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Let $\mathbb{1}$ be the indicator function, e.g

$$\mathbb{1}^x_{(0,1)} = \begin{cases} 1, \ x \in (0,1) \\ 0, \ \text{elsewhere}. \end{cases}$$

I've stumbled onto this integral, $\int_{0}^{1} \mathbb{1}^x_{(-y,y)}dy$. For some (obvious) reason, this equals $\int_{0}^{1} \mathbb{1}^y_{(|x|,1)}dy $. But I fail to see why. Any enlighten is much appreciated!

Oskar
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    Notice that $x,y\in (0,1)$ (this come from the bounds of the integral). So,$$\boldsymbol 1_{(-y,y)}^x=\begin{cases}1&|x|\leq y\leq 1\0&elsewhere\end{cases}=\boldsymbol 1_{(|x|,1)}^y$$ – Surb Sep 07 '23 at 16:04

1 Answers1

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$$0<y<1\text{ and }x\in(-y,y)\iff$$ $$0<y<1\text{ and }|x|<y\iff$$ $$0<y<1\text{ and }y\in(|x|,1).$$

Anne Bauval
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