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Let $f:[a,b] \to \mathbb{R}$ be bounded.

Is there always a Riemann integrable function $g:[a,b] \to \mathbb{R}$ such that for any partition $P$, we have $U(g,P) = U(f,P)$?

Of course, if $f$ is Riemann integrable, we can simply take $g \equiv f$, but that's not generally true.

The motivation to ask this question was the function $f(x) = \mathbf{1}_{\mathbb{Q} \cap [0,1]}(x)$ defined on $[0,1]$. For any partition $P$, $f$ shares its upper sum with $f \equiv 1$ on $[0,1]$, which happens to be integrable.

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[This is not a complete answer.]

I guess that a good candidate is the upper semicontinuous envelope $$ g(x) := \lim_{r\to 0+} \left( \sup_{y\in B_r(x)} f(y) \right). $$ Indeed, $g\geq f$ and, for every interval $[c,d]\subset[a,b]$ one has $$ \sup_{[c,d]} g = \sup_{[c,d]} f, $$ so that $U(g,P) = U(f,P)$ for every partition $P$.

The function $g$ is upper semicontinuous; on the other hand, an upper semicontinuous function can have a set of discontinuities of positive measure, hence need not be Riemann integrable. Such a function could be used to construct a counterexample.

In conclusion, I think that the answer to your question is negative.

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