Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question

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*}

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.