I don't understand: $$\int_{\Omega} \left( \int_{]0,\infty[}\chi_{\{f(\omega)>t\}} d\lambda_1(t)\right) \mu(d\omega)=\int_{\Omega} \left( \int_{]0,f(\omega)[} d\lambda_1\right) \mu(d\omega)$$
How do I get these integral limits $]0,f(\omega)[$?
I know that
$$\int_{\Omega} \left( \int_{]0,\infty[}\chi_{\{f(\omega)>t\}} d\lambda_1(t)\right) \mu(d\omega)=\int_{\Omega} \left( \int_{\{f(\omega)>t\}} d\lambda_1\right) \mu(d\omega)$$
and that $\{f(\omega)>t\}=f^{-1}((t,\infty))=E_t$ but how do I get $]0,f(\omega)[$ as the inner integral limits?
My second question is rather short and more general:
Does the notation $\lambda_2(x,y)$ mean that the integral should be done in order, meaning integrated with respect to $x$ first and then with respect to $y$ or does that just mean that we integrate over both without including an order?
Note that $1_{ \{ w | f(w) >s \} } (x) = 1_{ \{ t| f(x) > t \} } (s) = 1_{ [0,f(x)) } (s)$.