Suppose $(\Omega, \mu)$ is a Borel probability space and $f, g : \Omega \to \mathbb{R}$ are measurable functions such that $|\mu(f^{-1}(U)) - \mu(g^{-1}(U))| < \epsilon$ for all Borel subsets $U \subseteq \mathbb{R}.$ Does there exist a measure preserving map $\phi : \Omega \to \Omega$ such that $\|f - g \circ \phi\|_1 < c(\epsilon)$ with $c(\epsilon) \to 0$? If not, what can we get? I am ok with replacing $L^1$ with another kind of norm. If the claim is false, I'm also curious how general $\Omega$ can still be.
I first proved the easier problem that if $\mu(f^{-1}(U)) = \mu(g^{-1}(U))$ for all $U,$ then there exists $\phi$ such that $f = \phi \circ g^{-1}$ almost everywhere $(*).$ I tried the current problem for $\Omega = [0,1]$ and the solution was already long enough. My idea would still work for $\mathbb{R}$ and $\mathbb{R}^n,$ just with 1-2 more complications. But how do we approach the problem for general $\Omega$?
I considered the following: assume $f$ is injective, and replace $U$ with $f(U),$ we get $|\mu(U) - \mu(g^{-1}(f(U)))| < \epsilon$ so $h = f^{-1} \circ g$ is almost a measure-preserving map. Perhaps we can prove $\phi$ is close to $h$ for some $\phi : \Omega \to \Omega.$
We need to bound $f \circ \phi - g = f \circ (\phi - h),$ but there's no way to estimate the $L^1$ norm of a composition of functions. Even worse, $\phi - h$ doesn't make sense in general. Instead, from $\phi \approx h$ we should deduce $f \circ \phi \approx f \circ h = g,$ and from that deduce a bound on $\|f \circ \phi - h\|.$ But what kind of approximation should we use for $\approx$?
I have tried to come up with as much as possible on my own before asking here, and after getting stuck, I suspect I'm reinventing the wheel. For example, it might be the case $(*)$ is already a well-known result and so I should've just found a citation instead of spending an hour getting the idea and working out the details. Another idea: uncountable Borel spaces (and the countable are easy to handle) are isomorphic to the standard Borel space on $[0,1],$ so perhaps we can reduce to the case $\Omega = [0,1].$ For that case, here's my proof:
By composing with measure preserving maps, we may assume WLOG that $f, g$ are increasing. In the case $f, g$ are strictly increasing continuous and $f(0) = g(0) = a, f(1) = g(1) = b$ set $U = (a, x)$ to get $|f^{-1}(x) - g^{-1}(x)| < \epsilon$ for all $x,$ so $\|f - g\|_1 = \int_0^1 |f-g| = \int_a^b |f^{-1} - g^{-1}| < \epsilon(b-a).$ If $f(0) \ne g(0)$ or $f(1) \ne g(1),$ that's not a problem.
However, when $f$ or $g$ fail to be strictly increasing or fail to be continuous, there are jumps and flat lines on the graph. There can only be countably many jumps or flats, but they're still annoying to deal with. We can approximate $f$ with a strictly increasing continuous $r$ and even impose $\|f-r\|, \|f^{-1} - r^{-1}\| < \epsilon,$ then do the same for $g,$ but then a bound on $\mu(f^{-1}(U)) - \mu(r^{-1}(U))\|$ is needed. There is no bound at or near (although we can make "near" arbitrary small) any problems of $f,$ but we can just choose $U$ avoiding them. Similarly choosing $s$ for $g,$ we get an estimate for $\|\mu(r^{-1}(U)) - \mu(s^{-1}(U))\|$ for all $U$ avoiding the jumps, and need an estimate on $\|r-s\|$. However, $r^{-1}, s^{-1}$ may rise during the jumps, around which we don't have an estimate on $|r^{-1}(x) - s^{-1}(x)|$ and must go back to $f$ and $g.$ In the end it should all work out, but it's already gotten really messy.