Say that $(X,\mathcal{F}, μ)$ is a $\sigma$-finite measure space and $f:X→ℝ^+$ is $\mathcal{F}$-measurable and nonnegative. Given the Lebesgue measure $\lambda$ on $(ℝ,\mathcal{B}(ℝ))$, I want to show that
$∫_Xfd\mu = ∫_{ℝ^+}\mu(\{f>t\})d\lambda t$
applies directly when $\mu=\delta_0$ is the Dirac measure at $x=0$ and $X=ℝ$ and $\mathcal{F}=\mathcal{B}(ℝ)$.
As far as I'm concerned, I know that $∫_Xfd\delta_0 = f(0) $, moreover, if I'm not wrong, I know also that the right hand side of the above equality rewrites as $∫_{ℝ^+}\delta_0(\{f(0)>t\})d\lambda t=∫_{ℝ^+}_{(\{f(0)>t\})}d\lambda$ but at this point I'm stuck and I cannot figure it out how to procede. Someone can help me? Please consider that I'm not very skilled with measure theory, so a simple hint maybe not enough. Thank you.
$\delta_0 (f>t)=1$ if $f(0)>t$ and $0$ otherwise.Hence $\int \delta_0 (f>t) d \lambda(t)=\int_0^{f(0)} d\lambda(t)=f(0)$.