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I read the Wiki on Lebesgue integration ( and it says that the integral can be rewritten as the sum of volumes of equal-height contours:

  1. The volume of the (very thin) contour at height $t$ is defined as $f^*(t) = \mu(\{x | f(t) > t\})$.

  2. 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

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f*($1$)$ = $$0$ and f*($0.7$)$=$$0.3$ Isn't it? – Halil Duru Apr 4 '13 at 21:12
Note that in your version -- $f^*(t) = \mu(\{x | f(x) = t\})$. Many integrals would be $0$. – Halil Duru Apr 4 '13 at 21:47
+1 Thanks, my mistake! But it's strange that $\{x|f(x)>t\}$ is used-- it doesn't seem to correspond to the second panel here: – user Apr 4 '13 at 21:54
It exactly corresponds -- look at the width of the block at the bottom --isn't the width of the block same as the measure of values whose image is greater than the height of the block? – Halil Duru Apr 4 '13 at 21:59
Ah, thanks. I see. I was thinking of a non-function (e.g. an hourglass), whereby the width would be counted as too large. But if there is exactly one y per x, then it would be OK. – user Apr 4 '13 at 22:11
up vote 1 down vote accepted

$f^*(t) = \mu(\{x | f(x) > 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.

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