Is $f$ Riemann integrable if $\epsilon>1$ and if $\epsilon<1$?

Question

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

(1) If for each $\epsilon>1$ there is a partition $P$ of $[a,b]$ such that $U(f,P)-L(f,P)<\epsilon$ then does it follow that $f$ is integrable?

(2) If for each $\epsilon<1$ there is a partition $P$ of $[a,b]$ such that $U(f,P)-L(f,P)<\epsilon$ then does it follow that $f$ is integrable?

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From (1) it doesn't follow that $f$ isintegrable, because since $\epsilon$ is not small enough it doesn't necessarily follow that $\underline\int_a^bf=\overline\int_a^bf$ and so Riemann criterion is not satisfied and $f$ is not integrable.

Is that correct?

At (2) it follows that $f$ is integrable, or not?

Answer 1

Consider the Dirichlet function on $[0,1]$: $$f(x) = \begin{cases}1, &x\in\mathbb Q \\ 0, &x\in\mathbb R\setminus\mathbb Q\end{cases}.$$ The difference $U(f,P)-L(f,P) \leqslant 1$, but it is not a Riemann integrable function. (The inequality holds for any partition $P$ in this example).

The function $f = I_{[0,1]\cap\mathbb Q}$ is Lebesgue integrable, since $[0,1]\cap\mathbb Q$ is measurable (its measure is zero).

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(2) is true, because if $\varepsilon >0$ is arbitrary, we can by assumption for every $\varepsilon ' < 1$ find $P$ such that $U(f,P)-L(f,P) < \varepsilon ' \leqslant \varepsilon$. Equivalently, $f$ is Riemann integrable in $[a,b]$.