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Let $\mu$ be a nonatomic probability measure on a Banach space $X$. Is it true that for $\mu$ a.e. $x \in X$, the function $g_x: (0, \infty) \to \mathbb R$ given by

$$g_x (r) := \mu(B_r (x))$$

is continuous in $r$?

Note: Here $B_r (x)$ denotes the open ball of radius $r$ around $x$.

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No, consider $X=\mathbf{R}^2$ with the $\ell_\infty$ norm, and let $\mu$ be a non-atomic probability measure giving mass $\frac 1 2$ to both segments $I_0=[-1,1]\times\{0\}$ and $I_1=[-1,1]\times\{2\}$. Then for every $x$ in the support of $\mu$ (so in particular for a.e. $x$), $g_x$ is discontinuous at $2$: $g_x(2)=\frac 1 2$ but $g_x(2+\varepsilon) = 1$ for every $\varepsilon>0$.

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  • $\begingroup$ Ah, the $\ell^\infty$ norm! $\endgroup$
    – Nate River
    Commented Oct 15, 2022 at 20:57
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    $\begingroup$ @NateRiver or, in this case, the $\ell_1$-norm (just tilt your head) $\endgroup$
    – Yemon Choi
    Commented Oct 15, 2022 at 21:08
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    $\begingroup$ What is important about the $\ell_{\infty}$ norm is that the space is not strictly convex, that is there is a non-degenerate segment contained in the unit sphere. When this holds, the same construction works (take a measure supported in this segment and a parallel segment containing the origin). $\endgroup$
    – Mikael de la Salle
    Commented Oct 17, 2022 at 7:27

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