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If $f$ is a riemann integrable function over $[a,b]$ , is the floor function{f(x)} riemann integrable over $[a,b]$ ?

Here is what i've tried: i figured out that the discontinued points of {f(x)} don't always be the subset of f(x)'s. And i also can't construct a counterexample... So i don't have any ideas now:(

Idele
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    Oh i see...we can take https://en.wikipedia.org/wiki/Thomae's_function?wprov=sfsi1 with negative sign – Idele Aug 12 '17 at 05:33
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    You can use this for a continuous example, where $C$ is the fat Cantor set. In fact, I think the construction can be modified so as to create a $C^{\infty}$ example. – MathematicsStudent1122 Aug 13 '17 at 05:32

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Counterexamples have been discussed in the comments.

However, this does hold if we have the condition that $f:[a,b] \to \mathbb{R}$ is real-analytic.

If $f$ is constant, it's immediately clear. If $f$ is not constant, then note that $f$ is bounded, so the image of $f$ is a subset of $[-M,M]$ for $M>0$. There are finitely many integers in $[-M,M]$, so the set $S=\{x \in [a,b] : f(x) \in \mathbb{Z} \cap [-M,M] \}$ must also be finite. This is an immediate consequence of Bolzano–Weierstrass and the principle of permeance. But the discontinuities of $\lfloor f \rfloor$ are a subset of $S$ which means $f$ must have at most finitely many discontinuities, which implies Riemann integrability.

MathematicsStudent1122
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