If $g$ is bounded and measurable and $\mu$ is obtained by taking the lim inf of $g$ approaching each point, is $\mu$ Riemann integrable?

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Let $$g:[a,b] \to \mathbb{R}$$ be bounded and measurable. Define $$\mu = \lim_{\delta \to 0} \mu_\delta(x),$$ where $$\mu_\delta(x) = \inf \{g(y):y \in b_\delta(x) \cap [a,b]\}.$$ Is $\mu$ always Riemann integrable?

I know that $\mu$ is Lebesgue integrable but I'm not sure whether it is always Riemann integrable. I'm suspecting not, but I can't seem to find a counterexample to this. Is there any suggestion?

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Let $K\subset[a,b]$ be a fat Cantor set and let $g$ be the characteristic function of $[a,b]\setminus K$. Then $\mu=g$ is not Riemann integrable since it is discontinuous at every point of $K$ and $K$ has positive measure.

In fact, the assumption that $g$ is measurable is not really relevant. For any function $g$, $\mu$ will be a lower semicontinuous function (and thus is measurable, even if $g$ was not!), and if $g$ was already lower semicontinuous, then $\mu=g$. So you are really just asking whether every bounded lower semicontinous function is Riemann integrable.