Optimal Transport: Showing Map is Optimal (simple question: more of an analysis/measure theory question)

I'm reading these lecture notes (not for homework, for fun) http://www.maths.gla.ac.uk/~gbellamy/LMS/BourneLectures.pdf

I'm trying to figure out what it means for a transport means to be optimal. I'm mainly trying to figure out Exercise 2.1 in the notes.

First he writes that: Let $$X,Y \subset \mathbb{R}^d$$ and $$T: X \rightarrow Y$$. Let $$f$$ be a probability density on $$X$$ and $$g$$ be a probability density on $$Y$$. We say that $$g$$ is the push-forward of $$f$$ under $$T$$, and write $$g=T\#f$$, if $$\int_B g(y) dy=\int_{T^{-1}(B)} f(x) dx$$ for every $$B \subset Y$$.

Now, here's exercise $$2.1$$: Let $$T: \mathbb{R} \rightarrow \mathbb{R}$$ be the translation $$T(x)=x+1$$. Let $$f= \chi_{[0,1]}$$ and $$g=\chi_{[1,2]}$$ be probability densities on $$\mathbb{R}$$. Show that $$T\#f=g$$.

Now, I think I understand why its true for lebesgue integrals as $$\int \chi_{[0,1]} d\mu=\mu[0,1]=1$$. So the shift won't change the lengths so the measures will be the same, something like this. Not sure how to make this more rigorous.

However, the author of the lecture notes doesn't discuss measure theory or measures at all so I'm not sure if in the definition of push-forward, if he is talking about Lebesgue integrals or Riemann integrals.

Is there a way to show that $$T\#f=g$$ in exercise 2.1 for Riemann integrals? Or does the question not make sense for Riemann integrals. That, is is there an analog for Riemann integration? Also, is there a way to make my idea for Lebesgue integral more rigorous? Thank you.

• does anyone have any ideas? thank you very much
– kemb
Apr 13, 2019 at 0:30

It doesn't matter what sort of integrals you use, as long as you suitably restrict the nature of the set $$\ B\$$ for which the equation $$\ \int_B g(y) dy=\int_{T^{-1}(B)} f(x) dx\$$ has to hold. For Riemann integrals, this will only make sense if the indicator function $$\ \chi_B\$$ is Riemann integrable. If it is, you have $$\begin{eqnarray} \int_{T^{-1}(B)} f(x) dx&=&\int_{-\infty}^\infty \chi_{T^{-1}(B)}\left(x\right)\chi_{[0,1]}\left(x\right) dx\\ &=& \int_{-\infty}^\infty \chi_B\left(x+1\right)\chi_{[0,1]}\left(x\right) dx\\ &=& \int_{-\infty}^\infty \chi_B\left(y\right)\chi_{[0,1]}\left(y-1\right) dy\\ &=& \int_{-\infty}^\infty \chi_B\left(y\right)\chi_{[1,2]}\left(y\right) dy\\ &=& \int_B \chi_{[1,2]}\left(y\right) dy\\ &=&\ \int_B g(y) dy\ . \end{eqnarray}$$ If you're using Lebesgue integration, the same proceure works for all Lebesgue measurable $$\ B\$$.