Paradoxical sums of non-integrable measurable functions Are there reals $\alpha_1,\alpha_2$ and Lebesgue-measurable functions $f_1,f_2:\mathbb R/\mathbb Z\to\mathbb R$ such that
$$f_1(x)+f_2(x)>f_1(x+\alpha_1)+f_2(x+\alpha_2)$$
for almost all $x\in\mathbb R/\mathbb Z$? Here $\mathbb R/\mathbb Z$ has its usual Lebesgue measure coming from the unit interval.
This would be a functional relative of paradoxical decompositions, where moving sets around can make them bigger. The famous example is the Banach-Tarski decomposition of the 2-sphere, which can be achieved with Baire measurable pieces [1].
Feel free to use more functions if it helps with a positive answer, or to add $1$ to the right-hand-side if it helps with a negative answer.

Neither of the offsets $\alpha_1,\alpha_2$ can be rational. If $\alpha_i$ is a rational $p/q,$ we can sum (*) over $x=z,z+\tfrac1q,\dots,z+\tfrac{p-1}q$ to eliminate $f_i.$ Then $\alpha_i$ is irrelevant, so we can replace it by any irrational and apply the following argument.
Neither $f_1$ nor $f_2$ can be integrable. To show this it suffices to consider the case that $f_2$ is integrable and $\alpha_1\not\in\mathbb Q.$ We can then find an integrable function $g$ satisfying $\int g>0$ and
$$f_1(x)+f_2(x)> g(x)+f_1(x+\alpha_1)+f_2(x+\alpha_2)$$
for almost all $x,$ for example $g(x)=\min(1,f_1(x)+f_2(x)-f_1(x+\alpha_1)-f_2(x+\alpha_2))/2.$
Averaging over $x=z,z+\alpha_1,\dots,z+(n-1)\alpha_1$ and cancelling and rearranging gives
$$\frac1n (f_1(z)-f_1(z+n\alpha_1))>\frac1n \sum_{k=0}^{n-1}(g(z+k\alpha_1)+f_2(z+\alpha_2+k\alpha_1)-f_2(z+k\alpha_1)).\tag{*}$$
Pick $C$ such that $\{x:f_1(x)>C\}$ has positive measure. By the pointwise ergodic theorem $f_1(z+n\alpha_1)>C$ for infinitely many integers $n\geq 0,$ for almost all $z.$ So $$\liminf_{n\to\infty}\frac1n (f_1(z)-f_1(z+n\alpha_1))\leq 0$$ for almost all $z.$ By the pointwise ergodic theorem again, the right-hand-side of (*) tends to $\int g>0$ for almost all $z\in\mathbb R/\mathbb Z.$ This is a contradiction.
[1] Marks, Andrew; Unger, Spencer, Baire measurable paradoxical decompositions via matchings, Adv. Math. 289, 397-410 (2016). ZBL1335.54035.
 A: Here is one possible construction. We shall need a pair of positive numbers $\alpha,\beta$ such that for every positive integer $N$, the numbers $k\alpha+\ell\beta\mod 1$ with $1\le k,\ell\le CN^{2/3}$ ($C>0$ is some absolute constant) form a $1/N$-net in $[0,1]$. I'll postpone the choice of such $\alpha,\beta$ for a while (it turns out that most pairs are OK) and show how to build a function $f>0$ such that $2f(x)\le \max(f(x+\alpha),f(x+\beta))$ for almost all $x$. Then we can put $f_1=f_2=-f$ in the original problem.
We shall start with $f=1$ and the interval $I=[0,1]$.
Suppose that we already have a bounded positive function $f\le M$ and an interval $I$ such that the desired inequality holds for all $x\notin I$. Change the values of $f$ on $I$ to $M$. Now split $I=J\cup K$ where $J$ is the left two thirds of $I$ and $K$ is the right third. By our condition on $\alpha,\beta$, we can choose some $k,\ell\in[1,C|I|^{-2/3}]$ such that $K+(k\alpha+\ell\beta)\subset J$ (everything is $\mod 1$, of course). Now to get things right for $K$, we can lift $f$ to $2M$ on $K_1$, which we can choose to be either $K+\alpha$ or $K+\beta$. This will fix $K$ but create a problem of $K_1$. It can be fixed by raising $f$ to $4M$ on $K_2$, which can be chosen to be either $K_1+\alpha$ or $K_1+\beta$, etc. We proceed this way choosing $k$ $\alpha$-steps and $\ell$ $\beta$-steps and end up with a bounded function that is problematic only on $J$ (the last problematic interval we created is a subset of $J$).
In this process, we have changed our function on a set $E$ of measure $\le |I|+C|I|^{-2/3}|I|\le C|I|^{1/3}$ and reduced the length of the problematic interval to $\frac 23|I|$. Taking $J$ to be new $I$, repeating this upgrade infinitely many times and observing that the length of the corresponding sets $E$ decay like a geometric progression (so the sum of lengths is finite), we see that $f$ stabilizes at almost every point of $I$. Define it in any way you want on the remaining set of measure $0$ (where it escapes to $+\infty$) and you are done.
It remains to choose $\alpha,\beta$. Fix $N>1$ and put $M=[CN^{2/3}]$. Consider an interval of length $1/N$ and build the usual triangular function $\psi$ on it of height $2N$ (so that $\int\psi=1$). The sum of all Fourier coefficients of $\psi$ is then $2N$. Now write
$$
\sum_{1\le k,\ell\le M}\psi(k\alpha+\ell\beta)
\\
=
M^2+\sum_{m\ne 0}\widehat\psi(m)\left(\sum_{1\le k\le M}e^{2\pi i km\alpha}\right)\left(\sum_{1\le \ell\le M}e^{2\pi i \ell m\beta}\right)
\\
\ge M^2-\sum_{m\ne 0}|\widehat\psi(m)|\min\left(\frac 1{\|m\alpha\|},M\right)\min\left(\frac 1{\|m\beta\|},M\right)>0
$$
unless
$$
\sum_{m\ne 0}|\widehat\psi(m)|\min\left(\frac 1{\|m\alpha\|},M\right)\min\left(\frac 1{\|m\beta\|},M\right)\ge M^2.
$$
Note also that $\widehat\psi(m)$ depend on the location of the interval of length $1/N$ we considered but $|\widehat\psi(m)|$ do not. Now it just remains to observe that the integral of the LHS over all $\alpha,\beta$ is at most $C'N\log^2 M\le 2C'N\log^2 N$, so the measure of the exceptional set of pairs is bounded by $c N^{-1/3}\log^2 N$ with as small $c>0$ as we want, provided that we take $C$ large enough. Since it suffices to check the property for $N=2^n$ ($n=1,2,\dots$), say, we can make the measure of the union of these exceptional set for all $N$ less than $1$, finishing the story.
I'm still curious about a Borel measurable example with the inequality holding everywhere, so don't hurry to accept this answer :-)
