$\int_{[a,b]} z \mu(dz)=\int_a^b \mu([x,b])dx$

Question

Let $\mu$ be a measure on $[0,1]$. Show that $$\int_{[0,1]} z \mu(dz)=\int_0^1 \mu([x,1])dx.$$ It looks simple, but I am unable to prove it.

Answer 1

$\int_0^{1} \mu [x,1]dx=\int_0^{1}\int_{[x,1]} 1d\mu (z) dx=\int \int_0^{z}dx d\mu(x)=\int zd\mu(z)$. The second equality follows by Fubini's Theorem.