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$$\int_0^x\int_z^1 f(y)dy dz=\int_0^1 \text{min}(x,y)f(y)dy,$$ where $x,y,z\in [0,1].$

Is this equation true? I wonder how to prove it. Thanks.

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1 Answer 1

up vote 3 down vote accepted

We have denoting characteristic functions by $\chi$, using Fubini: \begin{align*} \int_0^x \int_z^1 f(y)\,dy \,dz &= \int_0^1 \chi_{[0,x]}(z)\int_0^1 \chi_{[z,1]}(y)\, dy\,dz\\ &= \int_0^1 \int_0^1 \chi_{[0,x]}(z)\chi_{[z,1]}(y)f(y)\, dy\,dz\\ &= \int_0^1 f(y)\int_0^1 \chi_{[0,x]}(z)\chi_{[0,y]}(z)\, dz\,dy\\ &= \int_0^1 f(y) \int_0^1 \chi_{[0,y] \cap [0,x]}(z)\, dz\,dy\\ &= \int_0^1 f(y) \int_0^1 \chi_{[0, \min\{x,y\}]}(z)\, dz\,dy\\ &= \int_0^1 f(y) \cdot \min\{x,y\}\, dy. \end{align*}

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