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I'm calculating the measure $\mu(A)$, where $\mu = L^1 \otimes \sum_{k = 1}^{\infty}k\delta_k $ and $A = \{(x, y) \in \mathbb{R}^2: 3 - |x| > y > 2|x|\}$. I've drawn this set on the plane:

set A

Then I've proceeded to calculating this measure as an integral and using Fubini's theorem:

$\mu(A) = \int_{A} d\mu = \int_{A} d (L^1 \otimes \sum_{k = 1}^{\infty}k\delta_k) = \int_{[-1, 1]} \left ( \int_{3 - |x| > y > 2|x|} d(\sum_{k = 1}^{\infty}k\delta_k)(y) \right ) d L^1 (x)$

I have a question about the next step: what's the value of that inner integral? Can I just say that it's "visible" from the plot above that the measure $(\sum_{k = 1}^{\infty}k\delta_k)(3 - |x| > y > 2|x|) = 1 + 2 + 3 = 6$?

Then finishing the calculations would be quite easy, as $\int_{[-1, 1]}6dL^1(x) = 6 \times 2 = 12$, so $\mu(A) = 12$.

1 Answers 1

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I think you're forgetting that x is fixed in the inner integral.

For example if $x = -1$, then $\{y \in \mathbb{R}: 3-|x| > y > 2|x|\} = \{y \in \mathbb{R}: 2 > y > 2\} = \emptyset$. So the inner integral is $0$.

So look at the set $\{y \in \mathbb{R}: 3-|x| > y > 2|x|\}$ depending on x:

$x = -1: \int_{3-|x| > y > 2|x|} d\left(\sum_{k=1}^\infty k \delta_k\right)(y) = 0$ (see above) $-1 y > 2|x|\} \cap \mathbb{N} = \{2\}$, so $\int_{3-|x| > y > 2|x|} d\left(\sum_{k=1}^\infty k \delta_k\right)(y) = 2$ and so on. Then the outer integral will finally look like this

$\int_{[-1,1]} 2*\mathbf{1}_{(-1,-\frac{1}{2}]}(x)+... dL^1(x)$