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