Monge formulation of Optimal Transport

Question

I'm reading these notes trying to understand a bit more about optimal transport https://www.math.cmu.edu/~mthorpe/OTNotes

Can someone help me explain the following on pg 6

We say that $T: X \to Y$ transports $\mu \in P(x)$ to $v \in P(Y)$, and we call T a transport map if $$v(B) = \mu(T^{-1}(B))$$ for all $v$-measurable sets $B$

If the function T is injective then we can equivalently say that $ν(T(A)) = \mu(A)$ for all $\mu$-measurable A

Q1. Why is there equality when its injective? Shouldn't it be equal if bijective?

Q2.What is meant by $v$-measurable sets?

Thank you

Answer 1

Q1. It is also equal if it is bijective, since in that case is is injective. But only injectivity is needed to replace one formulation by the other.

Q2: You have measures defined on $X$ and $Y$ ($\mu$ and $\nu$ respectively). $\nu$-measurable means a subset of $Y$ that is in the sigma algebra of measurable subsets of $Y$.

The whole point of this construction is that $T$ preserves the measure of the sets, so it is some sort of isomorphism in the sense of measures.

Answer 2

For Q1, this is an instance of a general fact you will likely find handy to keep in the back pocket.

In general, for a function $T:X\to Y$ and a subset $B$ of $Y$, we can define the preimage of $B$ via $T$ as a set. You probably know the definition. We denote this set $T^{-1}(B)$.

If a function $T:X\to Y$ is injective, we can define an honest-to-goodness function $T^{-1}:T(X)\to X$ by defining $T^{-1}(y)$ to be the single element of the set $T^{-1}(\{y\})$. This is a slight abuse of notation because we're using $T^{-1}$ in two ways. Since $T$ is injective, $T^{-1}(\{y\})$ contains a single element. The function $T^{-1}$ is called a left inverse because $T^{-1}\circ T = 1_X$, i.e. $(T^{-1}\circ T)(x) = x$ for all $x\in X$.

Suppose $T$ is a transport map. Then $\nu(T(A)) = \mu(T^{-1}(T(A))$, and it is not hard to show that $T^{-1}(T(A)) = (T^{-1}\circ T)(A)$ where $T^{-1}$ is a left inverse of $T$. So $\mu(T^{-1}(T(A)) = \mu((T^{-1}\circ T)(A)) = \mu(1_X(A)) = \mu(A)$.

The function $T$ is bijective if it has a left inverse and a right inverse. But you only need a left inverse for this observation, so injective is enough.

For Q2, As GReyes says, a measure $\nu$ on $Y$ is a function $\nu:\Sigma \to \overline{\mathbb{R}} $ with special properties, where $\Sigma$ is a $\sigma$-algebra on $Y$. A probability measure is one where also $\nu(Y) = 1$, which you will almost certainly find to be an important assumption as you study optimal transport problems. So a $\nu$-measurable set $B$ simply means that $B\in\Sigma$.