Hi,

The isodiametric inequality tells us that, of all sets of diameter $r$, the one with the largest Lebesgue measure is the ball of radius $r/2$ - and this holds regardless of norm. Let $\tau(r)$ be the measure of that ball. I now have another set $S$ of diameter at most $r$, but its area is $\tau(r) (1- \epsilon)$, for some very small $\epsilon$.

I am interested in showing that this set is "almost" a ball, in the sense that there exists a ball $B$ such that the Lebesgue measure of the symmetric difference $B \triangle S$ is bounded above by some $f(\epsilon)$ that is also small.

I feel like this is some kind of application of Brunn-Minkowski or some other classical theorem, but I just can't seem to find it. Even if this can be done for the Euclidean norm on $\mathbb{R}^2$, it would be a great help [for the record, this is rather simple in the $\ell^\infty$ norm - just draw squares around the leftmost, rightmost, topmost, and bottommost points in $S$, and use the measure constraints. Dealing with a curved geometry requires a more subtle approach].

Thank you all very much!

-Matan

EDITED TO ADD: I'm not sure if this is required, but I'll also be fine if $S$ were restricted with rather strong regularity conditions - for example, connected, convex, $C^1$ boundary, or other such conditions.