I read the Wiki on Lebesgue integration (http://en.wikipedia.org/wiki/Lebesgue_integration) and it says that the integral can be rewritten as the sum of volumes of equal-height contours:
The volume of the (very thin) contour at height $t$ is defined as $f^*(t) = \mu(\{x | f(t) > t\})$.
Then you simply compute $\int_0^\infty f^*(t) \partial t$.
The second step makes perfect sense to me; however, the first step feels intuitively incorrect: the measure of the set $f^*(t)$ should be the width at height $t$ (otherwise it is not a true partition, and you are counting points multiple times). To me, it feels like then it would be $f^*(t) = \mu(\{x | f(t) = t\})$.
As defined in the Wiki, it feels like it would include the points at height $t=1$ when it computes $f^*(0.5)$. This would be a problem if you are computing the area under a right triangle of width 1:
$f(x) = x, x \in [0,1]$, otherwise $0$
$\int_0^1 f(x) \partial x = \frac{1}{2}$
As defined by the Wiki:
$f^*(t) = \mu(\{x | f(t) > t\})$
$\int_0^\infty f^*(t) \partial t = \int_0^1 f^*(t) \partial t$
But according to the Wiki $f^*(0)$ will be the measure of the set of $x$ points for which $f(x) > 0$, which is 1. In fact, this will be the case for all $t \in [0,1]$. So the integral will be 1 (when it should be 1/2).
Please let me know if you have any explanation for why I am incorrect about this.
Thanks a lot! Oliver
$f^*(t) = \mu(\{x | f(x) > t\}) $
$f^*(t)=$$1-t$
$\int_0^\infty f^*(t) dt = \int_0^1 f^*(t) dt$$=$ $\int_0^1 [1-t] dt$ $ =1/2$
Note that $f^*(t)$ corresponds to the width at level t.