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.