Vtyjkyuk

Nelson's proof of Liouville's theorem (in the case $n=2$) is as follows:

Consider a bounded harmonic function on Euclidean space. Since
it is harmonic, its value at any point is its average over any sphere,
and hence over any ball, with the point as center. Given two points,
choose two balls with the given points as centers and of equal radius.
If the radius is large enough, the two balls will coincide except for an
arbitrarily small proportion of their volume. Since the function is
bounded, the averages of it over the two balls are arbitrarily close,
and so the function assumes the same value at any two points. Thus
a bounded harmonic function on Euclidean space is a constant.

I tried to formalize it.

Let $f:mathbbCtomathbbC$ be a holomorphic bounded function. If $z,winmathbbC$ we have that
beginalign*
|f(z)-f(w)| &= frac1pi r^2left|int_D(z,r)f(x+iy):mathrmdx:mathrmdy - int_D(w,r)f(x+iy):mathrmdx:mathrmdyright|\
&= frac1pi r^2left|int_Af(x+iy):mathrmdx:mathrmdy - int_Bf(x+iy):mathrmdx:mathrmdyright| \
&leq frac2pi r^2(sup |f|)int_A 1:mathrmdx:mathrmdy \
&= frac2pi r^2(sup |f|) left(2r^2cos^-1left(fracd2rright)-fracd2sqrt4r^2-d^2, right)
endalign*

where $d=|z-w|$. Observe what are $A$ and $B$ in the following drawing:

I would expect that the right hand side tends to $0$ when $rtoinfty$. However, that is not the case. How should I formalize Nelson's proof?

Your computation of the area of $A$ doesn't look correct.

It would be easier to bound the integral as follows: the width of $A$ is never more than $d$, and its $y$-coordinates are bounded between $-r$ and $r$. Thus $int_A 1,dx,dy le int_-r^r d,dy = 2rd$.

Ok, let's calculate the volume $|A|$ of $A$. WLOG I let $z=0$ and $w=d > 0$. Then
$$
fracA2 = int_d/2^r+dsqrtr^2-(x-d)^2,dx - int_d/2^rsqrtr^2-x^2,dx = int_-d/2^d/2sqrtr^2-x^2,dx.
$$
Substituting $x = rsin t$ gives
$$
fracA2 = r^2int_-arcsin(d/2r)^arcsin(d/2r)cos^2t,dt = fracr^22left[cos tsin t + tright]_-arcsin(d/2r)^arcsin(d/2r).
$$
As $cos(arcsin(x)) = sqrt1-x^2$,
$$
fracA2 = r^2left(fracd2rsqrt1-fracd^24r^2+arcsinfracd2rright),le,r^2left(frac d2r + arcsinfrac d2rright).
$$
Now, for small positive $x$ we have $arcsin xle 2x$, hence $|A|le 3dr$ for large $r$.