- #1
laonious
- 9
- 0
Hi all,
I'm really banging my head on this problem:
Let f be a real-valued measurable function on the measure space (X,\mathcal{M},\mu).
Define
\lambda_f(t)=\mu\{x:|f(x)|>t\}, t>0.
Show that if \phi is a nonnegative Borel function defined on [0,infinity), then
\int_0^{\infty}\phi(|f(x)|)d\mu=-\int_0^{\infty}\phi(t)d\lambda_f(t).
A hint is given, which is to look at
\nu((a,b])=\lambda_f(b)-\lambda_f(a)=-\mu\{x:a<|f(x)|\leq b\},
and argue that it extends uniquely to a Borel measure.
This is straightforward, I think, as closed intervals form a semi-ring and \nu is a premeasure. I'm just not sure where to go from here. Any help would be greatly appreciated, thanks!
I'm really banging my head on this problem:
Let f be a real-valued measurable function on the measure space (X,\mathcal{M},\mu).
Define
\lambda_f(t)=\mu\{x:|f(x)|>t\}, t>0.
Show that if \phi is a nonnegative Borel function defined on [0,infinity), then
\int_0^{\infty}\phi(|f(x)|)d\mu=-\int_0^{\infty}\phi(t)d\lambda_f(t).
A hint is given, which is to look at
\nu((a,b])=\lambda_f(b)-\lambda_f(a)=-\mu\{x:a<|f(x)|\leq b\},
and argue that it extends uniquely to a Borel measure.
This is straightforward, I think, as closed intervals form a semi-ring and \nu is a premeasure. I'm just not sure where to go from here. Any help would be greatly appreciated, thanks!
