Is $f:\mathbb R\to\mathbb R$ concave everywhere if, within each interval on the domain, $\exists$ a sub-interval s.t. $f$ is concave locally?

Question

Let $f:\mathbb R^n\to\mathbb R$

Given that $\forall$ open ball $\mathcal B\subseteq\mathbb R^n$, $\exists$ a subset $\mathcal B_1\subseteq\mathcal B$ s.t. $f$ is locally concave on $\mathcal B_1$,

What can we imply about $f$? Is $f$ concave everywhere or almost everywhere?

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My (edit: questionable) approach: by utilizing the fat Cantor set we could find a $f$ that satisfies the condition but not almost everywhere concave. So I guess the best we can get is $f$ is concave everywhere expect a nowhere dense set.

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Answer 1

Let $q_k$ be an enumeration of $\mathbb{Q}$ and $\varepsilon_k=2^{-k}$. Define $B_k= B_{\varepsilon_k}(q_k)$ and $f$ on $\bigcup B_k$ as $f(x) = - x^2$ and $f= 1$ on the complement.

Note that $\bigcup B_k$ is dense but has finite volume.