Let $X$ be a closed convex subset of a Banach space, and let $A\subseteq X$ be a Borel set. Denote
$$
B\_r(A) :=\{y\in X:\exists x\in A \text{ such that }\|x-y\|\leq r\}
$$
and by $H(A)$ the convex hull of $A$. Is that true that 
$$
 B\_r(H(A)\cap X)\cap X = H(B\_r(A)\cap X)\cap X?
$$ 
Perhaps, there are some additional conditions needed to ensure the equality?



Commutativity of convex hulls and closed balls
This seems kind of trivially true (even without the assumption of $A$ Borel), unless I'm mistaken. First, let's see that $H(B\_r(A)) \subset B\_r(H(A))$. Any $y \in H(B\_r(A))$ can be written as $\sum \lambda\_i y\_i$, with $y\_1, \dots, y\_m \in B\_r(A)$, and $\lambda\_1, \dots, \lambda\_m$ non-negative reals adding to $1$. Let $x\_i \in A$ be such that $||x\_i - y\_i|| \leq r$. Then it's clear from the triangle inequality that $||y - \sum \lambda\_i x\_i || \leq r$. So $y \in B\_r(H(A))$, since $\sum \lambda\_i x\_i \in H(A)$. Conversely, let $y \in B\_r(H(A))$. Then $||y-z|| \leq r$ for some $z = \sum \lambda\_i x\_i$ with $x\_i \in A$ and $\lambda\_i$ non-negative reals adding to $1$. Let $t = y - z$, and $y\_i = x\_i + t$. Then $||y\_i - x\_i || = || y - z|| \leq r$ and $\sum \lambda\_i y\_i = \sum \lambda\_i x\_i + (\sum \lambda\_i) t = z + t = y$. That proves $y \in H(B\_r(A))$.




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Update: For the updated question (when you take intersections), there is indeed a counterexample. The first part of the proof still goes through, so the containment $H(B\_r(A) \cap X) \cap X \subset B\_r(H(A) \cap X) \cap X$ is still valid (since if the $y\_i \in X$ their convex combination $y$ is too, because $X$ is convex). For a counterexample to the opposite inclusion consider the following picture in $\mathbb{R}^2$ (maybe someone can actually draw it :-) )


Let $p = (0,0)$, $q = (1,0)$ and $z = \frac{1}{2}(1, \tan(\theta))$ for some small angle $\theta$. Then $p,q,z$ form an isosceles triangle with angles $\theta, \theta, \pi - 2 \theta$. Let $X$ be this triangle (including the interior), and $A = \{p,q\}$. Then if $r < \tan(\theta)/2$ is small, the convex hull of $A$ is the line segment joining $p$ and $q$ which is contained in $X$, and so $B\_r(H(A) \cap X) \cap X$ contains for instance the point $w = (1/2, r)$, since its distance from $(1/2,0)$ is $r$. But $B\_r(A) \cap X$ consists of two disjoint small sectors of circles of radius $r$ and angle $\theta$ around $p$ and $q$. So the every point of it has $y$-coordinate at most $r \sin(\theta) < r$. So the same holds for its convex hull too. Hence $w$ is not in $H(B\_r(A) \cap X) \cap X$. (On second thought, it seems that any $\theta < \pi/2$ will work.) 


